Applied Statistics Probability Engineers 3ed Montgomery Runger

Student Workbook with Solutions Heecheon You Pohang University of Science and Technology

Applied Statistics and Probability lor Engineers

Third Edition

Douglas C. Montgomery Arizona State University

George C. Runger Arizona State University

JOHN WILEY & SONS, INC.

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Copyright O 2003 by John Wiley & Sons, Inc.

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ISBN 0-47 1-42682-2

Printed in the United States of America.

Printed and bound by Courier Kendallville, Inc.

This book is dedicated to:

Jiyeon, Sophia, Jaechang, Sookhyun, Young, and Isik

Preface Objectives of the Book

As a supplement of Applied Statistics and Probability for Engineers (3rd edition) authored by Drs. Douglas C. Montgomery and George C. Runger (abbreviated by MR hereinafter), this book is developed to serve as an effective, learner-friendly guide for students who wish to establish a solid understanding of probability and statistics. First, this study guide outlines learning goals and presents detailed information accordingly; this goal-oriented presentation is intended to help the reader see the destination before learning and check histher achievement after learning. Second, this book helps the reader build both scientific knowledge and computer application skill in probability and statistics. Third, this book provides concise explanations, step-by-step algorithms, comparison tables, graphic charts, daily-life examples, and real-world engineering problems, which are designed to help the reader to efficiently learn concepts and applications of probability and statistics. Lastly, this book covers essential concepts and techniques of probability and statistics; thus, the reader who has interest in the advanced topics indicated as CD only in MR should refer to the MR CD.

Features of the Book

The reader will enjoy the following learner-friendly features of the book: Learning Goals: Each section starts with a list of learning goals which states topics covered and competency levels targeted. Organized, Concise Explanation: According to the learning goals defined, topics are presented with concise, clear explanations. Learning Aids: Tables, charts, and algorithms are provided wherever applicable to help the reader form integrated mental pictures by contrasting or visualizing concepts and techniques. Easy-to Understand Examples: For topics requiring problem-solving practices, daily-life examples (which are readily understandable) are used to help the reader better see the application process of statistical techniques. Engineering-Context Exercises: Together with daily-life examples, engineering problems (selected from exercises in the MR textbook; the original exercise numbers are indicated) are used to have the reader explore the application of statistical techniques to the real-world engineering domain. Complete Solutions of Exercises: At the end of each chapter, complete solutions of exercises are provided so that the reader checks hidher own answers with ease. Software Demonstration: To help students learn use of Excel and Minitab (student version) in solving probability and statistics problems, respectively, step-by-step directions and screen snapshots are provided for examples as applicable.

Organization of the Book

This book presents various topics of probability and statistics with the structure displayed in the figure on next page. Chapters 2 through 5 describe probability and probability distributions. After explaining fundamental concepts, rules, and theorems of

vi Preface

probability, this book presents various discrete and continuous probability distributions of single and multiple random variables. Then, chapters 6 through 15 explain descriptive and inferential statistics. Descriptive statistics deals with measures and charts to describe collected data, whereas inferential statistics covers categorical data analysis, statistical inference on singleltwo samples, regression analysis, design of experiment, and nonparametric analysis.

Multiple Discrete Probability Distributions 1 Ch.5

Multiple Continuous J Probability Distributions

Ch. 11

Ch. 12

Ch. 13

Ch. 14

Acknowledgements

I would like to express my considerable gratitude to Jenny Welter for her sincere support, trust, and encouragement in developing this book. I also wish to thank Drs. Douglas C. Montgomery and George C. Runger at Arizona State University for their pleasant partnership and helpful advice in producing this book. Finally, I greatly appreciate the insightful feedback and useful suggestions from the professors and students who reviewed the various parts of the manuscript: Dr. Yasser M. Dessouky at San Jose State University; Major Brian M. Layton at U.S. Military Academy; Dr. Arunkumar Pennathur at University of Texas at El Paso; Dr. Manuel D. Rossetti at University of Arkansas; Dr. Bruce Schmeiser at Purdue University; Dr. Lora Zimmer at Arizona State University; Christi Barb, Shea Kupper, Maria Fernanda Funez-Mejia, Tram Huynh, Usman Javaid, Thuan Mai, Prabaharan Veluswamy, Cheewei Yip, and Ronda Young at Wichita State University. Without their invaluable contributions, this book may not have been completed. Peace be with them all.

Heecheon You

Contents

Chapter 1 The Role of Statistics in Engineering 1-1

1 - 1 The Engineering Method and Statistical Thinking 1 - 1 1-2 Collecting Engineering Data 1-2 1-3 Mechanistic and Empirical Models 1-3 1-4 Probability and Probability Models 1-3

Chapter 2 Probability 2-1

2-1 Sample Spaces and Events 2-1 2-2 Interpretations of Probability 2-4 2-3 Addition Rules 2-5 2-4 Conditional Probability 2-6 2-5 Multiplication and Total Probability Rules 2-6 2-6 Independence 2-8 2-7 Bayes' Theorem 2-9 2-8 Random Variables 2- 10 Answers to Exercises 2-1 1

Chapter 3 Discrete Random Variables and Probability Distributions 3-1

3- 1 Discrete Random Variables 3- 1 3-2 Probability Distributions and Probability Mass Functions 3-1 3-3 Cumulative Distribution Functions 3-3 3-4 Mean and Variance of a Discrete Random Variable 3-4 3-5 Discrete Uniform Distribution 3-5 3-6 Binomial Distribution 3-6 3-7 Geometric and Negative Binomial Distributions 3-8 3-8 Hypergeometric Distribution 3-13 3-9 Poisson Distribution 3- 16 Summary of Discrete Probability Distributions 3-20 EXCEL Applications 3-2 1 Answers to Exercises 3-23

vii

viii Contents

Chapter 4 Continuous Random Variables and Probability Distributions 4-1

4- 1 Continuous Random Variables 4- 1 4-2 Probability Distributions and Probability Density Functions 4-1 4-3 Cumulative Distribution Functions 4-3 4-4 Mean and Variance of a Continuous Random Variable 4-4 4-5 Continuous Uniform Distribution 4-5 4-6 Normal Distribution 4-6 4-7 Normal Approximation to the Binomial and Poisson

Distributions 4-8 4-9 Exponential Distribution 4-1 1 4- 10 Erlang and Gamma Distributions 4- 14 4-1 1 Weibull Distribution 4-1 8 4- 12 Lognormal Distribution 4-20 Summary of Continuous Probability Distributions 4-21 EXCEL Applications 4-22 Answers to Exercises 4-25

Chapter 5 Joint Probability Distributions 5-1

5-1 Two Discrete Random Variables 5-1 5-2 Multiple Discrete Random Variables 5-5 5-3 Two Continuous Random Variables 5-7 5-4 Multiple Continuous Random Variables 5- 12 5-5 Covariance and Correlation 5- 13 5-6 Bivariate Normal Distribution 5-1 5 5-7 Linear Combinations of Random Variables 5- 16 5-9 Moment Generating Functions 5-18 5- 10 Chebyshev's Inequality 5-20 Answers to Exercises 5-2 1

Chapter 6 Random Sampling and Data Description 6-1

6- 1 Data Summary and Display 6- 1 6-2 Random Sampling 6-3 6-3 Stem-and-Leaf Diagrams 6-4 6-4 Frequency Distributions and Histograms 6-8 6-5 BOX Plots 6-12 6-6 Time Sequence Plots 6-13 6-7 Probability Plots 6-14 MINITAB Applications 6- 1 5 Answers to Exercises 6-20

Contents ix

Chapter 7 Point Estimation of Parameters 7-1

7- 1 Introduction 7- 1 7-2 General Concepts of Point Estimation 7-2 7-3 Methods of Point Estimation 7-6 7-4 Sampling Distributions 7-7 7-5 Sampling Distributions of Means 7-7 Answers to Exercises 7-10

Chapter 8 Statistical Intervals for a Single Sample 8-1

8- 1 Introduction 8- 1 8-2 Confidence Interval on the Mean of a Normal Distribution, Variance

Known 8-2 8-3 Confidence Interval on the Mean of a Normal Distribution, Variance

Unknown 8-5 8-4 Confidence Interval on the Variance and Standard Deviation of a

Normal Population 8-7 8-5 A Large-Sample Confidence Interval for a Population

Proportion 8-9 8-6 A Prediction Interval for a Future Observation 8-1 1 8-7 Tolerance Intervals for a Normal Distribution 8- 13 Summary of Statistical Intervals for a Single Sample 8-14 MINITAB Applications 8- 1 5 Answers to Exercises 8- 1 8

Chapter 9 Tests of Hypotheses for a Single Sample 9-1

9-1 Hypothesis Testing 9-1 9-2 Tests on the Mean of a Normal Distribution, Variance Known 9-6 9-3 Tests on the Mean of a Normal Distribution,

Variance Unknown 9- 1 0 9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal

Population 9- 13 9-5 Tests on a Population Proportion 9-15 9-7 Testing for Goodness of Fit 9- 17 9-8 Contingency Table Tests 9-22 MINITAB Applications 9-28 Answers to Exercises 9-37

X Contents

Chapter 10 Statistical Inference for Two Samples 10-1 ""

10-2 Inference for a Difference in Means of Two Normal Distributions, Variances Known 10- 1

10-3 Inference for a Difference in Means of Two Normal Distributions, Variances Unknown 10-6

10-4 Paired t-Test 10- 1 1 10-5 Inference on the Variances of Two Normal Populations 10- 14 10-6 Inference on Two Population Proportions 10- 1 9 MINITAB Applications 10-24 Answers to Exercises 10-30

Chapter 11 Simple Linear Regression and Correlation 11-1

1 1 - 1 Empirical Models 1 1 - 1 1 1-2 Simple Linear Regression 1 1-3 1 1-3 Properties of the Least Squares Estimators 1 1-8 1 1-4 Some Comments on Uses of Regression 1 1 - 10 1 1-5 Hypothesis Tests in Simple Linear Regression 1 1 - 1 1 1 1-6 Confidence Intervals 1 1 - 17 1 1-7 Prediction of New Observations 1 1-2 1 1 1-8 Adequacy of the Regression Model 1 1-23 1 1-9 Transformations to a Straight Line 1 1-29 1 1- 1 1 Correlation 1 1-30 MINITAB Applications 1 1-34 Answers to Exercises 1 1-37

Chapter 12 Multiple Linear Regression 12- 1

12- 1 Multiple Linear Regression Model 1 2- 1 12-2 Hypothesis Tests in Multiple Linear Regression 12-5 12-3 Confidence Intervals in Multiple Linear Regression 12- 13 12-4 Prediction of New Observations 12-15 12-5 Model Adequacy Checking 12- 16 12-6 Aspects of Multiple Regression Modeling 12- 18 MINITAB Applications 12-28 Answers to Exercises 12-35

Contents xi

Chapter 13 Design and Analysis of Single-Factor Experiments: The Analysis of Variance 13-1

1 3- 1 Designing Engineering Experiments 13- 1 13-2 The Completely Randomized Single-Factor Experiment 13-2 13-3 The Random Effects Model 13- 13 13-4 Randomized Complete Block Design 13- 15 MINITAB Applications 13-2 1 Answers to Exercises 13-26

Chapter 14 Design of Experiment with Several Factors 14-1

14-1 Introduction 14- 1 14-3 Factorial Experiments 14- 1 14-4 Two-Factor Factorial Experiments 14-4 14-5 General Factorial Experiments 14-1 1 14-7 2k Factorial Designs 14-1 1 14-8 Blocking and Confounding in the 2k Design 14-1 8 14-9 Fractional Replication of the 2k Design 14-2 1 MINITAB Applications 14-29 Answers to Exercises 14-32

Chapter 15 Nonparametric Statistics i 5-1

15- 1 Introduction 15- 1 15-2 Sign Test 15-2 15-3 Wilcoxon Signed-Rank Test 15-7 15-4 Wilcoxon Rank-Sum Test 15-12 15-5 Nonparametric Methods in the Analysis of Variance 15-14 MINITAB Applications 15- 18 Answers to Exercises 15-24

The Role of Statistics in Engineering

1-1 The Engineering Method and Statistical 1-3 Mechanistic and Empirical Models

Thinking 1-4 Probability and Probability Models

1-2 Collecting Engineering Data

1-1 The Engineering Method and Statistical Thinking

O Explain engineering problem solving process. 0 Describe the application of statistics in the engineering problem solving process. 0 Distinguish between enumerative and analytic studies.

" M * * X p ~ ~ ~ v m _ _ I ^ ^ I ~ _ - w- v4--aIIMI*ldXw v l lm.w~-XI~mm- --.-* UXmXIIIt --. * - /wM zXI,(Y -iiymw r " a- -a c

Engineering Engineers develop practical solutions and techniques for engineering problems by Problem applying scientific principles and methodologies. Existing systems are improved

Solving and/or new systems are introduced by the engineering approach for better Process harmony between humans, systems, and environments.

In general, the engineering problem solving process includes the following steps: 1 . Problem definition: Describe the problem to solve. 2. Factor identification: Identify primary factors which cause the problem. 3. Model (hypothesis) suggestion: Propose a model (hypothesis) that

explains the relationship between the problem and factors. 4. Experiment: Design and run an experiment to test the tentative model. 5. Analysis: Analyze data collected in the experiment. 6. Model modification: Refine the tentative model. 7. Model validation: Validate the engineering model by a follow-up

experiment. 8. Conclusion (recommendation): Draw conclusions or make

recommendations based on the analysis results. (Note) Some of these steps may be iterated as necessary.

1-2 Chapter 1 The Role of Statistics in Engineering

Application of Statistical methods are applied to interpret data with variability. Throughout the Statistics engineering problem solving process, engineers often encounter data showing

variability. Statistics provides essential tools to deal with the observed variability. Examples of the application of statistics in the engineering problem solving process include, but are not limited to, the following:

1. Summarizing and presenting data: numerical summary and visualization in descriptive statistics (Chapter 6).

2. Inferring the characteristics (mean, median, proportion, and variance) of single/two populations: z, t, x*, and F tests in parametric statistics (Chapters 7-10); signed, signed-rank, and rank-sum tests in nonparametric statistics (Chapter 15).

3. Testing the relationship between variables: correlation analysis (Chapter 11); categorical data analysis (Chapter 8).

4. Modeling the causal relationship between the response and independent variables: regression analysis (Chapters 1 1-12); analysis of variance (Chapters 13-14).

5. Identifying the sources of variability in response: analysis of variance (Chapters 13-14).

6. Evaluating the relative importance of factors for the response variable: regression analysis (Chapter 12); analysis of variance (Chapter 14).

7. Designing an efficient, effective experiment: design of experiment (Chapters 13-14).

These applications would lead to development of general laws and principles such as Ohm's law and design guidelines.

Enumerative Two types of studies are defined depending on the use of a sample in statistical vs. inference:

Analytic 1. Enumerative study: Makes an inference to the well-defined population from Studies which the sample is selected. (e.g.) defective rate of products in a lot

2. Analytic study: Makes an inference to a future (conceptual) population. (e.g.) defective rate of products at a production line

1-2 Collecting Engineering Data

0 Explain three methods of data collection: retrospective study, observational study, and designed experiment.

Data Three methods are available for data collection: Collection 1. Retrospective study: Use existing records of the population. Some crucial

Methods information may be unavailable and the validity of data be questioned. 2. Observational study: Collect data by observing the population with as

minimal interference as possible. Information of the population for some conditions of interest may be unavailable and some observations be contaminated by extraneous variables.

3. Designed experiment: Collect data by observing the population while controlling conditions on the experiment plan. The findings would obtain scientific rigorousness through deliberate control of extraneous variables.

1-3 Mechanistic and Empirical Models 1 -3

1-3 Mechanistic and Empirical Models

Explain the difference between mechanistic and empirical models. **_* j.i,. l _ " l U _ ~ ~ = ~ ~ - ~ ~ - ~ a ~ " ~ ~ " ~ ' ~ w " ~ . w ~ ..,ese.ss. w w . ~ w ~ m ~ ~ - ~ I _ 1 8 V 8 U ~ _ n _ P P ~ , X P ~ " a " ~ ml-(.s P I I I , I - I - L - ~ ~ " ~ * ~ ~ ~ " ~ 8 w ' w a , n ~ , ~ . a . P l i n l a " r * a x u ~ ~ a - r ~ a r a r ~ i a a n I nn"en/.ir~la.vrax**~~m~a~'~~.~~"r.nc~xlr"xr.~~awr~arrex"~~>~-.~~w

Mechanistic Models (explaining the relationship between variables) can be divided into two vs. categories:

Empirical 1. Mechanistic model: Established based on the underlying theory, principle, or Models law of a physical mechanism.

E (e.g.) I = - + E (Ohm's law)

R where: I = current, E = voltage, R= resistance, and E = random error

2. Empirical model: Established based on the experience, observation, or experiment of a system (population) under study.

(e.g.1 Y = P o + P I X + E

where: y = sitting height, x = stature, and E = random error

1-4 Probability and Probability Models

O Describe the application of probability in the engineering problem solving process. X.__ ___" *--* ~ ~ ~ w m ~ ~ n _ m w ~ B - ~ - ~ / - r r a a ~ m ~ " m x ~ x a ~ z m ~ ~ ~ - - - - u m-smaaxwurw --arrwar-= --m

Application of Along with statistics, the concepts and models of probability are applied in the Probability engineering problem solving process for the following:

1. Modeling the stochastic behavior of the system: discrete and continuous probability distributions.

2. Quantifying the risks involved in statistical inference: error probabilities in hypothesis testing.

3. Determining the sample size of an experiment for a designated test condition: sample size selection

Probability

2-1 Sample Spaces and Events 2-6 Independence

2-2 Interpretations of Probability 2-7 Bayes' Theorem

2-3 Addition Rules 2-8 Random Variables

2-4 Conditional Probability Answers to Exercises

2-5 Multiplication and Total Probability Rules

2-1 Sample Spaces and Events

0 Explain the terms random experiment, sample space, and event. Define the sample space and events of a random experiment.

0 Define a new joint event from existing events by using set operations. 0 Assess if events are mutually exclusive andlor exhaustive.

Use a Venn diagram and/or tree diagram to display a sample space and events.

Random When different outcomes are obtained in repeated trials, the experiment is called Experiment a random experiment. Some sources of the variation in outcomes are controllable

and some are not in the random experiment. (e.g.) Sources of variability in testing the life length of an INFINITY light bulb

( I ) Material (2) Manufacturing process (3) Production environment (temperature, humidity, etc.) (4) Measurement instrument (5) Drift of current (6) Observer

Sample A sample space is the set of possible outcomes of a random experiment (denoted Space as S). Two types of sample space are defined:

(9 1. Discrete Sample Space: Consists of finite (or countably infinite) outcomes. (e.g.) Tossing a coin: S = {head, tail)

2. Continuous Sample Space: Consists of infinite and innumerable outcomes. (e.g.) The life length of the INFINITY light bulb (X): S = {x; x 2 0)

2-2 Chapter 2 Probability

Event 0

Set Operations

Laws for Set Operations

Mutually Exclusive1

Exhaustive Events

An event is a subset of the sample space of a random experiment (denoted by E).

Three set operations can be applied to define a new joint event from existing events: 1. Union (E l u E2 ): Combines all outcomes of El and E2.

2. Intersection ( E l n E2 ): Includes outcomes that are common in El and E2.

3. Complement ( E' or E ): Contains outcomes that are in E. Note that (Ef)' = E .

The following laws are applicable to the set operations: 1. Commutative Law

E , n E 2 = E 2 n E l

E l u E 2 = E 2 u E l 2. Distributive Law

( E l n E 2 ) u E 3 = ( E l u E , ) n ( E , u E , )

(El u E,) n E, = (El n E,) u (E, n E,) 3. DeMorgan's Law

(E, ~ E , ) ' = E , ' u E ;

(E, u E2)'= E; n E;

A collection of events El, E2,. . . , E k is said to be mutuallv exclusive (see Figure 2-I), if the events do not have any outcomes in common, i.e.,

E, n E, = 0 for all pairs (i, j), i # j

Next, the collection of events El, E2,. . . , Ek is said to be exhaustive (see Figure 2- I), if the union of the events is equal to S, i.e.,

El u E , u . . . u E , = S

r Example 2.1

Figure 2-1 Mutually exclusive and exhaustive events.

Two dice are cast at the same time in an experiment. 1. (Sample Space) Define the sample space of the experiment. h The sample space is S = {(x, y); x = 1,. . . , 6, y = 1,. . . , 6 ) , where x and y

represent the outcome of each die. Since there are 6 possible outcomes with the first and second dice each, the sample space consists of a total of 6 x 6 =

36 outcomes. 2. (Event) Of the sample space, find the pairs whose sum is 5 (El) and the pairs

whose first die is odd (E2).

2-1 Sample Spaces and Events 2-3

Example 2.1 (cont.)

Exercise 2.1 (MR 2-33)

I 3. (Set Operation) Find El u E, , E, n E, , and E;

4. (Mutually Exclusive Events) Are El and E2 mutually exclusive?

h No, because El n E, # 0.

5. (Exhaustive Events) Are El and E2 exhaustive?

h No, because El u E, # S .

The rise times (unit: min.) of a reactor for two batches are measured in an experiment.

1. Define the sample space of the experiment.

2. Define El where the reactor rise time of the first batch is less than 55 min. and E2 where the reactor rise time of the second batch is greater than 70 min.

3. Find E, u E, , El n E, , and E; .

4. Are El and E2 mutually exclusive?

5. Are El and E2 exhaustive?

Diagrams Diagrams are often used to display a sample space and events in an experiment: 1. Venn Diagram: A rectangle represents the sample space and circles indicate

individual events, as illustrated in Figure 2-2.

Figure 2-2 Venn diagrams: (a) El u E, ; (b) E, n E, ; (c) E; . The areas

shaded indicate the results of union, intersection, and complement, respectively.

7- 2 4 Chapter 2 Probability

Diagrams 2. Tree Diagram: Branches represent possible outcomes (see Figure 2-3). The (cont.) tree diagram method is useful when the sample space is established through a

series of steps or stages.

HHH HHT HTH HTT THH THT TTH TTT

Figure 2-3 Tree diagram for outcomes of tossing three coins at the same time.

2-2 Interpretations of Probability

0 Explain the termprobability. 0 Determine the probability of an event.

Probability The probability of an event means the likelihood of the event occurring in a random experiment. If S denotes the sample space and E denotes an event, the following conditions should be met:

(1) P(S)=I

(2) 0 5 P(E) I 1

( 3 ) P(E, u E, u E, u . a = ) = P(E, ) + P(E,) + P(E,) + -.., where the events are mutually exclusive.

Probability If a sample space includes n outcomes that are equally likely, the probability of of Event each outcome is lln. Thus, the probability of an event E consisting of k equally

likely outcomes is 1 k

P ( E ) = k x - = - n n

where: n = number of possible outcomes in S k = number of equally likely elements in E

(Note) For any event E, P(E f ) = 1 - P(E)

2-3 Addition Rules 2-5

2-3 Addition Rules

0 Find the probability of a joint event by using the probabilities of individual events. I n _LLnj,m *._*~-_^n%"YLn"" ----.--" a.mm *-*," _ilajlm-lm_-,*n, nc-.re.r*r* nxl*ann.~a,m.*.nra nrren-.xlawa rm~am.-n *emarxa,i.Pxa - w ~ ~ . a ~ * . w . ~ d , a , * d * ' ~ ~ . m < w ~ a r*xriri w.-aariexnmi.i w'ww ,ar.i,;

Probability The probability of a joint event can often be calculated by using the probabilities o f Joint of the individual events involved. The following addition rules can be applied to

Event determine the probability of a joint event when the probabilities of existing events are known:

P(El u E , ) = P ( E , ) + P ( E , ) - P ( E l n E , )

= P ( E l ) + P ( E , ) if El n E, # 0 P(El W E , u E , ) = P ( E , ) + P ( E , ) + P ( E , )

-P(El n E , ) - P(El n E , ) - P(E, n E , ) + P(El n E, n E , ) r Example 2.2


(Probability o f Joint Event) An instructor of a statistics class tells students that the probabilities of earning an A, B, C, and D or below are t , 3 , &, and h ,

I respectively. Find the probabilities of ( I ) earning an A or B and ( 2 ) earning a B or below.

Ohr Let E l , E2, E3, and E4 denote the events of earning an A, B, C, and D or below, respectively. These individual events are mutually exclusive and exhaustive because

P(El u E, u E, u E , ) = P ( E l ) + P ( E , ) + P ( E , ) + P ( E 4 )

I (I) The event of earning an A or B is E, u E, . Therefore,

(2 ) The event of earning a B or below is E, u E, u E4 , which is equal to

E; . Therefore,

Test results of scratch resistance and shock resistance for 100 disks of polycarbonate plastic are as follows:

Shock Resistance High (4 LOW ( B ' )

Scratch High (A) 80 9 Resistance LOW ( A '2 6 5

m8* ww*-e -* *AwM--* --** m- *ww-wwBwws*+e m--w--* w WAww m m * *em-- A * ** a sw w#--re- * ~ = ~ ~ a Me r m

Let A and A ' denote the event that a disk has high scratch resistance and the event that a disk has low scratch resistance, respectively. Let B and B ' denote the event that a disk has high shock resistance and the event that a disk has low shock resistance, respectively.

1. When a disk is selected at random, find the probability that both the scratch and shock resistances of the disk are high. -


Exercise 2.2 (cont.)

2. When a disk is selected at random, find the probability that the scratch or shock resistance of the disk is high.

3. Consider the event that a disk has high scratch resistance and the event that a disk has high shock resistance. Are these two events mutually exclusive?

2-4 Conditional Probability

O Calculate the conditional probability of events.

Conditional The conditional probability P(B 1 A) is the probability of an event B, given an Probability event A. The following formula is used to calculate the conditional probability:

P(B ( A) = P ( A n B )

, where P(A) > 0 P(A) r Example 2.3

h Exercise 2.3

(Conditional Probability) A novel method to screen carpal tunnel syndrome (CTS) at the workplace is tested with two groups of people: 50 workers having CTS and 50 healthy workers without CTS. Let A and A' denote the event that a worker has CTS and the event that a worker does not have CTS, respectively. Let B and B ' denote the event that a CTS test is positive and the event that a CTS

1 test is negative. resnectivelv. The summarv of CTS test results is as follows:

I " I 1

Test Result Negative ( B ' ) Positive ( 8 )

CTS (A) 10 Group

Healthy ( A ' ) 45

Find the probability that a CTS test is positive (B) when a worker has CTS (A).

In Exercise 2.2, find the probability that a disk has high scratch resistance (A) when it has high shock resistance (B).

2-5 Multiplication and Total Probability Rules

R Apply a total probability rule to find the probability of an event when the event is partitioned into several mutually exclusive and exhaustive subsets.

2-5 Multiplication and Total Probability Rules 2-7

Multiplication From the definition of conditional probability, Rule P(B n A) = P(B I A)P(A) = P(A I B)P(B)

Total When event B is partitioned into two mutually exclusive subsets B n A and Probability B n A' (see Figure 2-4),

Rule P(B) = P(B n A) + P(B n A')

= P(B I A)P(A) + P(B I A')P(A')

Figure 2-4 Partitioning event B into two mutually exclusive subsets: B n A and ~n A'.

As an extension, when event B is partitioned by a collection of k mutually exclusive and exhaustive events El , E2,. . . , Ek,

P(B)= P ( B n E l ) + P ( B n E2)+. . .+ P ( B n E,)

=P(B I El )P(El ) + P(B I E2)P(E2) + - . a + P(B I E,)P(E,)

r Example 2.4

Figure 2-5 Partitioning event B into k mutually exclusive subsets: B n E l ,

(Total Probability Rule) In Example 2.3, the CTS screening method experiment indicates that the probability of screening a worker having CTS (A) as positive (B) is 0.8 and the probability of screening a worker without CTS ( A ') as positive (B) is 0.1, i.e.,

P(B I A) = 0.8 and P(B I A') = 0.1

Suppose that the industry-wide incidence rate of CTS is P(A) = 0.0017. Find the probability that a worker shows a positive test (B) for CTS at the workplace.

P(A) = 0.0017 and P(Ar) = 1 - P(A) = 0.9983 By applying a total probability rule,

P(B) = P(B I A)P( A) + P(B I A')P(Ar) =0.8~0.0017+0.1~0.9983=0.101


2-6 Independence

d

0 Explain the term independence between events. Assess the independence of two events.

Exercise 2.4 ( M R 2-97)

Independence Two events A and B are stochastically independent if the occurrence of A does of Events affect the probability of B and vice versa. In other words, two events A and B.

are independent if and only if (1) P(A I B) = P(A) (2) P(B I A) = P(B) ( 3 ) P(A n B) = P(A)P(B)

Customers are used to evaluate preliminary product designs. In the past, 95% of highly successful products, 60% of moderately successful products, and 10% of poor products received good reviews. In addition, 40% of product designs have been highly successful, 35% have been moderately successful, and 25% have been poor products. Find the probability that a product receives a good review.

(Derivation) P(A n B ) = P(A)P(B)

P(B I A) = P(A n B)

= P(B) if A and B are independent. P(A)

r Example 2.5

Eb Exercise 2.5

I (Independence) For the CTS test results in Example 2.3, the following probabilities have been calculated:

45 4 P(B) = - and P(B I A) = -

100 5

I Check if events A and B are independent,

I 4 45 Ikr Since P(B I A) = - # P(B) = -, A and B are not independent. This

5 100

I indicates that information from the CTS test is useful to screen workers having CTS at the workplace.

I For the resistance test results in Exercise 2.2, the following probabilities have been calculated:

I 89 40 P(A) = - and P(A I B) = -

100 43 I Assess if events A and B are independent.

2-7 Bayes' Theorem 2-9

2-7 Bayes' Theorem

p m m----m--7a----

Apply Bayes' theorem to find the conditional probability of an event when the event is partitioned into several mutually exclusive and exhaustive subsets.

Bayes' From the definition of conditional probability, Theorem

P(A I B ) = P(A n B) - P(B I A)P(A) - - - P(B I A)P(A)

P(B) P(B) P(B n A) + P(B n A')

The multiplication rule for a collection of k mutually exclusive and exhaustive events El , E2,. . . , Ek and any event B is

P(B) = P(B I E l )P(E,) + P(B I E2 )P(E2) + ... + P(B I Ek )P(Ek)

From the two expressions above, the following general result (known as Bayes' theorem) is derived:

Example 2.6 (Bayes' Theorem) In Examples 2.3 and 2.4, the following probabilities have been identified:

P(B I A) = 0.8, P(B I A') =0.1, P(A) = 0.0017, and P(B) = 0.101 Find the probability that a worker has CTS (A) when the test is positive (B). h By applying Bayes' theorem,

P(AI B) = P(B I A)P(A) - - P(B I A)P(A) P(B I A)P(A) + P(B I A')P( A') P(B)

- - 0.8 x 0.0017 = 0.013

0.101 Since the incidence rate of CTS in industry is low, the probability that a worker has CTS is quite small even if the test is positive.

Exercise 2.6 In Exercise 2.4, what is the probability that a new design will be highly I successful if it receives a good review?

2-1 0 Chapter 2 Probability

2-8 Random Variables

O Describe the terms random variable X and range o fX. R Distinguish between discrete and continuous random variables.

Random A random variable, denoted by an uppercase such as X, associates real numbers Variable with individual outcomes of a random experiment. Note that a measured value of

Xis denoted by a lowercase such as x = 10.

The set of possible numbers of X is referred to as the range of X. Depending on the type of range, two categories of random variables are defined: 1. Discrete Random Variable: Has a finite (or countably infinite) range.

(e.g.) Tossing a coin: X = 0 for head and X = 1 for tail 2. Continuous Random Variable: Has an interval of real numbers for its infinite

range. (e.g.) The life length of an INFINITY light bulb: X2 0

Answers to Exercises 2-1 1

Answers to Exercises

Exercise 2.1 1. (Sample Space) S = {x; x > 0), where x represents the rise time of the reactor for a certain batch.

2. (Event) E l = {x ;O<x<55) E2 = {x; x > 70)

3. (Set Operation)

E, u E , = { x ; O < x < 5 5 o r x > 7 0 )

E, n E, = 0 E; = (x; x 2 55)

4. (Mutually Exclusive Events)

Yes, because E, n E, = 0.

5. (Exhaustive Events)

No, because E, u E, + S .

Exercise 2.2 I (Probability of Joint Event)

I 80 3. Since P(A n B) = - # 0 , A and B are not mutually exclusive.

100

Exercise 2.3 I (Conditional Probability)

Exercise 2.4 (Total Probability Rule) Let El , E2, and E3 represent the event of being a highly successful product, the event of being a moderately successful product, and the event of being a poor product, respectively. Also, let G denote the event of receiving good reviews from customers.

P

2-1 2 Chapter 2 Probability

The events El , E2, and E3 are mutually exclusive and exhaustive because P(El u E, u E,) = P ( E l ) + P ( E 2 ) + P(E3) = 1 = P(S)


By applying a total probability rule,

P(G) = P(G I El )P(El) + P(G I E2 )P(E2 + P(G I E, )P(E3 = 0.95 x 0.40 + 0.60 x 0.35 + 0.10 x 0.25 = 0.62

Then, P(GIEl)=0.95, P(GIE2)=0.60, P(GIE,)=O.lO,

P(E l ) = 0.40, P(E2) = 0.35, and P(E3) = 0.25

Exercise 2.5 ( (Independence)

I 40 89 Since P(A I B) = - # P(A) = - , A and B are not independent. This

43 100

I indicates that the scratch resistance and shock resistance of disk are stochastically related.

Exercise 2.6 (Bayes' Theorem)

By applying Bayes' theorem,

P(E1 I G ) = P(G I El )P(E, )

P(G I E, )P(E, ) + P(G I E2 )P(E, + P(G I E, )P(E3

- - 0.95 x 0.40 = 0.62

0.95 x 0.40 + 0.60 x 0.35 + 0.10 x 0.25

Discrete Random Variables and Probability Distributions

3- 1 Discrete Random Variables

3-2 Probability Distributions and Probability

Mass Functions

3-3 Cumulative Distribution Functions

3-4 Mean and Variance of a Discrete Random

Variable

3-5 Discrete Uniform Distribution

3-6 Binomial Distribution

3-7 Geometric and Negative Binomial

Distributions

3-7.1 Geometric Distribution

3-7.2 Negative Binomial Distribution

3-8 Hypergeometric Distribution

3-9 Poisson Distribution

Summary of Discrete Probability Distributions

EXCEL Applications


3-1 Discrete Random Variables

See Section 2-8. Random Variables to review the notion of discrete random variable.

3-2 Probability Distributions and Probability Mass Functions

Distinguish between probability mass function and cumulative distribution function. 0 Determine the probability mass function of a discrete random variable.

Probability A probability distribution indicates how probabilities are distributed over possible Distribution values of X.

Two types of functions are used to express the probability distribution of a discrete random variable X 1. Probability Mass Function (p.m.f.): Describes the probability of a value of X,

i.e., P(X= xi).

3-2 Chapter 3 Discrete Random Variables and Probability Distributions

Probability Distribution

(cont. )

Probability Mass Function

(p.m.f.)

r Example 3.1

2. Cumulative Distribution Function (c.d.f.): Describes the sum of the probabilities of values ofXthat are less than or equal to a specified value, i.e., P(X I xi).

The probability mass function of a discrete random variable X, denoted asf(xi), is f (x i )= P ( X = x i ) ,xi=x1,x2 ,... ,xn

and satisfies the following properties (see Figure 3- 1): (1) f (xi) 2 0 for all xi

Figure 3-1 Probability mass function of casting a die.

(Probability Mass Function) The grades of n = 50 students in a statistics class are summarized as follows:

Grade (X) A B C D or below

(X = 1) (x = 2) (x = 3) (X = 4) No. of students 10 20 15 5

-m-*--(1ILn_--mm-"~-~m-7-m"-~~-"b-~%"~-~"~-~-a ~ X " P I I " ~ ~ X I I X T X E ~ , - , ~ ~ , ~ " b ~ ~ ~ ~ . ~ " ~ - ~ ~ " ~ ~ ~ ~ " ~ ~ . ~ ..llimi(."a .XmPI. X.,lVII,*I"" I-XXY. ".," -Cll"lln .*?". -lil

Let X denote a grade in statistics. Let the values 1,2, 3, and 4 represent an A, B, C, and D or below, respectively. Determine the probability mass function of X and plot f (xi) .

Probability mass function of X

3-3 Cumulative Distribution Functions 3-3

Exercise 3.1 The orders from n = 100 customers for wooden panels of various thickness (X) (MR 3-10) I are summarized as follows:

I Wooden Panel Thickness (X; unit: inch)

i + 3 - 8

No. of customer orders 20 70 10 * e_l*_m_Xm *-r_U-l_j___**jle-s-*~~x **-rx;xr inr mmmx#d^naea.dm rlaixrxi xiun r m m a a a a * a - r x u a s - x x

Determine the probability mass function of X and plot f (x, ) .


Determine the cumulative distribution function of a discrete random variable.

Cumulative The cumulative distribution function of a discrete random variable X, denoted as Distribution F(x), is

Function F(x)= P(X l x ) = f (x i ) (c.d.f.) X, SX

and satisfies the following properties (see Figure 3-2): (1) 0 I F(x) 51

(2) F(x) 5 F ( y ) ifx l y

Example 3.2

Figure 3-2 Cumulative distribution function of casting a die.

(Cumulative Distribution Function) In Example 3-1, the following probabilities have been calculated:

1 2 3 1 P ( X = l ) = - , P ( X = 2 ) = - , P ( X = 3 ) = - , and P ( X = 4 ) = -

5 5 10 10

Determine the cumulative distribution function of X and plot F(x)

h By using the probability mass function of X,


Example 3.2 (cont.)

Cumulative distribution function of X

I Determine the cumulative distribution function of X and plot F(x) .

Exercise 3.2

3-4 Mean and Variance of a Discrete Random Variable

In Exercise 3-1, the following probabilities have been calculated:

P ( X = $ ) = 0 . 2 , P (X=$)=0 .7 , and ~ ( ~ = $ ) = 0 . l

amMm #ww-mmmwm-*8-m-----m- ---=- -mmm-*-- BBm

Calculate the mean, variance, and standard deviation of a discrete random variable.

Mean The mean of X, denoted as p or E(X) , means the expected value o f X

Variance The variance ofX, denoted as o2 or V(X) , indicates the dispersion ofXabout p: of X (oZ)

02 = V ( X ) = C ( x -p) l f (x) = E x 2 f (X) -pi X X

The standard deviation of X is o = ,/V(X)

(Derivation) V(X) = Z x 2 f (x) - p2 (see page 67 of MR) X

Example 3.3 (Mean, Variance, and Standard Deviation) Suppose that, in an experiment of casting a single die, six outcomes (X) are equally likely. Determine the mean, variance, and standard deviation of X.

3-5 Discrete Uniform Distribution 3-5

Example 3.3 ( chr The probability mass function of X is

I Therefore,

3-5 Discrete Uniform Distribution

I

Describe the probability distribution of a discrete uniform random variable. 0 Determine the probability mass function, mean, and variance of a discrete uniform random

variable.

Exercise 3.3

Discrete A discrete uniform random variable X has an equal probability for each value in Uniform the range o f X = [a, b], a < b (see Figure 3-3). Thus, the probability mass Random function of X is

In Exercise 3-1, the following probabilities have been calculated:

P ( X = $ ) = 0 . 2 , P (X=$)=0 .7 , and P(X=;)=O.~

Determine the mean, variance, and standard deviation of X.

Variable 1 f (x) = , w h e r e x = a , a + l , ..., b

b - a + 1

Figure 3-3 A discrete uniform distribution.

The mean and variance of X are b + a p=- (b-a+112 -1

and o = 2 12


Example 3.4 Suppose that six outcomes are equally likely in the experiment of casting a single ld ie .

1. (Probability Mass Function; Discrete Uniform Distribution) Determine the probability mass function of the number (X) of the die.

h Since x = 1,2, . . . , 6, a = 1 and b = 6. Therefore, the probability mass function of X is

I 2. (Probability) Find the probability that the number (X ) of the die in the experiment is greater than three.

1 3. (Mean and Variance) Calculate the mean and variance of X (h Since a = 1 and b = 6, the mean and variance of X are

I (Note) Check that the values of p and o2 above are the same with the answers in Example 3.2.

( 2. Calculate the mean and variance o f X


3. Calculate the mean and variance of the number of letters in 100 product codes (Y = 1002'). Note that E(cX) = cE(X) and V(cX) = c2v(X) (refer to Section 5-7. Linear Combinations of Random Variables for further detail).

Suppose that product codes of 2, 3, or 4 letters are equally likely.

1. Determine the probability mass function of the number of letters (X) in a product code.

3-6 Binomial Distribution

Describe the terms Bernoulli trial and binomial experiment. O Describe the probability distribution of a binomial random variable.

Determine the probability mass function, mean, and variance of a binomial random variable.

Binomial A binomial experiment refers to a random experiment consisting of n repeated Experiment trials which satisfy the following conditions:

(1) The trials are independent, i.e., the outcome of a trial does not affect the outcomes of other trials,

(2) Each trial has only two outcomes, labeled as 'success' and 'failure,' and (3) The probability of a success (p) in each trial is constant.

3-6 Binomial Distribution 3-7

Binomial In other words, a binomial experiment consists of a series of n independent Experiment Bernoulli trials (see the definition of Bernoulli trial below) with a constant

(cont.) probability of success (p) in each trial.

Bernoulli A Bernoulli refers to a trial that has only two possible outcomes. Trial (e.g.) Bernoulli trials

(1) Flipping a coin: S = {head, tail) (2) Truth of an answer: S = {right, wrong) (3) Status of a machine: S = {working, broken) (4) Quality of a product: S = {good, defective) (5) Accomplishment of a task: S = {success, failure)

The probability mass function of a Bernoulli random variable X is

The mean and variance of a Bernoulli random variable are 2 p = p and o = p ( l - p )

(Derivation) p and 02 of a Bernoulli random variable

p = ~ x f ( x ) = o x ( l - p ) + l x p = p

Binomial A binomial random variable Xrepresents the number of trials whose outcome is a Random success out of n trials in a binomial experiment with a probability of success p Variable (see Table 3-1).

* If an item selected from a population is replaced before the next trial, the size of the population is considered infinite even if it may be finite. If the probability of success p is constant, the trials are considered independent; otherwise, the trials are dependent.

The probability mass function of X, denoted as B(n, p ) , is

n ! (Note) The number of combinations of x from n: C," =

The mean and variance of X are 2 p = np and o = np(1- p )


r Example 3.5


A test has n = 50 multiple-choice questions; each question has four choices but only one answer is right. Suppose that a student gives hisfher answers by simple guess. 1. (Probability Mass Function; Binomial Distribution) Determine the

probability mass function of the number of right answers (X) that the student gives in the test.

~ . r r Since a 'right answer' is a success, the probability of a success for each question is p = $ . Thus, the probability mass function of X is

I 2. (Probability) Find the probability that the student answers at least 30 questions correctly.

3. (Mean and Variance) Calculate the mean and variance of X.

~ . r r p = np = 50 x 0.25 = 12.5 right answers

o2 = np(1- p) = 50x 0.25 x 0.75 = 9.4 = 3.12

Because all airline passengers do not show up for their reserved seat, an airline sells n = 125 tickets for a flight that holds only 120 passengers. The probability that a passenger does not show up is 0.10. Assume that the passengers behave independently. 1. Determine the probability mass function of the number of passengers (X) who

show up for the flight. 2. Find the probability that the flight departs with empty seats.

3. Calculate the mean and variance ofX.

3-7 Geometric and Negative Binomial Distributions

3-7.1 Geometric Distribution

0 Describe the probability distribution of a geometric random variable. O Compare the geometric distribution with the binomial distribution. 0 Explain the lack of memory property of the geometric distribution.

Determine the probability mass function, mean, and variance of a geometric random variable. 111/ e--*w-*wa _iawaw- _w---m _*a_i_C_ "we- ___WnXa-I-- ammm--m- *Iw--Xlmt(P-Xldll-"' nm- m-MwA--m-H ",*--* 9-s w* a--**mw"w a -" -*,- "--*- -m+ -

3-7 Geometric and Negative Binomial Distributions 3-9

Geometric A geometric random variable Xrepresents the number of trials conducted until Random the first success in a binomial experiment with a probability of success p: Variable Fail Fail Fail Success u t

Fail (x - 1) times Succeed at the x' time

P ( X = x ) = ( ~ - ~ ) ~ - l X P

The probability mass function of X is

f ( ~ ) = ( l - ~ ) " - ' ~ , x = 1,2, ...

The mean and variance of X are 1 2 1 - P

p=- and o =- P p2

The probability of X decreases geometricallv as illustrated in Figure 3-4.

0 2 4 6 8 10 12

Figure 3-4 Geometric distributions with selected values ofp.

Geometric vs. In the geometric distribution the number of trials is a random variable and the Binomial number of successes is one, whereas in the binomial distribution the number of

Distributions trials is constant and the number of successes is a random variable (see Table 3- 2).

Table 3-2 The Characteristics of Binomial and Geometric Distributions

Distribution Population* Probability of No. of No. of i Succe~sfp)~ i Successes f

Geometric Infinite Constant i Variable (X) 1 -- wB- m ~ ~ ~ ~ - M m w d ~ ~ ~ ~ ~ ~ - - - a ~ ~ ~ ~ ~ ~ - - ~ - - ~ ~ ~ - * ~ - M - ~ m ~ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .?. . . . . . . . . .+ * If an item selected from a population is replaced before the next trial, the size of the population is considered infinite even if itmay be finite. If the probability of successp is constant, the trials are considered independent; otherwise, the trials are dependent.

3-1 0 Chapter 3 Discrete Random Variables and Probability Distributions

Lack of The lack of memory property means that the system does not remember the Memory history of previous outcomes so that the timer of the system can be reset at any Property time, i.e.,

P ( t < X < t + A t ) P ( X < t + A t I X > t ) = = P ( X < At)

P (X > t)

The lack of memory property of the geometric distribution implies that the value o f p does not change with trial. Due to the memoryless property of the geometric distribution, the number of trials (X) until the next success can be counted from any trial. The geometric distribution is the & discrete distribution having the lack of memory property.

(Proof) The lack of memory property The conditional probability P (X < t + At ( X > t) is

Note that the sum of the first n terms of the geometric sequence

a , ar , ar 2 , . . . , ar is n-1

k a - urn S, = C a r =- , where r # 1

k=O l - r

Accordingly,

(t+At-1)-1 t -1

C ~ ( 1 - p ) ~ - - C P ( ~ - P ) ~ - - k=O k=O

t-1 , where k = x - 1

1 - C P ( ~ - P)' k=O

t+At-l P - ~ ( 1 - P ) - P - ~ ( 1 - P)'

- - l - ( l - P ) l - ( l - p )

1 - P - ~ ( 1 -PIt l - ( l - P )

3-7 Geometric and Negative Binomial Distributions 3-1 1

Lack of Memory Property

(cont.) The probability P(X < At) is

At-I

P(X < At) = P(X 2 At - 1) = (1 - p)"-' p

(At-I)-]

= ~ ( I - ~ ) " - ~ ~ , where k = x - l

Therefore, P ( X < t + A t I X > t ) = P ( X < A t )

t Example 3.6


Suppose that the probability of meeting a 'right' person for marriage is 0.1 at a blind date and blind dates are independent (i.e., the probability of meeting a right person for marriage is constant).

1. (Probability Mass Function; Geometric Distribution) Determine the probability mass function of the number of blind dates (X) to make until succeeding at meeting a right person for marriage.

h Since meeting a right person for marriage is a success, the probability of success is p = 0.1. Thus, the probability mass function of X is

f (x)=(1-p)x-1p=0.9x- '0.1, x = 1 ,2 ,...

2. (Probability) Find the probability of meeting a right person for marriage within 10 blind dates.

I 3. (Mean and Variance) Calculate the mean and variance of X.

Suppose that the probability of a successful optical alignment in the assembly of an optical data storage product is 0.8. Assume that the trials are independent.

1 . Determine the probability mass function of the number of trials (X ) required until the first successful alignment.

2. Find the probability that the first successful alignment requires four trials maximum.

3. Calculate the mean and variance of X.


3-7.2 Negative Binomial Distribution

0 Describe the probability distribution of a negative binomial random variable. Compare the negative binomial distribution with the binomial distribution.

O Determine the probability mass function, mean, and variance of a negative binomial random variable.

Negative A negative binomial random variable X represents the number of trials conducted Binomial until r successes in a binomial experiment with a probability of success p: Random Fail Fail Success - - . Success Fail Success Variable v f

Succeed (r - 1) times during (x - 1) times Make the rth success (= Fail (x - r) times during (x - 1) times) at the xth time


The mean and variance of X are r 2 4 1 -PI p=- and =- P P

Negative In the binomial distribution the number of trials is constant and the number of Binomial vs. successes is a random variable, whereas the ovvosite becomes true in the negative

Binomial binomial distribution (this is how the 'negative binomial' is named) (see Table 3- Distributions 3).

Table 3-3 The Characteristics of Binomial and Negative Binomial Distributions -----

Distribution Population* Probability of I No. Success (p)t ;

r Example 3.7

Binomial Negative binomial

Infinite

Infinite

Constant j Constant (n) Variable (X) i Constant i Variable (X) Constant (r) j - ?"*-""es 8 e*me-e'ea#- # * w e ~ v ~ - - ~ - - a - - - ~ - m w *>----"--= .... . . . . . . . . . . . . . , . . . . . . . . . .r * If an item selected from a population is replac t trial, the size of the

population is considered infinite even if it may be finite. If the probability of success p is constant, the trials are considered inde~endent; otherwise. the trials are de~endent.

A travel company is recruiting new members by making calls for the membership program ECQNQTRAVEL. A cost analysis indicates that at least 20 new members per day be recruited to maintain the program. Suppose that the probability of recruiting a new member per call is 0.2.

1. (Probability Mass Function; Negative Binomial Distribution) Determine the probability mass function of the number of calls (4 to make to recruit 20 new members per day.

3-8 Hypergeometric Distribution 3-1 3

Example 3.7 (cont.)


h Since recruiting a new member is a success, the probability of success is p =

0.2 and the number of successes is r = 20. Thus, the probability mass function of X is

I 2. (Probability) Find the probability of recruiting 20 new members per day by making 100 calls.


r 20 P=-=- = 100 calls per day

p 0.2

An electronic scale at an automated filling operation stops the manufacturing line after three underweight packages are detected. Suppose that the probability of an underweight package is 0.01 and fills are independent. 1. Determine the probability mass function of the number of fills (X) before the

manufacturing line is stopped. 2. Find the probability that at least 100 fills can be completed before the

manufacturing line is stopped. 3. Calculate the mean and variance of X,

i 3-8 Hypergeometric Distribution

R Describe the probability distribution of a hypergeometric random variable. 0 Compare the hypergeometric distribution with the binomial distribution. R Determine the probability mass function, mean, and variance of a hypergeometric random

variable. 0 Apply the binomial approximation to a hypergeometric random variable.

---m-w--- -- rh # -- *- a a ** -- L

Hyper- A hypergeometric random variable X represents the number of successes in a geometric sample of size n that is selected at random without replacement from a finite

Random population of size N consisting of K successes and N - K failures. Since each Variable item selected from the population is not replaced, the outcome of a trial depends

on the outcome(s) of the previous trial(s); therefore, the probability of success p at each trial is not constant.


Hyper- The probability mass function of X is geometric

Random Variable f (XI = (:)(I I) , x=max{O,n-(N-K)) to min{K,n)

(cont.)

The mean and variance of X are

2 p = np and o = np(1- p ) - K ( z l) where p = - N

(Note) The variance of a hypergeometric random variable is different from the variance of a binomial random variable by (N - n)l(N - I), which is called finite population correction factor.

Hyper- In the hypergeometric distribution the population is finite and the probability of geometric vs. success is changing, whereas in the binomial distribution the population is

Binomial infinite and the probability of success is constant (see Table 3-4). Distributions

Table 3-4 The,.~h.a~~~~.c~i~~~c~..~fBin~m~.~!..~.qd Hypergeometric Distributions - : P-eters

Distribution i Population* Probability of i No. of Trials No. o f Success 03)' Successes

i Finite Hyper- ; (K successes; Changing ; Constant (n) Variable (X) geometric IN - K failuresL-

* m-,, *lan--.. - a # * b a r a - -xsx X

* If an item selected from a populatio-d before the next trial, thesize oit-he A

population is considered infinite even if it may be finite. If the probability of successp is constant, the trials are considered independent; otherwise, the trials are dependent. r Example 3.8 An instructor of a statistics class is planning to interview a sample of n = 10

students who are randomly selected from the class. The class has a total of 30 students, consisting of 20 traditional and 10 non-traditional students.

1. (Probability Mass Function; Hypergeometric Distribution) Determine the probability mass function of the number of non-traditional students (X) in the sample.

o-rr The size of the population is N = 30. Since a non-traditional student is a success, the number of successes in the population is K = 10.

To determine the range of the number of non-traditional students (X) in the sample, calculate the following:

max(0,n-(N - K))=max{O, 10 -(30 - 10)) =max(O, -10) = 0 min{K,n) = min{l0,10) = 10

( Therefore, the probability mass function of X is

3-8 Hypergeometric Distribution 3-1 5

I 2. (Probability) Find the probability that at least one non-traditional student is in the sample.

1 3. (Mean and Variance) Calculate the mean and variance of X.

I 1 p = np = 10 x- = 3.3 non-traditional students in the sample of 10 students

3

I

Exercise 3.8 A lot of 75 washers contains 5 defectives whose variability in thickness is (MR 3-90) unacceptably large. A sample of 10 washers is selected at random without

replacement.

1. Determine the probability mass function of the number of defective washers (X) in the sample.

2. Find the probability that at least one unacceptable washer is in the sample.

3. Calculate the mean and variance of X.

Binomial Approximation

r Example 3.9

The hypergeometric distribution approximates to the binomial distribution with p =KIN, if the sample size n is relatively small to the population size N. A rule of thumb is that the binomial approximation of a hypergeometric distribution is satisfactory if nlN < 0.1.

(Binomial Approximation; Hypergeometric Distribution) Suppose that X has a hypergeometric distribution with N = 200, n = 10, and K = 20. Find P(X = 2) by using the hypergeometric distribution and approximate binomial distribution. Is the binomial approximation reasonable?

thr The probability mass function of X is

(Note) max(0, n - (N - K)) = max{O, 10 - (200 - 20)) = max(0, -170) = 0

min{K, n) = min(20,lO) = 10

3-1 6 Chapter 3 Discrete Random Variables and Probability Distributions

Example 3.9 (cont.)

r$ Exercise 3.9

Thus,

The approximate binomial distribution of the hypergeometric random variable Xis

K 20 (Note) p = - = - = 0.1

N 200

Based on the binomial approximation,

' Since n is relatively small to N (nlN = 101200 = 0.05), the binomial 1 approximation of the hypergeometric random variable Xis reasonable.

Suppose that X has a hypergeometric distribution with N = 500, n = 50, and K = 100. Find P(X = 10) by using the hypergeometric distribution and approximate binomial distribution. Is the binomial approximation reasonable?

3-9 Poisson Distribution

- m m - m - * B w a s m - 8 w ~ ~ w - - ~ m . ~ ~ ~ - - ~ - ~ m v - ~ ~ ~ - - B ~ ~ ~ - ~ ~ - - ~ ~ ~ ~ ~ m - ~ b -wsws--"-

R Explain the term Poisson process. R Describe the probability distribution of a Poisson random variable. R Compare the Poisson distribution with the binomial distribution.

Determine the probability mass function, mean, and variance of a Poisson random variable.

Poisson Suppose that the occurrence of an event over an interval (of time, length, area, Process space, etc.) is countable and the interval can be partitioned into subintervals.

Then, a random experiment is defined as a Poisson process (see Figure 3-5) if (I) The probability of more than one occurrence in a subinterval is

infinitesimal (approximately zero), (2) The occurrences of the event in non-overlapping subintervals are

stochastically independent, and (3) The probability of one occurrence of the event in a subinterval is the same

throughout all subintervals and prouortional to the length of the subinterval.

3-9 Poisson Distribution 3-1 7

Poisson Process

(cont.)

Poisson Random Variable

Poisson vs. Binomial

Distributions

possible 0, with I - p . . . probabilities 1, with p

I I I

I , 1 2 . . . 1,

subintervals

Figure 3-5 Poisson process.

In other words, the Poisson process is a binomial experiment with infinite n trials. (e.g.) Poisson Process

(1) The number of defects of a product (2) The number of customers in a store (3) The number of automobile accidents (4) The number of e-mails received

A Poisson random variable Xrepresents the number of occurrences of an event of interest in a unit interval (of time, space, etc.) specified.



p = h and 0 2 = h

(Caution) Use consistent units to define a Poisson random variable X and the corresponding parameter h. For example, the following pairs of X and h are equivalent to each other:

X h (counts/unit interval) (average no. of countsfunit interval)

No. of flaws a disk 1 No. of flaws every 10 disks 10 No. of flaws e v e x 100 disks 100

---*-- ----Bzmm --mw-w---m - ---* e-d - a"a - - ,*-wamaad - -a,' - --- a In the Poisson distribution, the number of trials is infinite, whereas in the binomial distribution the number of trials is finite (see Table 3-5). In other words, the Poisson distribution with E(X) = h is the limiting form of the binomial distribution with E(X) = np:

n-x e-hhx lim B(n, p ) = lim ("Ip (1 - p ) n-x = 1im (")(:)' (1 - :) = - n-+- n+- x n+- x x !

(Proof) Poisson vs. binomial distributions Suppose that Xis a binomial random variable with parameters n and p , and let h = np.

Then,

lim (1 - n-x = lim (L)x(l-q)n-x n+- x n+- x!(n - x)! n


Poisson vs. Binomial

Distributions (conb)

r Example 3.10

3L" n(n- l ) . . . (n -x+1) -- - lim X! "+== nX (1 -:In (1 -

As n becomes large, n(n - l ) . . - (n - x + 1)

lim =1, n+- nx

Therefore,

Table 3-5 The Characteristics of Binomial and Poisson Distributions -------- 4 ----.

Distribution Population* Probability of i No. ofTrials i No. of Success ( P ) ~ f : Successes

Constant j Poisson Infinite . Infinite j Variable (4

(n = hln) I a W r n M W Y w e ~ ~ ~ d ~ - ~ ~ e a m w " s ~ ~ e d ~ - ~ , ~ * % ~ w'. m--'rn---. Am&-.- -kw P P l l r l w w m ~ ~ ~ ~ + --m=.m. -..-.----a

* If an item selected from a population is replaced before the next trial, the size of the population is considered infinite even if itmay be finite. If the probability of successp is constant, the trials are considered inde~endent; otherwise, the trials are dependent.

I The number of customers who come to the YUMMY donut store follows a Poisson distribution with a mean of 5 customers every 10 minutes.

1 . (Probability Mass Function; Poisson Distribution) Determine the probability mass function of the number of customers (X) per hour coming to the donut store.

I a For a consistent unit for Xand L, h = E(X) = 5 customers/lO min. x 60 min. = 30 customers per hour

I Therefore, the probability mass function of X is

I 2. (Probability) Find the probability that 40 customers come to the donut store in an hour.

3-9 Poisson Distribution 3-1 9

Example 3.10 (cont.)



~ - r r p = 3L = 30 customers per hour

o2 = I . = 30 = 5.52

The number of cracks that need repair in a section of interstate highway follows a Poisson distribution with a mean of two cracks per mile.

1. Determine the probability mass function of the number of cracks (X) in 5 miles of highway.

2. Find the probability that there are at least five cracks in 5 miles of highway that require repair.

1 3. Calculate the mean and variance of X.


Summary of Discrete Probability Distributions

A. Probability Distributions

Distribution Probability Mass F W o n Mebus §tx%i~n

Uniform

Binomial

Geometric

Negative Binomial

, K I N , n l N , nP N - n Hypergeometric K " ~ ( 1 - P)(-) 3-8

P = N N - 1

x = max (0, n - ( N - K)) to min{K, n )

Poisson e-' h"

, x=O, 1,2, ... h h 3-9 x !

B. Comparison

Geometric Infinite Constant Variable (X) 1

Negative Binomial Infinite Constant Variable (X) Constant (r)

Finite Hypergeometric (K successes; Changing Constant (n) Variable (X)

N - K failures)

Poisson Infinite Constant ( p = Wn)

Infinite Variable (X ) a - arrran"m 4a-;ean mi1 - **am-- *,a r r x a m ---nxwm r nn*. .n unusme-a-- r-n*euc no r-H xn -rrr n" w*uru .c -a^ a era

* If an item selected from a population is replaced before the next trial, the size of the population is considered infinite even if it may be finite. If the probability of successp is constant, the trials are considered indevendent; otherwise, the trials are dependent.

EXCEL Applications 3-21

EXCEL Applications

Example 3.5

Example 3.7

2. (Probability; Binomial Distribution)

(1) Choose Insert ) Function. ( 2 ) Select the category Statistical and the function BINOMDIST. (3) Enter the number of successes (Number-s; 3, number of trials (Trials; n),

probability of a success of interest (Probability-s; p), and logical value (Cumulative; true for the cumulative distribution function and false for the probability mass function).

(4) Find the result circled with a dotted line below.

= 0.999999836 Returns the individual term bmomlal dKtrbut~on probability,

Cumulative is a logical valus: for the cumulative distribut~on funct~on, use TRUE; For ther pobability mars function, use FALSE.

2. (Probability; Negative Binomial Distribution)

(1) Choose Insert > Function. (2) Select the category Statistical and the function NEGBINOMDIST. (3) Enter the number of failures (Number-f; x - r), number of successes

(Number-s; r), and probability of a success of interest (Probability-s; p) (4) Find the result circled with a dotted line below.


Example 3.8

Example 3.10

2. (Probability; Hypergeometric Distribution)

(I) Choose Insert N Function. (2) Select the category Statistical and the function HYPGEOMDIST. (3) Enter the number of successes in the sample (Sample-s; x), size of the

sample (Number-sample; n), number of successes in the population (Population-s; K), and size of the population (Numberqop; N )


2. (Probability; Poisson Distribution)

(1) Choose Insert N Function. (2) Select the category Statistical and the function POISSON. (3) Enter the number of events (X), mean (Mean; A), number of successes in

the population (Population-s; K), and logical value (Cumulative; true for the cumulative distribution function and false for the probability mass function).


Answers to Exercises 3-23


Exercise 3.1

Exercise 3.2

Exercise 3.3

(Probability Mass Function)

P ( X = f ) = 0 . 2 , P (X=f)=0 .7 ,and P(X=$)=0.1

Probability mass function of X

(Cumulative Distribution Function)

By using the probability mass function of X, ~ ( t ) = P ( X S 6 ) = 0 . 2 F($)=P(XS$)=0.2+0.7=0.9

F($) = P(X I $) = 0.9 + 0.1 = 1

x < $

0.2, F I X < $ 1 F (x) =

10.9, + x < $ 1

Cumulative distribution function of X

(Mean, Variance, and Standard Deviation)

F = z x f ( x ) = $ x 0 . 2 + f x0.7+$ x0.1=0.2375 inch X

o2 = X x 2 f ( x ) - ~ ~ =(f)2 ~ 0 . 2 + ( $ ) ~ x0.7+($12 ~ 0 . 1 - 0 . 2 3 7 5 ~ X

= 0.0045

o = ,/V(X> = .\!= = 0.067 inch

3-24 Chapter 3 Discrete Probability Distributions

Exercise 3.4

Exercise 3.5

Exercise 3.6

1 . (Probability Mass Function; Discrete Uniform Distribution) Since x = 2,3,4, a = 2 and b = 4. Therefore, the probability mass function of X i s

2. (Mean and Variance) Since a = 2 and b = 4, the mean and variance of X are

3. (Mean and Variance of a Linearly Combination) The mean and variance of Y = 1 OOX are

E(Y) = E(100X) = lOOE(X) = 1 0 0 ~ 3 = 300

1. (Probability Mass Function; Binomial Distribution) Since "a passenger showing up" is a success and the probability of a success i s p = 1 - 0.1 = 0.9. Thus, the probability mass function of X is

2. (Probability) The flight would have empty seats i f X < 120. Thus, the probability that the flight departs with empty seats is

3. (Mean and Variance) y = np = 125 x 0.9 = 112.5 passengers

o2 = np(1- p) = 125 x 0.9 x (1 - 0.9) = 1 1.2 = 3.32

1 . (Probability Mass Function; Geometric Distribution) Since a successful optical alignment is a success, the probability of success is p = 0.8. Thus, the probability mass function of Xis

f ( x ) = ( l - p ) x - 1 p = 0 . 2 " - ' 0 . 8 , x = l , 2 ,...

I 2. (Probability)


Exercise 3.6 ( 3. (Mean and Variance) (cont.) 1 1

p = - = - = 1.25 trials

Exercise 3.7 1 1. (Probability Mass Function; Negative Binomial Distribution)

Exercise 3.8

1 2. (Probability)

1 3. (Mean and Variance)

r 3 p=-=- = 300 fills

p 0.01

1. (Probability Mass Function; Hypergeometric Distribution) N=75, K = 5 , n = 1 0 max(0, n - (N - K)) = max(0, 10 - (75 - 5)) = max (0, -60) = 0 min{K,n) = min(5, 10) = 5

2. (Probability)

3. (Mean and Variance)

1 p = np = 10 x - = 0.67 unacceptable washers in the sample of 10 washers

15

3-26 Chapter 3 Discrete Probability Distributions

Exercise 3.9 (Binomial Approximation; Hypergeometric Distribution) I The probability mass function of Xis

(Note) max{O,n - (N - K)} = max{0,50 - (500 - 100)) = max(0, -350) = 0

min{K,n) = min{100,50) = 50

Thus,

The approximate binomial distribution of the hypergeometric random variable Xis

~ x ) = ( n ) p x ( - p ) n - x x =[~)0.2x0.8s0-x, x - 0 to 50

K 100 (Note) p = - = - = 0.2 N 500

Based on the binomial approximation,

P(X = 10) = f (10) = = 0.140

Since n is relatively small to N (n/N = 50/500 = 0. l), the binomial approximation of the hypergeometric random variable X is satisfactory.

Exercise 3.10 1. (Probability Mass Function; Poisson Distribution) I h = E(X) = 2 crackslmile x 5 miles = 10 cracks every 5 miles

1 2. (Probability)

3. (Mean and Variance) p = h = 10 cracks every 5 miles

o2 = h = 1 0 = 3 . 2 ~

4 Continuous Random Variables and Probability Distributions

4- 1 Continuous Random Variables

4-2 Probability Distributions and Probability

Density Functions


4-4 Mean and Variance of a Continuous

Random Variable

4-5 Continuous Uniform Distribution

4-6 Normal Distribution

4-7 Normal Approximation to the Binomial

and Poisson Distributions

4-9 Exponential Distribution

4- 10 Erlang and Gamma Distributions

4-1 0.1 Erlang Distribution

4- 10.2 Gamma Distribution

4-1 1 Weibull Distribution

4- 12 Lognormal Distribution

Summary of Continuous Probability

Distributions

EXCEL Applications


4-1 Continuous Random Variables

See Section 2-8. Random Variables to review the notion of continuous random variable.

4-2 Probability Distributions and Probability Density Functions

Distinguish between probability density function and cumulative distribution function. R Determine the probability of a continuous random variable by using the corresponding probability

density function.

Probability Like the probability distribution of a discrete random variable, two types of Distribution functions are used to express the probability distribution of a continuous random

variable X

\

4-2 Chapter 4 Continuous Random Variables and Probability Distributions

Probability 1. Probability Density Function (p.d.f.;J(x)): Describes the densities of Distribution probabilities over a range of X.

(cont.) 2. Cumulative Distribution Function (c.d.f.; F(x)): Represents the probability

ofXthat is less than or equal to a specified value, i.e., P(X I x).

Probability The probability density function f (x) of a continuous random variable X satisfies Density the following properties:

Distribution (1) f (x )>O (p.d. f.)

(2 ) 1 - f ( x W = 1

The integral off (x) from points a to b represents the probability that X belongs to the interval [a, b] (i.e., the area under the density function between a and b as shown in Figure 4-1).

(4) P(X = x) = 0 The probability of every single point is zero because the width of any point is zero. Thus,

P ( a < X I b ) = P ( a < X < b ) = P ( a I X < b ) = P ( a < X < b )

Figure 4-1 Probability density function. The probability P(a I X I b) (shaded) is determined by integrating f (x) from a to b.

Example 4.1 I g b a b i l i t y Density Function) Suppose that the probability density iunction of

f ( ~ ) = e - ~ , x > O = 0, elsewhere

Determine P(X < 2) , P(2 I X < 4) , and P(X 2 4)

I 2 O-w (I) P(X < 2) = &(x)& = e"& = - e-' = -e-2 + e-' = 0.86

l o

7 4-3 Cumulative Distribution Functions 4-3

I Exercise 4.1 Suppose that the probability density function of X is I

f (x)=3x2, O < x < l I

= 0, elsewhere I I

Determine P(X < f) , P($ I X < -$) , and P(X 2 3).

1


m - - - m - - M - v a ~ - - m - - m - - - * wm-m-L#a - O Determine the cumulative distribution function of a continuous random variable. --- * -e-----m-----mFmppwa-~-m-pm~- wm--awmam I

Cumulative The cumulative distribution function of a discrete random variable X is I I

Distribution Function F(x) = P(X l x) = 1 f (u)du

co

(~.d.~.) and satisfies the following properties: (1) 0 I F(x) l l

(2) F(x) l F(y) i fx 5 y t 1

dF(x) (Note) f (x) = -

t' Example 4.2

Exercise 4.2

(Cumulative Distribution Function) In Example 4- 1, the probability density function of X is

f(x)=e-", x > O = 0, elsewhere

Determine the cumulative distribution function of X. h By using the probability density function of X,

6; 1 F(x) = P(X I x) = f (u)du = e-"du = - e-" = -e-" + e-O = 1 - e-" I, Therefore,

x < o F(x) =

1-e-", O I x 1" In Exercise 4- 1, the probability density function of X is

f (x)=3x2, O < X < 1 = 0, elsewhere 6

Determine the cumulative distribution function of X. 1 I

i


4-4 Mean and Variance of a Continuous Random Variable

Calculate the mean, variance, and standard deviation of a continuous random variable.

Mean of X

(P)

Variance of X

(02)

Standard Deviation of

X

Y Example 4.3

The mean (expected value) of X is

\,/,

The variance of X is

The standard deviation of X is

O = JV(X>

1 (Mean, Variance, and Standard Deviation) In Example 4-1, the probability 1 density function of X is

I f (x)=e-" , x > O

= 0, elsewhere Determine the mean, variance, and standard deviation ofX.

(1) The mean of X is

p = xf(x)dx= [xe-"dx R Apply the method of integration by parts as follows:

Thus, - p = R x e - " ~ = - xe-x 1; + Re-"dx=-e-" =-e -= + e o = I

In

I Apply the method of integration by parts as follows:

4-5 Continuous Uniform Distribution 4-5

Example 4.3 1 Then, (cont.)

I Therefore,

x e dx-1=2-1=1 = R -x

1 (3) The standard deviation of X is a = 1 . a

4-5 Continuous Uniform Distribution

Exercise 4.3

0 Describe the probability distribution of a continuous uniform random variable. 0 Determine the probability density function, mean, and variance of a continuous uniform random

variable. .-.. ".'-<* ws,, ~."* -,,,-.. "".m.m>B ", *.* ,a.".,~~**d,a*".+e,<>B~"~-.--".Bw# **a~,d, -v*~-e--*B*%,.",-*w... ~ ~ w ~ , . ' a ~ ~ . ~ ~ " " . ~ ~ . ~ . ~ ~ " ~ ~ : ~ - ~ w ~ ~ - * ~ . ~ - a , ~ ~ ~ ~ , ~ * ~ - ~ - ~ m ~ ~ ~ ~ ~ ~ ~ . a a " . ~ . , - . w A ~ " . ~ ~ , ~ . w " ~ w.*.a # we-*"

In Exercise 4-1, the probability density function of X is

f (x )=3x2 , O < X < 1 = 0, elsewhere

Determine the mean, variance, and standard deviation of X.

Continuous A continuous uniform random variable X has a constant probability density Uniform function over the range o f X (see Figure 4-2): Random 1 Variable f (x)=- , a < x l b

b - a

Figure 4-2 A continuous uniform distribution.

The mean and variance of X are b + a p=- (b-a12

and a =--- ,where a l x l b 2 12

Example 4.4 Suppose that a random number generator produces real numbers that are uniformly distributed between 0 and 100.

1. (Probability Density Function; Continuous Uniform Distribution) Determine the probability density function of a random number (X) generated.


Example 4.4 1 h a = 0, b = 100 (cont.)

m Exercise 4.4 (MR 4-34)

1 - 1 - 1 f(x)=----- , 0 1 x 5 1 0 0 b-a 100-0 100

I 2. (Probability) Find the probability that a random number (X ) generated is between 10 and 90.

1 3. (Mean and Variance) Calculate the mean and variance ofX.

Suppose that the thickness of a flange is uniformly distributed between 0.95 and 1.05 mm. 1. Determine the probability density function of flange thickness (4. 2. Find the probability that the thickness of a flange (X) exceeds 1.02 mm. 3. Calculate the mean and variance of X.

4-6 Normal Distribution

--- Describe the characteristics of a normal distribution.

O Read the z table. 0 Standardize a normal random variable.

Normal A normal random variable with mean y and variance o2 has the probability Random density function Variable (x-p12

1 -- f(x)=-e 202 , - o o < X < o o

&(3

A normal distribution with mean y and variance 02, denoted as N(p, 02), is svmmetric about u and bell-shaved (see Figure 4-3). The symmetry of a normal curve implies

P (y< X) = P(X > p)= 0.5

The parameters p and o2 determine the center and shape of the normal curve, respectively. As illustrated in Figure 4-3, the larger the value of y, the more to the right the center of the normal curve is located; the smaller the value of 02, the sharper the normal curve.

4-6 Normal Distribution 4-7

Normal Random Variable

(cont.)

Probabilities of Normal

Distribution

Standard Normal

Random Variable

Figure 4-3 Normal curves with selected values of p and 02.

Selected probabilities of a normal distribution are displayed in Figure 4-4. The area under the normal curve beyond +30 is quite small (less than 0.01). Since 99.7% of possible values of Xare within the interval (p - 30, p + 30), 6 0 is referred to as the width of a normal distribution.

Figure 4-4 Probabilities of a normal distribution.

The standard normal random variable (denoted as Z) is a normal random variable with mean p = 0 and variance o2 = 1. The cumulative distribution function of Z is denoted as

@(z) = P(Z 1 z )

The z table (see Appendix Table I1 in MR) provides values of @(z) . In the table, the z column shows z values with tenth digits and the column headings indicate hundredth digits of z values.

(e.g.) Reading the z Table (1) @(-1.57) = P(Z 5 -1.57) = 0.058 (2) @(1.57)=P(Z 51.57)=0.942 (3) P(Z > 1.57) = 1 - P(Z 1 1.57) = 1 - 0.942 = 0.058 (4) P(-1.57 1 Z 11.57) = P(Z 11.57) - P(Z 1 -1.57)

= Q(1.57) - @(-I .57) = 0.942 - 0.058 = 0.884


Standardization An arbitrary normal random variable X with p and o2 can be transformed into Z (with p = 0 and o2 = l), as illustrated in Figure 4-5, by using the algebraic relationship between X and Z:

x-c1 z=- (3

Figure 4-5 Standardizing a normal random variable.

After standardizing the normal random variable X, probabilities of X are determined by using the z table.

(Standardization) Suppose that the life length of an INFINITY light bulb follows a normal distribution with p = 750 hours o2 = 402. Calculate P(740 1 X 5 770).

Exercise 4.5 The sick-leave time of employees in a firm in a month is normally distributed (MR 4-57) with a mean of 100 hours and a standard deviation of 20 hours. Find the

probability that the sick-leave time of an employee in a month exceeds 130 hours.

4-7 Normal Approximation to the Binomial and Poisson Distributions

R Apply the normal approximation to the binomial, hypergeometric, and Poisson random variables.

Normal The binomial distribution B(n, p ) approximates to the normal distribution with Approximation p = np and o2 = np(1 - p) if n p > 5 and n(l - p ) > 5 (see Figure 4-19 on page 119

to of MR). However, whenp is so close to 0 or 1 that np < 5 or n(l -p ) < 5, Binomial corresponding binomial distribution is significantly skewed (see Figure 4-20 on

Distribution page 120 of MR); thus, the normal approximation is inappropriate.

4-7 Normal Approximation to the Binomial and Poisson Distributions 4-9

Normal When normal approximation is applicable, the probability of a binomial random Approximation variable X with p = np and o2 = np(1 -p ) can be determined by using the standard

to normal random variable Binomial - X - u X - n o

L=-- Distribution -

(cant.)

Normal Recall that the binomial approximation is applicable to a hypergeometric Approximation distribution if the sample size n is relatively small to the population size N, i.e.,

to nlN < 0.1 (see Section 3-8. Hypergeometric Distribution). Consequently, the Hypergeometric normal approximation can be applied to the hypergeometric distribution withp =

Distribution KIN (K: number of successes in N) if nlN < 0.1, np > 5. and n(1 - p) > 5.

Thus, when normal approximation is applicable, the probability of a hypergeometric random variable Xwith p = np and o2 = np(1 - p)(N - n)/(N - 1) can be determined by using the standard normal random variable

x -P- z=- - X - np K , where p = -

o ,/np(l - P) (N - n) /(N - I) N

Normal Recall that a Poisson distribution with k = h and o2 = h is the limiting form of a Approximation binomial distribution with p = np and o2 = np(1 -p ) (see Section 3-9. Poisson

to Distribution). Therefore, the normal approximation is applicable to a Poisson Poisson distribution if m.

Distribution Accordingly, when normal approximation is applicable, the probability of a Poisson random variable X with p = h and oZ = h can be determined by using the standard normal random variable

X - p X - h z=- - -- o J x

Example 4.6 i i

1. (Normal Approximation; Binomial Distribution) Suppose that X is a binomial random variable with n = 100 andp = 0.1. Find the probability P(X I 15) based on the corresponding binomial distribution and approximate normal distribution. Is the normal approximation reasonable?

~ . r r The probability mass function, mean, and variance of the binomial random variable X are

I Thus,

4-1 0 Chapter 4 Continuous Random Variables and Probability Distributions

Example 4.6 (cont.)

By applying the normal approximation to the binomial random variable X,

---) = P(Z I 1.6'7) = 0.953 3

Since np = 10 > 5 and n(l - p ) = 90 > 5, the normal approximation to the binomial random variable X i s satisfactory.

2. (Normal Approximation; Hypergeometric Distribution) Suppose that X has a hypergeometric distribution with N = 1,000, K = 100, and n = 100. Find the probability P ( X I 15) based on the corresponding hypergeometric distribution and approximate normal distribution. Is the normal approximation reasonable?

FF The probability mass function, mean, and variance of the hypergeometric random variable X are

(Note) max (0, n - (N - K)) = max (0, 100 - ( 2 000 - 100)) = max{O, -800) = 0

min{K,n) = min(100, 100) = 100

Thus,

By applying the normal approximation to the hypergeometric random variable X,

P ( X 6 1 5 ) = P - ----) = P(Z 6 1.75) = 0.960 2.85

Since nlN = 0.1, np = 10 > 5 and n(l - p ) = 90 > 5, the normal approximation to the hypergeometric random variable Xis satisfactory.

. (Normal Approximation; Poisson Distribution) Suppose that X has a Poisson distribution with A. = 10. Find the probability P(X I 15) based on the corresponding Poisson distribution and approximate normal distribution. Is the normal approximation reasonable?

4-9 Exponential Distribution 4-1 1

Example 4.6 The probability mass function, mean, and variance of the Poisson random (cont.) I variable X are


I Thus,

I By applying the normal approximation to the Poisson random variable X,

I Since h =10 > 5, the normal approximation to the Poisson random variable X is satisfactory.

1. Suppose that X is a binomial random variable with n = 200 andp = 0.4. Find the probability P ( X I 70) based on the corresponding binomial distribution and approximate normal distribution. Is the normal approximation reasonable?

2. Suppose that X has a hypergeometric distribution with N = 2,000, K = 800, and n = 200. Find the probability P ( X I 70) based on the corresponding hypergeometric distribution and approximate normal distribution. Is the normal approximation reasonable?

3. Suppose that X has a Poisson distribution with h = 80. Find the probability P(X 1 70) based on the corresponding Poisson distribution and approximate normal distribution. Is the normal approximation reasonable?

4-9 Exponential Distribution

R Describe the probability distribution of an exponential random variable. R Explain the relationship between exponential and Poisson random variables. R Explain the lack of memory property of an exponential random variable. R Determine the probability density function, mean, and variance of an exponential random variable.

-ms*.?"+,w"M %* * s.B-d *-- -- -. *- " d--* -< <-* * * Mm -* .-M *- **--* -a a *w - - B wwB " * "* A s- A *> " * a * .. - * w * a s

Exponential An exponential random variable represents the length of an interval (in time, Random space, etc.) from a certain point until the next success in a Poisson process. The Variable exponential distribution parallels the geometric distribution by being concerned

with the probability of the next success.


Exponential The probability density function and cumulative distribution function of Random exponential random variable X with parameter h (average number of successes Variable per unit interval) are

(cont.) f ( x ) = ~ e - ~ and ~ ( x ) = l - e - k , x 2 0 , h > 0

The mean and variance of X are 1 2 1

p=- and o =- h h2

(Caution) Like the Poisson distribution, use consistent units to define an exponential random variable X and the corresponding parameter h.

An exponential curve is skewed to the right, decreasing exponentially (see Figure 4-6; as h increases, the skewness of the curve increases.

Figure 4-6 Exponential curves with selected values of h.

Relationship The exponential and Poisson distributions are interchangeable. Let an exponential between random variable X with parameter 3L (average number of successes during the unit

Exponential length of time) denote the time until the next success. Let N denote the number of and Poisson successes during a specific time interval x, i.e., N is a Poisson random variable

Distributions with parameter h .

Then, the following relationship exists between X and N:

P(X > x) = P(N = 0) = e-k(h)O -h

= e 0 !

Therefore,

F ( X ) = P ( X < X ) = ~ - P ( X > X ) = ~ - ~ - ~

By differentiating F(x), the probability density function of the exponential random variable X i s

4-9 Exponential Distribution 4-1 3

Exponential In the exponential distribution the number of successes is fixed as one and the vs. length of a time interval between successes is a random variable, whereas in the

Poisson Poisson distribution the number of successes is a random variable and the length Distributions of a time interval is fixed (see Table 4-1).

Table 4-1 The Characteristics of Poisson and Ex~onential Distributions

Parameters - . . - - . - --..LI"UU.Y"U.-.*H.---..C;

Distribution Population Rate of i Length of No. of Success Time Interval ~uccesses i

Poisson Infinite Constant (h) Constant Variable (X) I Exponential Infinite Constant (A) i Variable (X) Constant (1) .................................................... :" %-,~~".m~",z~,~~." -< me.'" -~%>d.*as,w.~s-m mm.~~."d~*~.d~%~w*m.m.m.~m.-"w~ ~ . ~ . ~ . ~ 4 ~ ~ " ~ , w m r ~ , ~ a - e . ~ * ~ , ~ . ~ ~ ~ ~ " ~ ~ m ~ , ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ , " ~ m r " . ~ . ~ . ~ , ~ ~ w.".v8a.".B.". ?>~.,".*d.B .,"A:me"".

Lack of As introduced in Section 3-7.1 Geometric Distribution, the lack of memory Memory property means that the system does not remember the history of previous Property outcomes so that the timer of the system can be reset at any time, i.e., - -

P ( t < X < t + A t ) P ( X < t + A t I X > t ) = = P ( X < At)

P(X > t)

The lack of memory property of an exponential distribution implies that the value of h does not change with X. For example, in a reliability study of failure of a device, if the failure rate h is constant (not increasing with time due to wear), the time to failure of the device has the lack of memory property. The exponential distribution is the only continuous distribution having the lack of memory. (Recall that the geometric distribution is the only discrete distribution having the lack of memory property.)

(Proof) The lack of memory property The conditional probability P ( X < t + At I X > t) is

-kt e-h(t+At) 1 - ,-h(t+At) - (1 -,-kt) e - - - - - =1-e- hat

1 - (1 - e-"1 e -ht

The probability P(X < At) is

P(X < At) = F(At) = 1 - e-lAt

Therefore, P ( X < t + A t I X > t ) = P ( X < A t )

Example 4.7 The number of customers who come to the YUMMY donut store follows a Poisson process with a mean of 5 customers every 10 minutes.

1. (Probability Density Function; Exponential Distribution) Determine the probability density function of the time (X; unit: min.) until the next customer arrives.

4-1 4 Chapter 4

Example 4.7 (cont.)

Continuous Random Variables and Probability Distributions

rm For a consistent unit for Xand h, h = 5 customers/l0 min. = 0.5 customers/min.

Therefore, the probability density function of X is

f (x) = he-h" = 0.5e-0.5x , x 2 0 (unit: min.)

2. (Probability) Find the probability that there are no customers for at least 2 minutes by using the corresponding exponential and Poisson distributions.

rm By using the exponential distribution,

P ( x > ~ ) = I - ~ ( ~ < 2 ) = ~ 2 ) = l - ( 1 - e - ' ) = e - ' = 0 . 3 6 8 i \

The parameter of the corresponding PoiSson distribution is lx = 0.5 customers/min. x 2 min. = 1

I Therefore,

As shown above, the probability from the exponential distribution is equal to the probability from the Poisson distribution.

I 3. (Mean and Variance) Find the mean and variance of X 1 1

rm p = - = - = 2 min. until the next customer arrival I h 0.5

4-1 0 Erlang and Gamma Distributions


4-10.1 Erlang Distribution

The lifetime (X; unit: hour) of a mechanical assembly in a vibration test is exponentially distributed with a mean of 1 failure every 400 hours.

1. Determine the probability density hnction of the operating time (X) of the assembly before failure.

2. Find the probability that the assembly fails the vibration test in less than 100 hours by using the corresponding exponential and Poisson distributions.

3. Find the mean and variance of X.

LI Describe the probability distribution of an Erlang random variable. 0 Explain the relationship between Erlang and exponential random variables.

Explain the relationship between Erlang and Poisson random variables. random variable.

---"w#*--m*.*"w----m"-".,"*r

Erlang Random Variable

Relationship between

Erlang and Poisson

Distributions

4-10 Erlang and Gamma Distributions 4-1 5

An Erlang random variable represents the length of an interval (in time, space, etc.) until the next r (positive integer) successes in a Poisson process. IfX,, X2,. . . , Xr are mutually independent exponential random variables and each represents the length of an interval for the ith success after the (i - l)th success, the Erlang random variable Xis the sum of the r exponential random variables:

X = X , + X 2 +...+ X r

The Erlang distribution parallels the negative binomial distribution by being concerned with the probability of the next r successes. (Recall that the exponential distribution parallels the geometric distribution by being concerned with the probability of the next success.)

The probability density function of an Erlang random variable X with parameters h (average number of successes per unit interval) and r (number of successes) is

The mean and variance of X are r 2 r

p=- and o =- h h2

As illustrated in Figure 4-7 an Erlang curve is skewed to the right like an exponential curve; an Erlang curve with r = 1 becomes an exponential curve.

" ' " L I L 6 1 1 I 1 . L . I 3 b -

0 2 5 6 7 8 9 I0 A. . - Figure 4-7 Erlang curves with selected values of h and r.

Like the exponential distribution, the Erlang and Poisson distributions are interchangeable. Let an Erlang random variable X with parameter h (average number of successes during the unit length of time) denote the time until the next r successes. Let N denote the number of successes during a specific time interval x, i.e., N is a Poisson random variable with parameter hx.

Then, the following relationship exists between X and N:


Erlang vs. Poisson

Distributions

In the Erlang distribution the number of successes is fixed as r and the length of a time interval until the next r successes is a random variable, whereas in the Poisson distribution the number of successes is a random variable and the length of a time interval is fixed (see Table 4-2).

r Example 4.8

ariable (3 Constant (r) i **w"-s ' l ls l * I IM*X I**,*ll'l.H.l.*f. , l ( l W . l n , W ~ . * ~

The number of customers coming to the YUMMY donut store follows a Poisson process with a mean of 5 customers every 10 minutes.

1. (Probability Density Function; Erlang Distribution) Determine the probability density function of the time (X; unit: min.)) until r = 3 customers come to the donut store.

RW For a consistent unit for X and A, h = 5 custorners@~\0.5 customers/min.

\ , 1 Therefore, the probability density function of X with r = 3 customers is

y x r - 1 -Ax 0.53 x3-1e-0.5x e f (x) =

- - = 0.0625x~e-O'~~, x > 0 (unit: min.) ( r - I)! (3 - I)!

2. (Probability) Find the probability that three customers come to the donut store in less than 10 min by using the corresponding Erlang and Poisson distributions.

O-W By using the Erlang distribution, 0

P(X < 10) = 0.0625x~e-~.~"dx = 0.875 (this integral is solved by

I integration by parts, which is not shown here)

The parameter of the corresponding Poisson distribution is hx = 5 customers/lO min. x 10 min. = 5

( Therefore,

The probability from the Erlang distribution is equal to the probability from the Poisson distribution.

( 3. (Mean and Variance) Find the mean and variance of X. r 3

p = - = - = 6 min. for r = 3 customers I 2. 5/10

4-10 Erlang and Gamma Distributions 4-1 7


The time between process problems in a manufacturing line is exponentially distributed with a mean of 30 days.

1. Determine the probability density function of the time (X; unit: day) until the fourth problem (r = 4).

2. Find the probability that the time until the fourth problem exceeds 120 days by using the corresponding Erlang and Poisson distributions.


4-1 0.2 Gamma Distribution

Explain the relationship between gamma and Erlang random variables.

Gamma A gamma random variabl an Erlang random variable by Random having a parameter r of a that an Erlang random Variable variable has a parameter

distribution with a

The probability density function of a gamma random variable X with parameters 3L and r (> 0) is

(Note) Gamma function r ( r )

A gamma function has the following properties: (1) r ( r ) = ( r - l ) r ( r - 1)

(2) T(r) = ( r - I)! if r is a positive integer; r(1) = O! = I

(e.g.) Gamma function ( I ) T(~)=(v-1)!=(4-1)!=3!=6

(2) r(1.5) = ( r - 1)r(r - 1) = (1.5 - l)r(1.5 - 1) = 0.5r(0.5) = 0 . 5 h = 0.886

The mean and variance of X are r 2 r

p=- and o =- 3L x2

Gamma vs. While the Erlang distribution has a positive integer r, the gamma distribution has Erlang a positive real number integer r (see Table 4-3).

Distributions


Gamma vs. Table 4-3 The Characteristics of Erlang and Gamma Distributions Erlang Pammcters .

Rate of Distributions Distribution Population Lengthof ] No. of (cont.) Success Interral f Successes i

4-1 1 Weibull Distribution

Cl Describe the probability distribution of a Weibull random variable. Explain the relationship between Weibull and exponential random variables.

O Determine the probability density function, mean, and variance of a Weibull random variable. a a " L -A"-"- - mm ass - % m -me r* *------*w---v------ --

Weibull A Weibull random variable represents the length of an interval (in time, space, Random etc.) until the next success for various types (increasing, decreasing, or constant) Variable of success rates. Therefore, the Weibull random variable is more flexible than the

exponential and gamma random variables by dealing with various types of success rates.

The probability density function and cumulative distribution function of a Weibull random variable X with shape parameter P (> 0) and & parameter 6 (> 0) are

When the shape parameter P = 1, the becomes the exponential distribution with h = 118

f (4 1

P 6

A. 0 1 2 3 4 5 6 7 8 9 1 0

Figure 4-8 Weibull curves with selected values of fJ and 6.

4- 1 1 Weibull Distribution 4-1 9

Weibull The mean and variance o f X are Random Variable

(cont.)

Weibull vs. Exponential

Distributions

r Example 4.9


In the Weibull distribution the rate of success can vary, whereas in the exponential distribution the rate of success is constant (see Table 4-4).

Table 4-4 The CharacMstics of Ex~onential and Weibull Distributions .- . . . .

of Success i No. of 1 Intmal Succesws

(A) Variable (X) Constant ( I ) : Variable (3 Constant (1)

~ w ~ ~ # ~ . - * w - ~ ~ ~ - - s ~ ~~waMw..vwza-"

Assume that the time to failure of a hair braiding machine follows a Weibull distribution with a = 0.5 and P = 500 hours.

1. (Probability Density Function; Weibull Distribution) Determine the probability density function of the time (3 until failure of the braiding machine.

2. (Probability) Find the probability that the braiding machine lasts longer than 300 hours without failure.

3. (Mean and Variance) Find the mean and variance ofX.

b p = ST 1 + - = 500r 1 + - = 500l?(3) = 500 x 2! = 1,000 hrs until failure ( i) i d 5 )

Suppose that the life of a recirculating pump follows a Weibull distribution with parameters a = 2 and p = 700 hours. 1. Determine the probability density function of the time to failure (4 of the

Pump. 2. Find the probability that the pump lasts less than 700 hours. 3. Find the mean and variance of X.


4-12 Lognormal Distribution

Ll Describe the probability distribution of a lognormal random variable. O Determine the probability density function, mean, and variance of a lognormal random variable.

Lognormal A random variable X is called lognormally distributed if its natural logarithm, W Random = In X, follows a normal distribution with mean 8 and variance m2. Like the Variable Weibull distribution, the lognormal distribution is used to model the lifetime of a

product whose failure rate is increasing over time.

The probability density function (see Figure 4-28 in MR) and cumulative distribution function of X with parameters 8 and m2 are

(~n x-8)' 1 --

f(x)=- 2a2 ,X>O x d G e

The mean and variance of X are e+w2 1 2 and o2 = e 2e+m2 k = e - 1)

I 2. (Probability) Find the probability that the device lasts longer than 100 hours without failure.

Example 4.10

I 3. (Mean and Variance) -\F , ind the mean and variance of X.

Assume that the time (unit: hours) to failure of a newly developed electronic device follows a lognormal distribution with parameters 8 = - 1 and m2 = 22. 1. (Probability Density Function; Lognormal Distribution) Determine the

probability density function of the time (X) until failure of the device. (lnx-e)* (ln x+l)*

1 -- 1 -- M f(x)=-e 2@2 =----e , x > O

x& 2 G x

e+w2 I 2 2 1 h p = e = e-1+2 = 2.7 hour

/ 2 2e+w2 o = e -4 = e 2 (-1)+2* (e 2' - 1 )=19 .9~


T

Suppose that the length of time (in seconds) that a user views a page on a Web site before moving to another page is a lognormal random variable with parameters 8 = 0.5 and m2 = 1.

1. Determine the probability density function of the time (X) until a user moves from the page.

2. Find the probability that a page is viewed for more than 10 seconds.


Summary of Continuous Probability Distributions 4-21

Summary of Continuous Probability Distributions

A. Probability Distributions /' I'

Distribution Probability Density Function Mean \ Variance Section

Uniform

Normal

y X r - l -hx e r Erlang , x > O , k > O , r = 1 , 2 ,... -

( r - I)! h

B. Comparison

Parameters Distribution Population aate of

Success Len*h of Time NO. of ~uccesses Intervd

Poisson Infinite Constant (A) Constant Variable (X)

Exponential Infinite Constant (A) Variable (X) Constant (1 )

Erlang Infinite Constant (A) Variable (X) Constant (positive

integer r)

Gamma Infinite Constant (A) Variable (X) Constant (positive

real number r)

Weibull Infinite Changeable (P, 6) Variable (X) Constant ( I )


EXCEL Applications

(Reading the z Table)

(1) Choose Insert > Function. (2) Select the category Statistical and the function NORMSDIST. (3) Enter the value of Z (Z). (4) Find the result circled with a dotted line below.

Example 4.5

WRJ&DIST ---- - - I

i? = -1.57 I --- _ -- -

= 0.058207562 Returns the standard normal cumdat~ve dinbution (has a mean of zero and a standard deviation of one).

1 is the value For whlch you want the distr~butcon.

"- - " a -- -- __* - - - - -_ - -

Formula result = ::- Q,USSM~S@:; - - -___- - - Helo on th~s funct~n

(Standardization)

(1) Choose Insert % Function. (2) Select the category Statistical and the function NORMDIST. (3) Enter the value of X (X), mean (Mean; p), standard deviation

(Standard-dev; o), and logical value (Cumulative; true for the cumulative distribution function and false for the probability density function).


I = 0.691462467 Returns the normal cumulat~ve datr~butcon for the speclhed mean and standard deviatm.

Cumulative IS a bgical value: f o r t h cumulative dlstribut~on funct~on, use TRUE; for the probainy mass funct~on, use FALSE.

--- -- . -" *--- -.- -. m m - --" -------- m- ---- -A- - .. - Formda result = < -0.69146246::: - - _ _ _ _ _ - Hekl on ttus functlorl cancel 1

EXCEL Applications 4-23

Example 4.7

Example 4.8

2. (Probability; Exponential Distribution)

(1) Choose Insert * Function. (2) Select the category Statistical and the function EXPONDIST. (3) Enter the value of X (X), average number of success per unit interval

(Lambda; k), and logical value (Cumulative; true for the cumulative distribution function and false for the probability density function).

(4) Find the results circled with a dotted line below.

-_-----i - -

Returns the exponent14 cktnbutcon

Cumulative IS a bgicd value for the Function to return: the cumulatcve dKtrcbutcon funct~on - TRUE; the prabatdcty densky funct~on =FALSE.

- * - - *- - - -- - + _ - - - - - - - _ _

F o r d result = (*0.36767944!-> --..___-- Helo an thcs Functcot)

2. (Probability; Erlang Distribution)

(1) Choose Insert > Function. (2) Select the category Statistical and the function GAMMADIST. Note that

the gamma distribution is a general form of the Erlang distribution. (3) Enter the value of X (X), number of successes (Alpha; r), average time

until the next event (Beta; Ilk), and logical value (Cumulative; true for the cumulative distribution function and false for the probability density function). Note that a difference in notation exists between Excel and MR.


-" ----- --- -" - - * - -- a- = 0.875347981

eturns the gamma dcslrhutcon.

Cumulative s a laglcal value: return the cumulatcve dstrlbutian Functlon = TRUE; return the probability mass functcon =FALSE or omttted,

a C _ - - - - - - _ _ _

rmuta result = :. -0.875347981 : - -_____ - - - -


Example 4.9

Example 4.10

2. (Probability; Weibull Distribution)

( I ) Choose Insert > Function. (2) Select the category Statistical and the function WEIBULL. (3) Enter the value of X (X), shape parameter (Alpha; P), scale parameter

(Beta; ti), and logical value (Cumulative; true for the cumulative distribution function and false for the probability density function). Note that a difference in notation exists between Excel and MR.


Returns the Weibull dtstribut~on.

Cumulative 16 a log~cal value: for the cumulattve dlstrlbut~on function, use TRUE; for the probabw mass fumhon, use FALSE.

-" - - -- --- - ----a- --------------a ----- - --- - -- - - -- - - - --- ---------

He l~ on thrj functwn

2. (Probability; Lognormal Distribution)

( I ) Choose Insert > Function. (2) Select the category Statistical and the function LOGNORMDIST.

(3) Enter the values of X (X) and two parameters (Mean, 8; Standard-dev, a) of the lognormal distribution.


X =. 100 I I

Mean - -1 1 1

i Standard-dev = 2 j 1 I--_ ,- --

':4-[;.99746519$ Returns the cumufakive lognmal distnbubon of x, where In(x) IS n&%llydis&ib'ded with parameters Mean and Standard-dev.

Standard-dew IS the standard d e W m of In(x), a poskive numbsc.

-- --- __-----.._ Fornula result = ;:!I-, WES~_?~::

Help on this funct~on



Exercise 4.1 (Probability Density Function) 3

1 (I) P ( X < t ) = ly(x)dx= f i 3 3 x 2 d ~ = x 3 1 ~ 3 =(f) - o = - 27

19

Exercise 4.2 (Cumulative Distribution Function) By using the probability density function of X,

F ( x ) = P ( X I x ) = [f(u)du= 13u2du=u31x = x 3 -O=x 3 0

Therefore, x < 0

Exercise 4.3 (Mean, Variance, and Standard Deviation) I (1) The mean of X is

1 (2) The variance of X is

(3) The standard deviation of X is

o= JV(X>= J0.038=0.19

4-26 Chapter 4 Continuous Probability Distributions

Exercise 4.4 1. (Probability Density Function; Continuous Uniform Distribution) a = 0.95, b = 1.05

1 f(x)=-=

I - 1 - 10, 0.95 1 x 11.05

b - a 1.05-0.95 0.1

2. (Probability) .05 1.05

P(X >1.02)= f f(x)dx= =10(1.05-1.02)=0.3 .02


b + a 1.05+0.95 p=- - - =1.0 2 2

(b - a12 (1.05 - 0.95)~ (3 = - - = 8.3 x = 0.029~

12 12

Exercise 4.5 1 (Standardization)

Exercise 4.6 1. (Normal Approximation; Binomial Distribution) The probability mass function, mean, and variance of the binomial random variable X are

Thus,

By applying the normal approximation to the binomial random variable X,

Since np = 80 > 5 and n(1 - p) = 120 > 5, the normal approximation to the binomial random variable X i s satisfactory.

Exercise 4.6

(cont.)


2. (Normal Approximation; Hypergeometric Distribution) The probability mass function, mean, and variance of Xare

K N-K 800 2000-800 800 1200

f (XI = ( x ) ( n - x ) - - ( x )( 2 0 - x ) - - ( . )(200-I) , x = 0 to 200

(Note) max(0, n - (N - K)) = max{0,200 - (2000 - 800)) = max (0, - 1000) = 0

min{K,n) = min{800,200) = 200

Thus,

By applying the normal approximation to X

Since n/N = 0.1, np = 80 > 5 and n(l - p) = 120 > 5, the normal approximation to the hypergeometric random variable Xis satisfactory.

3. (Normal Approximation; Poisson Distribution) The probability mass function, mean, and variance of X are

Thus,

By applying the normal approximation to X

Since h = 80 > 5, the normal approximation to Xis satisfactory.

4-28 Chapter 4

Exercise 4.7

Exercise 4.8

Continuous Probability Distributions

1. (Probability Density Function; Exponential Distribution)

For a consistent unit for X and h, h = 1 failure1400 hrs = 0.0025 failurelhr

Therefore, the probability density function of X is

f (x) = he-h" = 0 . 0 0 2 5 e - ~ . ~ ~ ~ ~ " , x 2 0 (unit: hr)

2. (Probability) By using the exponential distribution,

P(X < loo)= F(100)=1 -e-O.:'= 0.221

The parameter of the corresponding Poisson distribution is x = 100 hrs, ?a = 0.0025 failurelhr x 100 hrs = 0.25

Therefore,

P(NZ1)=1-P(N=0)=1- e-)" (hx)' aJ - 1 - e-0.25 = 0.22 1 = l - e - -

O!

As shown above, the probability from the exponential distribution is equal to the probability from the Poisson distribution.


1 1 p = - = - = 400 hrs until failure h 11400

1. (Probability Density Function; Erlang Distribution)

For a consistent unit for X and ?L, , h = 1 problem130 days= 1/30 problemlday

Therefore, the probability density function of X with r = 4 problems is

where x 2 0 (unit: day)

2. (Probability) By using the Erlang distribution,

P (X > 120) = 2.06 x lo-' x3e-0.033xdr = 0.433 (this integral is solved 20

by integration by parts, which is not shown here)

The parameter of the corresponding Poisson distribution is hx = 1/30 problemfday x 120 days = 4


Exercise 4.8 1 Therefore,

The probability from the Erlang distribution is equal to the probability from the Poisson distribution.

(cont.)

I 3. (Mean and Variance)

e - h x ( ~ ) k P ( N < ~ ) = X = t$= 0.433

k=O k ! k=O

r - 120 days P=-=-- I i 1 / 3 0

Exercise 4.9

Exercise 4.10

1. (Probability Density Function; Weibull Distribution)

2. (Probability)


= 700 x 0.5 x & = 700 x 0.886 = 620.4 hrs until failure

1. (Probability Density Function; Lognormal Distribution) (In X - € I ) ~ (ln x-0.5)'

1 1 -- f(X)=-----e 2w2 =- , x 2 O xm,lG

2. (Probability)

P (X>lO)= l -P (XI lO)= l -0 (In l o 1 0'5) = 1 - 0(1 .go)= 1 - 0.96 = 0.04

3. (Mean and Variance) 0+w2 i 2 0.5+112 p = e = e = 2.7 sec.

2 20+w2 2.0.5+1 o = e (em' -1)=e (e ' -1 )=3 .6~

5 Joint Probability Distributions

5-1 Two Discrete Random Variables 5-5 Covariance and Correlation

5-2 Multiple Discrete Random Variables 5-6 Bivariate Normal Distribution

5-2.1 Joint Probability Distributions 5-7 Linear Combinations of Random

5-2.2 Multinomial Probability Variables

Distribution 5-9 Moment Generating Functions

5-3 Two Continuous Random Variables 5- 10 Chebyshev's Inequality

5-4 Multiple Continuous Random Variables Answers to Exercises

5-1 Two Discrete Random Variables

0 Determine joint, marginal, and conditional probabilities for two discrete random variables X and Y by using corresponding probability distributions.

Calculate the mean and variance of X (or Y) by using the corresponding marginal probability distribution.

0 Calculate the conditional mean and conditional variance of X given Y = y (or Y given X = x) by using the corresponding conditional probability distribution.

0 Assess the independence of X and Y.

Probability Distributions of

Two Random

Variables

Three kinds of probability distributions are used to describe the stochastic characteristics of two random variables Xand Y:

(1) Joint probability distribution (2) Marginal probability distribution (3) Conditional probability distribution

Joint Probability

Mass Function

The joint probability mass function (p.m.f.) of two discrete random variables X and Y, denoted as f, (x, y) , satisfies the following conditions:

5-2 Chapter 5 Joint Probability Distributions

Marginal The marginal p.m.f.'s ofXand Y with the joint p.m.f. f, (x, y) are -

Probability Mass f x ( x ) = p ( ~ = x ) = C f ~ ( x , Y )

Function Rx

fu(Y)=p(y=Y)=Cf , (x ,Y) RY

where Rx and R, denote the set of all points in the range of (X, Y ) for which X = x and Y = y, respectively.

The marginal p.m.f. f, (x) satisfies the following:

(1) f , (x)=P(X=x)>O


Conditional Recall that P(B ( A) = P(A n B) I P(A) (see Section 2-4. Conditional Probability). In parallel, the conditional p.rn.f. of X given Y = y, denoted as

Mass fxl, (x) , with the joint p.m.f. f, (x, y) is Function

f X l y ( x ) = P ( X = x I Y = y ) = P ( X = x , Y = y ) - - f,(x,y)

P(Y = Y) f r (Y)

The conditional p.m.f. fxl, (x) satisfies the following:

The conditional mean and conditional variance of X given Y = y are

E ( X I Y ) = PA, = C xfxly (XI

Independence Two random variables X and Yare independent if knowledge of the values of X does not affect any probabilities of the values of Y and vice versa. In other words, X and Yare not stochastically related.

Recall that P(A I B) = P(A) , P(B I A) = P(B) , and P(A n B) = P(A)P(B) for

two independent events A and B (see Section 2-4. Conditional Probability). Likewise, two independent discrete random variables X and Y satisfy any of the following:

(1 f Xiy (XI = f x (XI

Independence (cont.)

r Example 5.1

5-4 Chapter 5

Example 5.1 (cont.)

Joint Probability Distributions

The mean and variance of X are 4

E ( X ) = p x =~xfx(r)=lfx(1)+2.fx(2)+3.fx(3)+4fx(4) x=l

= 1 x 0.21 + 2 x 0.445 + 3 x 0.26 + 4 x 0.085 = 2.22

3 . (Discrete Conditional p.m.f.) Determine the conditional probability distribution of statistics grade (Y) given that ergonomics grade is an A ( X = 1 ) . Find also the conditional mean and conditional variance of Y given X = 1 .

h The conditional marginal probabilities of Y given X = 1 are

The conditional mean and conditional variance of Y given X = 1 are 1

4 . (Independence) Check if ergonomics grade (X) and statistics grade (Y) are independent.

mr Check if

f , ( 1 , 1 > = f x (1) . fY ( 1 )

By using the probabilities f , (1,l) = 0.12, f x ( 1 ) = 0.21, and f y (1) = 0.245,

f , (1,l) = 0.12 + f , ( l ) f y (1) = 0.21 x 0.245 = 0.052

Thus, it is concluded that ergonomics grade (3 and statistics grade (Y) are not independent; in other words, ergonomics grade and statistics grade are related.

5-2 Multiple Discrete Random Variables 5-5


The number of defects on the front side (X ) of a wooden panel and the number of defects on the rear side (Y) of the panel are under study.

1. Suppose that the joint p.m.f of X and Y is modeled as f x r ( x , y ) = c ( x + y ) , x = 1 , 2 , 3 and y = 1 , 2 , 3

Determine the value of c. 2. Determine the marginal p.m.f. of X. Find also the mean and variance ofX.

3. Determine the conditional p.m.f. of Y given X = 2. Find also the conditional mean and conditional variance of Y given X = 2.

4. Check if the number of defects on the front side (X) of a wooden panel and the number of defects on the rear side (Y) of the panel are independent.

5-2 Multiple Discrete Random Variables

5-2.1 Joint Probability Distributions

Explain the joint, marginal, and conditional probability distributions of multiple discrete random variables.

Explain the independence of multiple discrete random variables. _ _ j ~ m _ _ _ ~ - P l _ a _ l ( n ~ - w - w ~ ~ - ~ ~ ~ ~ m . ~ ~ h n ~ ~ ~ ~ m - ( r ~ ~ M B I ~ w B a w*ua--*l~i-*-a~-"s

Joint A joint p.m.f. of discrete random variables X I , X2 , . . . , X,, which describes the Probability probabilities of simultaneous outcomes ofXl, X2,. . . , X,, is

Mass f R ( ~ I , ~ 2 , . . . , x p ) = P ( x 1 = x ~ , ~ ~ = X 2 , . . . , X p =x,) Function

Next, the joint p.m.f. of K = { X I , X2, . . . , Xk), a subset of R = { X I , X2, . . . , Xp) with the joint p.m.f. fR (x , , x, , . . . , x, ) , is

Marginal Probability

Mass Function

Conditional Probability

Mass Function

f x x ~ f R ( x 1 2 x 2 , . . . 9 x p ) Rx,x2 ...Xk

where RX,X2...Xk denotes the set of all points in the range of R for which X , = x, ,

A marginal p.m.f. of& which belongs to R = { X I , X2,. . . , Xp) with the joint p.m.f. f R ( x l , x 2 ,..., x , ) , is

f x L ( x i ) = P ( X i = ~ i ) = C f R ( x l , x ~ , . . . , X p )

where R , denotes the set of all points in the range of R for which X i = xi .

A conditional p.m.f. of K = {x,, X2, . . . , Xk) given Kr={Xk+l , Xk+2,. . . , Xp) describes the probabilities of outcomes of K at the given condition of K' . Note that K and K' are mutually exclusive.

As an extension of the bivariate conditional p.m.f., the conditional p.m.f. of K


Conditional given K' with the joint p.m.f. f , ( x , ,x , , . . . , x , ) is Probability P ( X , = x l , X 2 = x 2 ; - , X p = x P )

Mass fKIK' ( X I X 2 , . . -, X k ) =

Function P(Xk+l = X k + ~ X k + 2 = Xk+2 ..., Xp = X p ) (cont.)

Independence As an extension of the bivariate independence conditions, multiple independent discrete random variables R = { X I , X2,. . . , Xp) , consisting of two mutually exclusive subsets K = { X I , X2,. .. , Xk) and Kf={Xk+,, Xk+2,. .. , Xp) , satisfy the following:

(1) fKIKr(x1,x2 ,... r X k Ixk+l , X ~ + ~ , . . . ~ X ~ ) = ~ K ( X ~ , X ~ , . . . , X ~ )

5-2.2 Multinomial Probability Distributions

O Determine the marginal probability distribution of multinomial random variables.

Multinomial Random

Variables

Suppose that ( 1 ) A random experiment consists of n trials. (2 ) The outcome of each trial can be categorized into one of k classes. (3 ) The multinomial random variables XI, X2, . . . , Xk represent the number of

outcomes which belong to class 1 , class 2,. . . , class k, respectively. (4) pl, p2, . . . , pk denote the probability of an outcome belonging to class 1,

class 2,. . . , and class k, respectively. (5) pl,p2,. . . ,pk are constant over n repeated trials.

Then, X l , X2,. . . , Xk have a multinomial distribution with the joint p.m.f.

where x, + x , +. . .+xk = n and p, + p , + . . .+pk = 1

Note that the marginal p.m.f. of &,, i = 1 to k, is a binomial distribution with mean and variance

E ( X i ) =px, = npi and v ( x , ) =o& =npi(l - p i )

Example 5.2 The maximum grip strength of an adult is categorized into one of three groups: 'low' for < 50 Ibs., 'moderate' for 50 to 100 lbs., and 'high' for > 100 lbs. Suppose the probabilities that an adult belongs to the low, moderate, and high groups are 0.1, 0.7, and 0.2, respectively. A group of n = 50 people is measured for grip strength.

5-3 Two Continuous Random Variables 5-7

Example 5.2 (cont.)

1. (Joint p.m.f.; Multinomial Distribution) Find the probability that 6 people have low grip strength, 36 have moderate grip strength, and 8 people have high grip strength.

h Let XI, X2, X3 denote the number of people that have low, moderate, and high grip strength, respectively. Since n = 50,pl = 0 . 1 , ~ ~ = 0.7, andp3 = 0.2, the joint p.m.f. ofXl, XZ, X3 is

where x, + x 2 + - . . + x , = n

I Thus,

2. (Marginal p.m.f.; Multinomial Distribution) Find the probability that 36 people in the group have moderate grip strength.

~ . r r The marginal p.m.f. of X2 follows a binomial distribution with n = 50 andp2 = 0.7, i.e.,

I Thus,

Exercise 5.2 In the transmission of digital information, the probabilities that a bit has high, (MR 5-30) moderate, and low distortion are 0.01, 0.04, and 0.95, respectively. Suppose that

i L n = 5 bits are transmitted and the amount of distortion of each bit is independent. 1 1 1. Find the probability that one bit has high distortion, one has moderate

distortion, and three bits have low distortion.

2. Find the probability that four bits have low distortion.

5-3 Two Continuous Random Variables

w..we"~.a.'w*e,~", "s~m*BaA.""A"~~~*~""~ea.B"B w ~ ~ v * ~ , ' ~ . ~ " m ~ ~ a . " . ~ . - v e

inuous random variables X and Y by ;sing corresponding probability distributions.

Calculate the mean and variance of X (or Y) a continuous random variable by using the corresponding marginal probability distribution.

C1 Calculate the conditional mean and conditional variance of X given Y = y (or Y given X = x) by using the corresponding conditional probability distribution.

C1 Assess the independence of X and Y.


Joint The joint probability density function (p.d.f.) of two continuous random variables Probability X and Y satisfies the following:

Density (1) f m ( x 7 Y ) 2 0 Function

(2) [ f , ( ~ ~ ~ ) m Q = l

Marginal The marginal p.d.f.'s of X and Y with the joint p.d.f. f , ( x , y ) are Probability

Density f x ( x ) = L f x r ( x , Y ) ~ Function

The marginal p.d.f. of X satisfies the following:

( 1 ) f x ( x ) 2 0


E ( X ) = P x = [xfx ( x ) A

Conditional The conditional p.d.f. of X given Y = y7 denoted as f x , ( x ) , with joint p.d.f. Probability

Distribution f m ( x 7 Y > is fxr (x7 Y )

f X i Y (XI = f Y ( Y )

The conditional p.d.f. f,,, ( x ) satisfies the following:

(1) f X l Y ( x ) 2 0

The conditional mean and conditional variance of X given Y = y are

E(X I Y ) = Pxly = [x fx ly ( x ) A

Independence Two continuous random variables X and Yare independent if any of the following is true:

( 1 ) f X l Y ( X I = f x (XI

(2) fYiX(Y) = f y ( Y )

(3) f , ( 4 Y ) = f x ( x ) f y ( Y )

ff Example 5.3

5-3 Two Continuous Random Variables 5-9

The sizes of college books are under study in terms of width (X) and height (Y). 1. (Continuous Joint p.d.f.) Suppose that the joint p.d.f of Xand Y is modeled as

fxr(x,y)=cxy, 5 < x < 9 and x + 1 <y<x+3(unit : inch) Determine the value of c.

h The ranges of X and Yare displayed below.

The joint p.d.f. of X and Y should satisfy the following: (1) f,(x,y) =cxy 2 0

I For the first condition, c 2 0 because x > 0 and y > 0.

I Next, for the second condition,

Therefore, the joint p.d.f. of X and Y is fxr(x,y)=&xy, 5 < x < 9 and x + l < y < x + 3

2. (Marginal Mean and Variance; Continuous Marginal p.d.f.) Determine the marginal probability distribution of X. Find also the mean and variance of X.

h The marginal p.d.f. of X is

= L x . l 515 2 [(X + 3)2 - (X + 112]

=&(x+2)x, 5 < x < 9

I The mean and variance of X are

5-1 0 Chapter 5 Joint Probability Distributions

Example 5.3 (cont.) =&($x41: +f x31:)=&x1886.7=7.33 in.

3. (Conditional Mean and Conditional Variance; Continuous Conditional p.d.f.) Determine the conditional probability distribution of Y given X= 8. Find also the conditional mean and conditional variance of Y given X= 8.

hr The conditional p.d.f. of Y given X = 8 is

The conditional mean and conditional variance of Y given X = 8 are

= 0.05. f ( l l3 - 93) = 10.03 in. 1 1

V(Y 18) =O$ = y' fyi8(Y)& - p & = 0.05 [ y3& -10.03'

4. (Independence; Multiple Continuous Random Variables) Check if the width (3 and height (Y) of a college book are independent.

rn Check if

fxr(x,y> - k x y f Y l x (Y) = - - --- - fy(y)

f x (x) & x(x + 2) 2(x + 2)

The marginal p.d.f. of Y is

515

By the way, the range of Xvaries along y (see the graph on next page) as follows:

5 < x < y - 3 for 6 < y < 8 (area I) y - 3 < x < y - 1 for 8 < y < 10 (areaII) - 3 < x < 9 for 1 0 < y < 12 (areaII1)

Thus, the marginal p.d.f. of Y is

5-3 Two Continuous Random Varla6fes 5-i'f

Example 5.3 (cont.)


-- 2 - , A o y(-y + 6 y + 7 2 ) = & y ( y + 6 ) ( 1 2 - y ) , for 1 0 < y < 12

In summary,

# f y ( y ) the width (X) and height (Y) of a college Since f y l x ( y ) = - 2(x + 2)

book are not independent.

Two measurement methods are used to evaluate the surface smoothness of a paper product. Let X and Y denote the measurements of each of the two methods.

1 . Suppose that the joint p.d.f of X and Y is modeled by f x r ( x , y ) = c , O < x < 4 , 0 < y and x - 1 < y < x + 1

Determine the value of c.

2. Determine the marginal probability distribution of X. Find also the mean and variance of X.

3. Determine the conditional probability distribution of Y given X= 2. Find also the conditional mean and conditional variance of Y given X = 2.

4. Check if the measurements of the two methods Xand Yare independent.


5-4 Multiple Continuous Random Variables

~ - ~ w ~ s m e M ~ - - m # A - - * w - s -

Explain the joint, marginal, and conditional probabilities of multiple continuous random variables. 0 Explain the independence of multiple continuous random variables.

*- _Xw-- Mm--_I-s-M- -#Avam-"am- mww-- ~ m - w w w ~ ~ m B m - . ~ w " l * " n a a a x x a a r a a " - ~ ~ ~ m ~ ~ ~ ~ ~ w w ~ ~ w * ~ a ~

Joint As an extension of the bivariate joint p.d.f., the joint p.d.f. of multiple continuous Probability random variables R = {XI, X2,. . . , Xp) satisfies the following:

Density (1) fR(x17x2,...,xp)20 Function

(2) j f . . j f , (x , , x , ,..., x p ) m , m2 = I R

The joint p.d.f. of K = {XI, X2,. . . , Xk), a subset of R = {XI, X2,. . . , X,) with the joint p.d.f. fR (x, , x2 , . . . , x,) , is

~ K ( x I , x ~ , . . . , x ~ ) = I I.' / fR(xl , x2,...,xp)dxk+ldxk+2 "'dxp R,,,,. .,

where R+,xI.,.xK denotes the set of all points in the range of R for which X , = x, ,

X 2 = x 2 , ..., X k =Xk.

Marginal A marginal p.d.f. ofX,, which belongs to R = {XI, X2,. . . , X,) with the joint p.d.f. Probability fR(xl ,x2 ,..., x p ) , i s

Distribution fx , 5 f " ~fR(xI ,X2, . . .7xp)dXI &2 .. 'dxl-ldX,+~ "'ibp

Rx'

where R , denotes the set of all points in the range of R for which X, = x, . 1

Conditional As an extension of the bivariate conditional p.d.f., the conditional p.d.f. of K =

Probability {XI, X2,. . . , Xk) for a given condition of K'= {&+I, Xk+2,. . . , Xp) with the joint Distribution p.d.f. fR (x, , x, , . . . , x, ) is

Independence As an extension of the bivariate independence conditions, multiple continuous random variables R = {XI, X2,. . . , Xp), consisting of two mutually exclusive subsets K = {XI, X2,. . . , Xk) and K'= {Xk+,, Xk+2,. . . , X,), are independent if

(I) fKJKf(x1,x2 , . . - , ~ ~ ) = f K ( ~ 1 > ~ 2 , . . . , ~ k )

5-5 Covariance and Correlation 5-1 3

5-5 Covariance and Correlation

O Explain the terms covariance and correlation between two random variables X and Y. Calculate the covariance and correlation of X and Y.

Covariance The covariance between two random variables X and Y (denoted as cov(X, Y) or (oxr) om) indicates the linear relationship between X and Y:

0, =E[(X-Px)(Y-~y)l=E(fl)-pXPY, -oo<om < = Note that the covariance may not be an appropriate measure to represent a relationship between random variables if they are related nonlinearly.

The positive, negative, and zero values of o m represent positive, negative, and zero linear relationships between Xand Y, respectively (see Figure 5-1).

Figure 5-1 Covariance and correlation between X and Y: (a) positive linearity (om > 0; 0 < p n < 1); (b) zero linearity (om = 0; pm = 0); (c) negative linearity (o,< 0; -1 < p,< 0).

(Derivation) E[(X - px)(Y - p,)] = E ( X Y ) - p x p y

E[(X - p x )(Y - PY 11 - PxY- PYX + PXPY I = E(f l ) - PxE(Y) - PYE(X) + PxPv

= E ( m ) - PXPY - PyPx + PXPY

= E ( m ) - PXPY

Correlation The correlation between X and Y (denoted as pH) represents the normalized @xu) linear relationship between X and Y (0, normalized by ox and oy):

cov(X,Y) 0, 7 -l<Pxy

While o m depends on the units of X and Y, p, is dimensionless. Like OH, the positive, negative, and zero values of p ~ y represent positive, negative, and zero linear relationships, respectively (see Figure 5-1).

Joint Probability Distributions 5-1 4 Chapter 5

Independence

r Example 5.4


If X and Yare independent (not related), then 0, =p, = 0

However, o, = p, = 0 is a necessarv (not sufficient) condition for the

independence ofXand Y. In other words, even if o, = p, = 0 , we cannot immediately conclude that X and Y are independent.

(Covariance and Correlation) In Example 5-1, determine the covariance and correlation between ergonomics grade (X ) and statistics grade (Y). h The mean and variance of X are

E(X) = px = 2.22 and V(X) = 0% = 0.87'

The mean and variance of Yare 4

E(Y)=Py =C~fY(~)=l.fY(1)+2.fY(2)+3.fY(3)+4.fY(4)

( The mean of XY is

Therefore, CJ, = E(XY) - PxP = 5.52 - 2.22 X 2.3 1 = 0.39

0, - pxr I-- 0.39 = 0.47

o,oy 0.87 x 0.95

I Since GXY and p, > 0, it can be concluded that ergonomics grade (X ) and statistics grade (Y) have a positive linear relationship.

In Exercise 5-1, the joint p.m.f. ofX(the number of defects on the front side of a wooden panel) and Y (the number of defects on the rear side of the panel) is

fxr(x ,y)=$(x+y) , x = 1 , 2 , 3 and y = 1 , 2 , 3

The means and variances of X and Yare

I E(X) = E(Y) = 2.17 and V(X) = V(Y) = 0.80' Determine the covariance and correlation of X and Y.

5-6 Bivariate Normal Distribution 5-1 5

5-6 Bivariate Normal Distribution

0 Explain the joint probability distribution of bivariate normal random variables. Determine the joint, marginal, and conditional probabilities of bivariate normal random variables.

-jimn*l(*^.^.wm xmsw ma rs-w# --* M*mrut a .n- - mrxsariarxarva MM+B - a\xnna -# .a urw*awm in*a u-w -nannnenana --- xlxr n r a m aa x ;."*ware xa* m

Bivariate The joint probability density function of two normal random variables Xand Y 2 2 Normal with means p~ and py, variances ox and o y , and correlation p, (-1 < pxu < 1)

Random is Variables

1 f, (x7 Y) =

2 n o x o y ~ l - p ,

Bivariate normal distributions with different values of pxy are shown in Figure 5- 2. The contour plots of the binomial curves are elliptical. The center of each ellipse is located at the point (px, py) in the xy plane. If p, > 0 (pxy < O), the major axis of each ellipse is positive (negative); if pxr = 0, the ellipses become circles.

Probability density function

(a) (b) (c)

Figure 5-2 Bivariate normal distributions with different values of p ~ : (a) 0 < pxr< 1; (b) pxr=O; (c) -1 < p,yy< 0.

The marginal probability distributions of X and Yare normal with means px 2 and py and variances ox2 and o y , respectively.

The conditional probability distribution of Y given X = x is normal with mean and variance

2 E ( Y x ) = p , + p , ~ ( x - p x ) and V(Ylx)=o,(l-pH) o x


r Example 5.5


(Bivariate Normal Distribution) The heights (X; unit: in.) and weights (Y; unit: lbs.) of adult males in the US are normally distributed with means = 68.3 and p y = 162.8, variances 0x2 = 2.3' and or2 = 23.02, and correlation pm = 0.49. Determine the probability that the weight (Y) of an adult male is between 160 and 180 Ibs given that his height (3 is 70 in.

D-rr The conditional probability density function of Y given X = 70 is normal with mean and variance

I Therefore,

Let Xand Y represent two dimensions of an injection molded part. Suppose that X and Y have a binomial distribution with means k= 3.00 and py = 7.70, and variances 02 = 0.04' and or2 = 0.08~. Assume that Xand Yare independent, i.e., p ~ = 0. Determine P(2.95 < X < 3.05, 7.60 < Y < 7.80).

5-7 Linear Combinations of Random Variables

0 Explain the term linear combination of random variables. 0 Determine the mean and variance of a linear combination of random variables.

Linear A random variable Y is sometimes defined by a linear combination of several Combination random variables X I , X2,. . . , Xp:

Y = c, XI + c2 X 2 + - a . + cp Xp , where ti's are constants

5-7 Linear Combinations of Random Variables 5-1 7

Rules The following rules are useful to determine the mean and variance of a linear for Linear combination of X and Y:

Combination 1. Rules for Means (1) If b is a constant, then E(b) = 0 . (2) If a is a constant, then E(aX) = aE(X) . (3) E(aX f by) = aE(X) f bE(Y)

2. Rules for Variances (1) V(b) = 0 .

(2) V(aX) = a 2 v ( x ) .

= a V(X) + b2 V(Y) , if X and Y are independent.

(Note) cov(X, Y) = o, = p,oxoy

The rules 1.3 and 2.3 can be extended for Y = c, X1 + c2X2 + .-. + cpXp as

follows: E(Y) =clE(Xl)+ c2E(X2) + cpE(Xp)

= ~ ? V ( X , ) + c;v(x~) + + c i V(Xp) , i f z s are independent r Example 5.6

I Exercise 5.6 v (MR 5-91) i

(Linear Combination) For adult males in the US, the length of the shoulder- elbow (X; unit: in.) and the length of the elbow-fingertip (Y; unit: in.) have a bivariate normal distribution with means px= 14.4 and y y = 18.9, variances 0x2 =

or2 = 0.g2, and correlation p, = 0.737. Determine the mean and variance of the length of the shoulder-fingertip (X+ Y).

D-rr The mean and variance of the length of the shoulder-fingertip Z = X + Y is E(X + Y) = E(X) + E(Y) = 14.4 + 18.9 = 33.3

Note that cov(X, Y) = p,oxoy = 0.737 x 0.8 x 0.8 = 0.472 . Thus,

V(X + Y) = 0 . 8 ~ + 0 . 8 ~ + 2 x 0.472 = 1 .49~

The width of a casing (X; unit: in.) and the width of a door (Y; unit: in.) are normally distributed with means px = 24 and y, = 23 +, and standard

1 I deviations ox = and o, = &, respectively. Assume that the width of the

casing (X) and the width of the door (X) are independent. Determine the mean and standard deviation of the difference between the width of the casing (X) and the width of the door (Y).


5-9 Moment Generating Functions

R Explain the term moment generating function. Determine the moment generating function of a random variable X and the mth moments of X.

R Find the mean and variance of X by using the first and second moments of X.

Moment The moment generating function of a random variable X (denoted as M(t)) is the Generating expected value of elx, i.e.,

Function elx f (x), if X is a discrete random variable

~ ( t ) = E(etX ) = " I r etx f (x)A, if X is a continuous random variable

The moment generating function of X is unique if it exists and completely determines the probability distribution of X. Thus, if two random variables have the same moment generating function, they have the same probability distribution.

If ~ ( ~ ) ( t ) denotes the mth derivative of M(t), the rnth moment of X about the

origin (zero) (denoted by E(X")) is

xm f (x), if X is a discrete random variable

E(x") = M(")(o) = " (r xmf(x)dx, if X is a continuous random variable

(Derivation) E ( r ) = M(") (0)

The mth derivative of M(t) is

xmetx f (x), if X is a discrete random variable

~ ( ~ ) ( t ) = ' dt" [[ xmetx f (x)dr, if X is a continuous random variable

By setting t = 0,

xm f (x), if X is a discrete random variable

M (") (0) = If xrn (x)A, if X is a continuous random variable

Application of The mean and variance ofXcan be determined by using the first and second Moments moments of X:

p = E(X) = M'(0)

o2 = E(x*) - [E(x)]~ = M"(0) - [M'(0)12

5-9 Moments Generating Functions 5-1 9

Example 5.7 Suppose that X has an exponential distribution 1 f ( x ) = ~ e - * . x > o

1. (Moment Generating Function) Find the moment generating function of X. h From the definition of moment generating function,

- -- ' f o r t - h < 0 h - t

I 2. (Application of Moments) Determine the mean and variance of X by using the first and second moments of Xabout zero.

I h The first moment of X about the origin is

I The second moment of X about the origin is

I Therefore, the mean and variance of X are

and 2 1 1 o2 = E ( x ~ ) - [E(x)]~ = M"(0) - [Mr(0)12 = - - - = -

h2 h2 h2

Exercise 5.7 The geometric random variable X has probability distribution (MR S5-15) f ( ~ ) = ( l - ~ ) ~ - ' ~ , x = 1 , 2 , ..., n

1. Find the moment generating function of X. 2. Determine the mean and variance of X by using the first and second moments

of X about zero.


5-1 0 Chebyshev's Inequality

Explain the use of Chebyshev's inequality rule. Bound the probability of a random variable by using Chebyshev's inequality rule and compare the bound probability with the corresponding actual probability.

-9 w - m ' - * e e ~ . ~ - ~ > ~ a m v ~ ~ ~ ~ ~ ~ - - - ~ --*

Chebyshev's A relationship between the mean and variance of a random variable X having a Inequality certain probability distribution is formulated by Chebyshev as follows:

~ ( ( ~ - p ( 2 c o ) l l l c ~ , c > o

By using Chebyshev's inequality rule, a bound probability of any random variable can be determined. For example, Table 6-1 presents bound probabilities of a normal random variable X with p and o2 (by using Chebyshev's inequality rule) and corresponding actual probabilities; the table indicates that a bound probability always includes the corresponding actual probability.

gi Example 5.8

Exercise 5.8 (MR S5-25)

Table 6-1 Bound Probabilities and Actual Probabilities of a Normal Random Variable X

Probability Condition c eauodwili' A@ RobWity

( 1/c9

(Chebyshev's Inequality) The power grip forces (X) of adult females in the US are normally distributed with a mean of 62.7 and a standard deviation of 17.1 lbs. Bound the probability that the grip strength of an adult female differs from the mean more than 2.5 times standard deviation. Compare also the bound probability with its actual probability.

By applying Chebyshev's inequality rule,

~ ( 1 ~ - p 1 2 2 . 5 0 ) l l / 2 . 5 ~ =0.16

The actual probability is

P(( X-p122.50)=P ( I - xi I 2 2.5) = P(\ Z 12 2.5)

The actual probability is less than the bound probability.

Suppose that the photoresist thickness (X) in semiconductor manufacturing has a continuous uniform distribution with a mean of 10 pm and a standard deviation of 2.3 1 pm over the range 6 < x < 14 pm. Bound the probability that the photoresist thickness is less than 7 or greater than 13 prn. Compare also the bound probability with its actual probability.



Exercise 5.1 1 . (Discrete Joint p.m.f.) I The joint p.m.f. f, (x, y) , x = l , 2 , 3 and y = l , 2 , 3, should satisfy the following:

(1) f m ( x , y ) = c ( x + y ) 2 0

I For the first condition, c 2 0 because x > 0 and y > 0.

I Next, for the second condition,

Therefore, the joint p.m.f. of X and Y is

f,(x,y)=$(x+y), x = 1 , 2 , 3 and y = 1 , 2 , 3

I 2. (Discrete Marginal p.m.f.) The marginal probabilities of X are

I The mean and variance of X are

V ( X ) = 0; = C x2 fx (x) - p; X

= 1 2 . f x ( l ) + 2 2 . fx (2)+32 . fx(3)-2.172 2 = 12x i + 2 2 ~ f + 3 2 ~ -2.17 =0.80 2


Exercise 5.1 1 3. (Discrete Conditional p.m.f.)

(cont.) I The conditional marginal probabilities of Y given X = 2 are

Exercise 5.2

I The conditional mean and conditional variance of Y given X = 2 are

4. (Independence)

Check if

fxr (191) = fx (1). fY (1)

I Thus, the number of defects on the front side (4 of a wooden panel and the number of defects on the rear side (Y) of the panel are not independent.

1. (Joint p.m.f.; Multinomial Distribution)

Let XI, X2, X3 denote the number of bits that have high, moderate, and low distortion, respectively. Since n = 5,pl = 0 . 0 1 , ~ ~ = 0.04, andp, = 0.95, the joint p.m.f. ofXl, X2, X3 is

Thus, 5!

fx,x2,3 (1,1,3) =-0.01xO.O4~0.95~ =0.007 l! I! 3!


Exercise 5.3


2. (Marginal p.m.f.) The marginal p.m.f. ofX3 follows a binomial distribution with n = 5 and p3 =

0.95, i.e.,

I Thus,

1 . (Continuous Joint p.d.f.)

The ranges of X and Yare displayed below. Note that integration for 0 < x I 1 (area I) and that for 1 < x < 4 (area 11) should be conducted separately because they have different low limits.

The joint p.d.f. of X and Y should satisfy the following:

(1) f , ( x , y ) = c ~ O

For the first condition, c 2 0 because x > 0 and y > 0.

Next, for the second condition,

Therefore, the joint p.d.f. of X and Y is fH(x,y)=i?j-, 0 < x < 4 , 0 < y and x - 1 < y < x + l



E(X)=p, = kfx(x)dr= l & x ( x + l ) d r + f ~ x d r

Exercise 5.3

(conL)

I 3. (Continuous Conditional p.d.f.)

The conditional p.d.f. of Y given X = 2 is

2. (Continuous Marginal p.d.f.)

The marginal p.d.f. of X is

0 < x l l

The conditional mean and conditional variance of Y given X = 2 are

I 4. (Independence)

Check if

Note that the range of X for 1 < y < 3 is y - 1 < x < y + 1 (see the graph on next page). Thus, tthe marginal p.d.f. of Y for 1 < y < 3 is

f h ) = [ f d x , r ) d x = l l&h=&(xl : r : )=&, l < y < 3

Since fr12 (y) = 3 # f y (y) = 5 , the measurements of the two methods X and

Yare not independent.



Exercise 5.4

Exercise 5.5

(Covariance and Correlation) The mean of XY is

Therefore, a, = E(XY)-pxpr =4.67 -2 .17~2.17 =-0.04

There is a weak, negative correlation between the number of defects on the front side (3 of a wooden panel and the number of defects on the rear side (Y) of the panel.

(Bivariate Normal Distribution) Since X and Yare independent,

P(2.95 < X < 3.05,7.60 < Y < 7.80) = P(2.95 < X < 3.05)P(7.60 < Y < 7.80)

By standardizing X and Y each,

= P(- 1.25 < Z < 1.25) = 0.789

Thus, P(2.95 < X < 3.05,7.60 < Y < 7.80) = 0.789x0.789 = 0.623


Exercise 5.6

Exercise 5.7

(Linear Combination) 2 7 1 2

X - N(24, ) , Y - N(23 8 , ) , and COV(X, Y) = 0 because X and Yare

independent. 1

Therefore, E ( X - Y ) = E ( x ) - ~ ( ~ ) = 2 4 - 2 3 + = $

1 . (Moment Generating Function) From the definition of moment generating function,

m w

M(t) = x e g f (x) = X e " ( 1 - p)"-' p = pet [(1 - p)e'lX-'

Note that the sum of the infinite geometric sequence ( a, ar, ar2, . . . ) is m

a s = C a r n " =- , where lr) < 1 n=l 1-r

Thus,

2. (Application of Moments) The first moment of X about the origin is

d{pet [1 - (I - p)et]-' dt Y .=.

The second moment of X about the origin is

Answers to Exercises 5-27 I Exercise 5.7

(cont.) I

Therefore, the mean and variance of X are p = E(X) = Mr(0) = np

and

Exercise 5.8 (Chebyshev's Inequality) I The probability density function of the uniform random variable X are

By applying Chebyshev's inequality rule, P(X<7)+P(X>13)=P(X-10<7-10)+P(X-10>13-10)

= ~ ( ~ ~ - 1 o ~ > 3 ) = ~ ( ~ ~ - 1 o ~ > c o ) < l l ~ ~

Therefore, the bound of P(X < 7 ) + P(X > 13) is

P ( ( X - l 0 1 > 3 ) = ~ ( 1 X-10)>1 .30x2 .31 )<1 /1 .3~ =0.59

I The actual value of P(X < 7 ) + P(X > 13) is

I The actual probability is less than the bound probability, which supports Chebyshev's inequality rule.

Random Sampling and Data Description

6-1 Data Summary and Display 6-6 Time Sequence Plots

6-2 Random Sampling 6-7 Probability Plots

6-3 Stem-and-Leaf Diagrams M~NITAB Applications

6-4 Frequency Distributions and Histograms Answers to Exercises

6-5 Box Plots

6-1 Data Summary and Display

Determine the mean, variance, standard deviation, minimum, maximum, and range of a set of data.

Mean The arithmetic mean of a set of data indicates the central tendency of the data.

1 . Sample Mean (denoted as T ): The average of all the observations in a sample, which is selected from larger population of observations.

- - k x i i=l

n

2. Population Mean (denoted as p): The average of all observations in the population.

T x i 1=1 p=- , for a finite population with size N

N

Variance and The variance of a set of data indicates the dispersion of the data along the mean: Standard the larger the variance, the wider the spread of the data about the mean. Deviation

6-2 Chapter 6 Random Sampling and Data Description

Variance and 1. Sample Variance Standard Deviation g ( x i -TI2 g x

(cant.) 2 = i=l - - i=l

n -1 n - 1

2. Population Variance N

C ( x i - ~ ) ~ t x : 02 = i=l - i=l p 2 , for a finite population with size N

N N

Note that the divisor for the sample variance is n-1, while that for the population variance is N.

The sample standard deviation and population standard deviation are

s = @ and o=@

Minimum The minimum and maximum of a set of data are the smallest and largest values of and Maximum the data set, respectively.

Sample Range The difference between the minimum and maximum of the sample. R =max(xi)-min(x,), i = 1 , 2 ,... , n

v Example 6.1 (Mean, Variance, and Range) The scores (X; integers) of n = 30 students in a statistics test are as follows (arranged in ascending order):

65 68 70 72 72 73 74 75 75 79 8 1 83 84 84 85 , 86 87 88 88 88 88 88 90 92 92 94 96 96 97 98

Determine the (1) sample mean, (2) sample variance, (3) minimum, (4) maximum, and (5) sample range of the test scores. The following summary quantities have been calculated:

n = 30, E x i = 2,508, and x,? = 212,194

- - 212,194 - 2,508~ 130 = 87.1 = 9.3 2

30 - 1 (3) minimum = 65 (4) maximum = 98 ( 5 ) R = 9 8 - 6 5 = 3 3

6-2 Random Sampling 6-3


The numbers of cycles (X; integers) to failure of n = 70 aluminum test coupons under repeated alternating stress at 21,000 psi and 18 cycleslsec. are as follows (arranged in ascending order):

Determine the (1) sample mean, (2) sample variance, (3) minimum, (4) maximum, and (5) sample range of the test results. The following summary

1 quantities have been calculated: . -

n = 70, E x i = 98,256, and Ex,? = 149,089,794 I

6-2 Random Sampling

O Distinguish between population and sample. O Describe the terms random sample and statistic.

Explain why selecting a representative random sample is important in research. O Describe the characteristics of a random sample.

- * - , - ~ B ~ > w . " ~ ~ " ~ % % - - - * ~ w w * - - - . ~ , ~ ~ ~ v ~ d - ~ - ~ w ~ ~ - ~ " ~ ~ m ~ - - ~ ~ ~ m e ~ - w a m m . w ~ , ~ ~ e ~ . B ~ w " ~ ~ - . ~ - ' ~ " v a ~ - w ~ u * * ~ . ~ ~ . ~ " # ",-B,,wB"w-""w*,- ~,--wm""*-,-'*-" ". *"-""" ""M,"w,..-a -,*"

Population 1. Population: The collection of all the elements of a universe of interest. The vs. Sample size of a population may be either (1) small, (2) large but finite, or (3) infinite.

(e.g.) Population size (1) Students in a statistics class: Small (2) College students in the United States: Large but finite (3) College students in the world: Infinite

2. Sample: A set of elements taken from the population under study, i.e., a subset of the uouulation (see Figure 6-1). Since it is impossible and/or impractical to examine the entire population in most cases, a sample is often used for statistical inference about the population.

Population Sample

Figure 6-1 Population versus sample.


Random A random sample refers to a sample selected from the population under study by Sample certain chance mechanism to avoid bias (over- or underestimation). Thus,

selecting a random sample that properly represents the population is important for valid statistical inference about the population.

A random sample of size n is denoted by random variables XI, X2,. . . , Xn where X, represents the i'h observation of the sample. The random sample XI, X2,. . . , X, has the following characteristics:

(I) The X 's are independent of each other, and (2) Every has the same probability distribution f (x) .

In other words, XI, X2,. . . , Xn are independently and identically distributed (Lid.) . Thus, the joint probability density function of XI, X2,. . . , Xn is

f x , , ~ ,,..., X, ( x I , x z , . . . , x ~ ) = ~ (xl)f(x2).. .f(xn)

h Even if the telephone numbers are randomly chosen, the survey sample is 8

biased because most working class people are not at home during the survey time period. Therefore, the survey result is most likely an over- or underestimate.

Example 6.2 (Random Sample) A public opinion poll on contraception is conducted by 1 calling people's homes, which are randomly selected, between 8 AM and 5 PM i on weekdays. Discuss this study design. i I

Statistic A statistic is a function of random variables X I , X2,. . . , Xn; therefore, a statistic is also a random variable.

(e.g.) Statistic n

(I) Sample mean X = X, I n i=I

n

(2) Sample variance s2 = (Xi - 2 ) ' /(n - 1) i=l

Exercise 6.2

6-3 Stem-and-Leaf Diagrams

Defects on sheet metal panels in three categories of sizes (small, medium, and large) are under study. The metal panels are made of the same material. The analyst randomly selects 20 large-size panels (not including small- and medium- size panels) and draws a conclusion regarding the average number of defects/ft2. Discuss this study design.

0 Explain the use of a stem-and-leaf diagram. 0 Construct a stem-and-leaf diagram to visualize a set of data. 0 Determine the 100kth percentiles, quartiles, interquartile range, median, and mode of a set of data. 0 Compare the variability measures variance, range, and interquartile range with each other.

6-3 Stem-and-Leaf Diagrams 6-5

Stem-and-Leaf A stem-and-leaf diagram is a graphical display that presents both the individual Diagram values and corresponding freauencv distribution simultaneously by using 'stems'

and 'leaves.'

The following procedures are applied to construct a stem-and-leaf diagram (see Figure 6-2):

Step 1: Determine stems and leaves. The digits of each data is divided into two parts: (I) stem for the 'significant' digits (one or two digits in most cases) and (2) leaf for the 'less significant' digit (last digit usually). The analyst should exercise histher own discretion to determine which digits are most significant with consideration of the range of data. The number of stems between 5 and 20 is commonly used. A stem can be further subdivided as necessary. As an example, the stem 5 can be divided into two stems 5L and 5U for the leaves 0 to 4 and for the leaves 5 to 9, respectively.

Step 2: Arrange the stems and leaves. The stems are arranged in ascending (or descending) order. Then, beside each stem, corresponding leaves are listed next to each other in a row.

Step 3: Summarize the frequency of leaves for each stem.

Example 6.3

Figure 6-2 A stem-and-leaf diagram.

(Stem-and-Leaf Diagram) In Example 6-1, construct a stem-and-leaf diagram of the test scores.

Step 1 : Determine stems and leaves. The test scores range from 65 to 98. Each test score consists of two digits and the first digit has higher significance. Thus, the first and last digits of a test score are defined as the stem and leaf of the data, respectively: 6 to 9 for stems and 0 to 9 for leaves. To better represent the distribution of the test scores, the stems are further split into

6U, 7L, 7U, 8L, 8U, 9L, and 9U where L is for the lower half leaves (0 to 4) and U is for the upper half leaves (5 to 9).


Example 6.3 (cont.)

I Step 3:

6U 5 8 2 7L 0 2 2 3 4 5 7 u 5 5 9 3 8L 1 3 4 4 4 8U 5 6 7 8 8 8 8 8 8 9L 0 2 2 4 4 i 9U 6 6 7 8 4 I

Exercise 6.3 1 In Exercise 6-1, construct a stem-and-leaf diagram of the test results.

Percentile The 1 OOkth percentile (0 < k I 1 ; denoted as pk) of a set of data means the 1 OOkth @k) largest value of the data set.

The procedures to findpk from n observations, x, , x,, . . . , x, , are as follows:

Step 1 : Arrange data in ascending order, say x(,) , x(,) , . . . , x(,, . Step 2: Calculate the position number r by using n and k

r = .Ink3 if nisodd (n + l)k, if n is even

Step 3: Determine p, based on r

x(,, , if r is an integer P k =I x(lr) + (x(l,.l) - xIlr))(r - Lrb, if r is not an integer

(Note) The symbols [rl (called ceiling) and Lv] (called floor) denote 'round-

up' and 'round-down' r to the nearest integer, respectively.

Quartile The following three quartiles divide a set of data into four equal parts in (qk) ascending order:

(1) First (lower) quartile: q, =

(2) Second quartile (median): 9, = p,,,,

(3) Third (upper) quartile: 9, = p,,,,

Interquartile The interquartile range (denoted as IQR) of a set of data is the difference between Range the upper and lower quartiles of the dataset, i.e., (IQR) *QR = 93 - ql = p . 7 5 - ~ 0 . 2 5

6-3 Stem-and-Leaf Diagrams 6-7

Comparison The variance, range, and interquartile range of a data set have different sensitivity of Variability (stability): variance is most sensitive (least stable) and interquartile is least

Measures sensitive (most stable). For example, Figure 6-3(a) shows that two different data sets with the same range can have different variances; and Figure 6-3(b) displays that two different data sets with the same interquartile range can have different ranges and variances. These illustrations imply that variance, range, and interquartile range have high (low), moderate, and low (high) sensitivities (stabilities), respectively.

(a) (b)

Figure 6-3 Sensitivities of variance, range, and interquartile range: (a) R1 = R2, while 012 # 0 z 2 ; (b) IQRl = IQR2, while R1 # R2 and ol2 # 0 2 ~ .

Of the three variability measures, the appropriate variability measure is selected depending on the analyst's judgment on which measure best represents the variability of the data.

Median The median of a set of data is the value that divides its ranked data set into two equal halves, i.e., p,,,, = 9 , .

Mode The mode of a set of data is the value that is observed most frequently. A data set can have no mode, one mode (unimodal), or multiple modes (bimodal, trimodal, etc.), as illustrated in Figure 6-4.

Figure 6-4 Modes of data distributions: (a) no mode; (b) unimodal; (c) bimodal.

Example 6.4 (Percentile, Quartile, Interquartile Range, Median, and Mode) In Example 6- 1, determine the (1) 8oth percentile, (2) first quartile, (3) interquartile range, (4) median, and (5) mode of the test scores. Note that n = 30 and 9, = 90.5.


Example 6.4 (cont.)

h ( 1 ) Since n = 30 is even, r = (n + l)k = 3 1 x 0.8 = 24.8, which is not an integer. Therefore,

PO 8 = X(lrj) + ( ~ ( r r ) - x ( [ r ~ )(' - 1'9 1 j

= X(124.8j) + ( ~ ( l 2 4 8) - X(j24 81) )(24.8 - 124.8B i = x ( ~ ~ ) + ( x ( ~ ~ ) - ~ ( ~ ~ ) ) ( 2 4 . 8 - 24) = 92 + (92 - 92) X 0.8 = 92 i

I

(2) Since r = (n + l)k = 3 1x0.25 = 7.75 is not an integer,

91 = PO 25 = X([rl) + ( ~ d r ) -'([r]) -Lrb = ~ ( 1 7 7 5 ~ + ( ~ ( r 7 751) - ~ ( 1 7 7 5 ~ l(7.75 - L7.75b

= x ( ~ , + ( x ( ~ ) - ~ ( ~ ) ) ( 7 . 7 5 - 7 ) = 74 + (75 - 74) X 0.75 = 74.75

(3) IQR = q3 - q1 = pR7, - ~ 0 . 2 ~ = 90.5 - 74.75 = 15.75

(4) Since r = (n + l)k = 3 1x0.5 = 15.5 is not an integer,

~ 0 . 5 = '(1.1) + (413 -xirr1))(r-rr)

= X ( i l 5 . 5 n + ('(l15.5l) - ~ ( [ l ~ . ~ l ) I(l5.5 - 115.5)

= x ( , ~ ) + ( x ( ~ ~ ) - ~ ( , ~ ) ) ( 1 5 . 5 - 15) = 85 + (86 - 8 5 ) ~ 0.5 = 85.5

(5) Mode = 88 (unimodal)

6-4 Frequency Distributions and Histograms

Exercise 6.4

0 Explain the termsfrequency, cumulative frequency, relativefrequency, and relative cumulative frequency.

0 Construct a frequency table of a set of data. 0 Construct a bar graph and histogram to visualize a set of data. O Explain the term skewness and describe the relationships between the central tendency measures

mean, median, and mode depending on the skewness of data. wi--*-~Y".~Ln -a* --n_i ~ ~ - a ~ ~ ~ M _ X _ X M _ - n i m - ~ ~ ~ ~ s m ~ l ~ - ~ * - ~ _ x ( ~ _ X ~ ~ ~ " n _ a n ~ ~ . ; r a i r r o - ~ - . u n a r - a ~ - ~ ~ r u - x

In Exercise 6-1, determine the ( 1 ) 9oth percentile, (2) third quartile, (3) interquartile range, (4) median, and ( 5 ) mode of the test results. Note that n = 70 and q, = 1,097.75.

Frequency A frequency indicates the number of observations that belong to a bin (class interval, cell, or category). The frequency of the fh bin is denoted by .

The sum of the frequencies of bins is called cumulative frequency (denoted by 6. ):

i

4 =z f , , i = 1,2, ..., k j=1

6-4 Frequency Distributions and Histograms 6-9

Relative A relative frequency is the ratio of a frequency to the total number of Frequency observations (n). The relative frequency of the i~ bin (denoted by pi ) is

p . =- A , n = total number of observations n

The sum of the relative frequencies of bins is called cumulative relative frequency (denoted by P, ):

r Example 6.5

d Exercise 6.5

(Frequency Measures) In Example 6-1, prepare a frequency table (including frequency, relative frequency, and cumulative relative frequency) of the test scores with five class intervals: below 60,60 to 69, 70 to 79, 80 to 89, and 90 or above. [kr Relative Cumulative

Class Interval (X ) Frequency Relative Frequency Frequency

In Exercise 6-1, prepare a frequency table (including frequency, relative frequency, and cumulative relative frequency) of the failure test results with five class intervals: below 500, 500 to 999, 1,000 to 1,499, 1,500 to 1,999, and 2,000 or above.

Bar Graph A bar graph is a graphical display with bars, the length of each bar being proportional to the frequency andlor relative frequency of the corresponding category. A bar graph is used to provide a visual comparison of the frequencies and/or relative frequencies of multiple categories.

Histogram A histogram is a bar graph in which the length and width of each bar are proportional to the frequency (or relative frequency) and size of the corresponding bin, respectively. The shape of a histogram of a small set of data may vary significantly as the number of bins and corresponding bin width

I change. As the size of a data set becomes large (say, 75 or above), the shape of the histogram becomes stable.

Different histograms can be constructed for the same data set depending on the analyst's judgment. The following procedures may be of use to construct a histogram with bins of an equal width for a data set with size n:

Step 1 : Determine the number of bins (k). The following formula can be used for k:

k G &I

(Note) The number of bins between 5 and 20 is used in most cases.

6-1 0 Chapter 6 Random Sampling and Data Description

Histogram Step 2: Determine the width of a bin (w). (cont.) The bin width is selected based on the range of the data set and the

number of bins: R max- min

W>-= k k

Step 3: Determine the limits of the histogram. The lower limit (I) of the first bin and the upper limit (u) of the last bin are slightly lower than the minimum and larger than the maximum, respectively, i.e.,

Step 4: Prepare a frequency table based on the bins determined. Count the number of observations that belong to each bin and calculate relative frequencies accordingly.

Step 5: Develop a histogram. Construct a bar graph by using the frequencies (frequency histogram) and/or relative frequencies (relative frequency histogram) of the bins.

(Bar GraphEIistogram) In Example 6-1, construct a histogram of the test scores by using the following summary quantities:

n = 3 0. minimum = 65. and maximum = 98

h Step 1: Determine the number of bins (k). Since & = = 5.5, k = 6.

I Step 2: Determine the width of a bin (w). R max- min 98 - 65

Since - = - - 5.5 , w = 6. k k 6

I Step 3: Determine the limits of the histogram.

6-4 Frequency Distributions and Histograms 6-1 1

Example 6.6 Step 5: Develop a histogram. (cont.)

I Test Scores

Skewness The skewness (denoted by y) of a set of data indicates the degree of asymmetry of the data set around the mean.

Exercise 6.6

As illustrated in Figure 6-5, the distribution with a long tail to the left and distribution with a long tail to the right have negative and positive values of skewness, respectively; the symmetric curve has a zero skewness.

In Exercise 6-1, construct a histogram of the test results by using the following summary quantities:

n = 70, minimum = 375, and maximum = 2,265

Figure 6-5 Distributions with different values of skewness (y): (a) y < 0 (left- skewed); (b) y =O (symmetric); (c) y > 0 (right-skewed).

Comparison The mean, median, and mode of a data set can be different depending on the of Central skewness of the data distribution, as illustrated in Figure 6-6. Both the mean and Tendency mode shift toward the direction of skewness with high sensitivity, whereas the Measures median moves with low sensitivity.

Of the three central tendency measures, the appropriate measure is selected depending on the analyst's judgment on which measure best suits the purpose of analysis: mean for arithmetic average, median for relative standing, and mode for frequency.


Comparison f ( x ) I ( x ) I f ( x )

of Central Tendency Measures

(cont.) iA 1 i- --- ___L L-L , +. iA -- - ---u n x" X X =x X z x

mean < median < mode mean = mode = median mode < median < mean (a) (b) (c)

Figure 6-6 Comparisons of means, medians, and modes of various distributions: (a) left-skewed; (b) symmetric; (c) right-skewed.

6-5 BOX Plots

Explain the use of a box plot. O Construct a box plot to visualize a set of data.

_"- ___L1_X_ IIII*mYeV *-.XX -.% ~ ~ * ~ - - * - ) U ( m - e i - - * a ~ - ~ ~ m , ~ ~ v ~ - m - ~ * m - - ~ - x 4 a -

Box Plot A box plot is a graphical display that presents the center, dispersion, symmetry, and outliers of a set of data simultaneously. Note that outliers are observations that are unusually low or high compared to most of the data.

A box plot consists of three components (see Figure 6-7): (1) Box: The left edge, middle line, and right edge of the box indicate the first

(q,), second (q2), and third (q3) quartiles, respectively; therefore, the length of the box indicates the interquartile range (lQR) of the data set.

(2) Whiskers: The left whisker extends from the left edge of the box to the smallest data point within 1.5 1QR from ql; and the right whisker extends from the right edge of the box to the largest data point within 1.5 IQR

'i 1 from 93.

I

i (3) Outliers: Points beyond the whiskers are outliers.

outliers 4.l 92 q3 outlier

1.5IQR IQR & 1.SIQR

Figure 6-7 A box plot.

Example 6.7 (Box Plot) In Example 6-1, construct a box plot of the test scores by using the following summary quantities:

minimum = 65, maximum = 98, q, = 74.75, q2 = 85.5, q3 = 90.5, and IQR = 15.75

- -

6-6 Time Sequence Plots 6-1 3

Exercise 6.7

uv Since the minimum = 65 is greater than ql - 1.5 x IQR = 5 1.1 and the maximum = 98 is less than q 3 + 1.5 x IQR = 1 14.1, the left and right whiskers extend to 65 and 98, respectively. Thus, there are no outliers in the test scores which exceed 1.5 IQR from the edges of the box.

Test Scores

I I I I

In Exercise 6-1, construct a box plot of the test results by using the following summary quantities:

minimum = 375, maximum = 2,265, q l = 1,097.75, q2 = 1,436.5, q3 = 1,735, and lQR = 637.25

6-6 Time Sequence Plots

60 70 80 90 100

O Explain the use of a time sequence plot. I -,-=

i

I I I I I I I I I

Time A time sequence plot is a graphical display that presents a time sequence (time Sequence series; data recorded in the order of time) to show the trend of the data over time.

Plot As an example, Figure 6-8 shows two time sequence plots of a company's sales of three years by year and by quarter. Each horizontal axis denotes time (years or quarters; t) and each vertical axis denotes sales (in year or quarter; x). Figure 6- 8(a) displays the annual sales are increasing over the years and Figure 6-8(b) indicates that the annual sales have a cyclic variation by quarter-the first- and second-quarter sales are higher than the sales of the other quarters.

I I I I 111111111l111l I I I I

sales (XI

(a) (b)

Figure 6-8 Time sequence plots of a company' sales: (a) annual sales; (b) quarterly sales.

,,*

*------. --e--*

sales + (4 3,

#' I,

,a, 9 ;" i I,, / 'w

0 : ,' , 1 0 , : .\; .* year (0

A U ~ A A I ~ ~ quarter ( r ) 1 2 3 4 1 2 3 4 1 2 3 4


6-7 Probability Plots

0 Explain the use of a probability plot. -"9".wa*w,.--a-~eeM ~ ~ , w ~ ~ " " ~ ~ " ~ ~ " * , " . ~ ~ ~ ~ ~ . - ~ b " ~ ~ ~ w ~ " ~ , ~ , ~ ~ ~ ~ . w ~ ~ ~ ~ ~ ~ , " ~ ~ . ~ ~ ~ ~ . ~ " ~ ~ ~ " w ~ ~ ~ B , ~ ~ . w ~ ~ ~ ' ~ ~ ~ ~ . ~ ~ " ~ ~ ~ ~ ~ ~ . ~ . ~ ~ ~ ~ ~ ~ - . ~ ~ " ~ ~ ~ " ~ " ~ " ~ . ~ ~ . ~ ~ " " ' ~ ~ ~ ~ " ~ " ~ ~ ~ . ~ ~ . ~ . ~ ~ " ~ -",---,.-as*

Probability A probability plot is a graphical display that presents the cumulative relative Plot frequencies of ordered data x(l), xcz), . . . , xc,, on the probability paper of a

hypothetical distribution (such as normal, Weibull, and gamma distributions) to examine if the hypothesized distribution fits the data set.

For example, Figure 6-9 shows a normal probability plot of 20 ordered observations xu) against corresponding cumulative relative frequencies (j - 0.5)ln. Statistical software is usually used to produce a probability plot. As the plotted points closely lie along a straight line (chosen subjectively), the adequacy of the hypothesized distribution is supported.

The probability plot method is useful to examine if the data conforms to a hypothetical distribution, but lacks objectivity by relying on subjective judgment. A formal method to test the goodness-of-fit of a probability distribution is presented in Section 9-7.

60 70 80 90 100

Observation (xu))

Figure 6-9 A normal probability plot.

MINITAB Applications


Examples 6.1 (Descriptive Statistics)

and 6.4 I (1) Choose File % New, click Minitab Project, and click OK.

:he worksheet.

1 (3) Choose Stat % Basic Statistics % Display Descriptive Statistics. In


Examples 6.2 and 6.4 (cont.)

Example 6.3

I Descriptive Statistics I

I 80 82 84 86 88 95% Confdence Interval tor Slgma I 74316 125444

95% Confidence Interval for Medlen

79 4574 88 0000

(Stem-and-Leaf Diagram)

(1) Choose Graph > Stem-and-Leaf. In Variables, select Score. In Increment, enter the difference between the smallest values of adiacent stems. Then click Ok.

Character Stem-and-Leaf Display

I Stem-and-leaf of Score N = 30 Leaf Unlt = 1.0 I

Example 6.5

-

MINITAB Applications 6-1 7

I (Frequency Measures)

(1) Choose Graph > Histogram. In Graph variables, select Score. Then

(2) Under Type of Histogram, check Frequency or Cumulative Frequency. Under Type of Intervals, check Cutpoint. Under Definition of Intervals, check Midpointlcutpoint positions and enter the set of cut

Score


Example Graph/Histogram)

1 ) Choose Graph k Histogram. In Graph variables, select Score.

6.6

Then

(Bar

(

Jnder Type of Histogram, check Frequency or Cumulative Frequency. Under Type of Intervals, check Cutpoint. Under Definition of Intervals, check Midpointlcutpoint positions and enter the set of cut

64 70 76 82 88 94 100

Score

Example 6.7


I (Box Plot)

Score



Exercise 6.1 (Mean, Variance, and Range) I

= 161,913.9 = 402.4~ (3) Minimum = 375 (4) Maximum = 2,265 (5) R = 2,265 - 375 = 1,890

Exercise 6.2 1 (Random Sample)

Since the same material is used for the sheet metal panels in various sizes, the number of defects per unit panel area will not be significantly different depending on the size of the panel. Thus, the random sample selected only from the large-size panels will properly represent the entire roup of the B sheet metal panels to study the average number of defectslft .

Exercise 6.3 (Stem-and-Leaf Diagram)

Step 1 : Determine stems and leaves. The test results range from 375 to 2,265. For simplicity, round each data at the 10s place and use only the 1,000s and 100s digits for plot. Since the 1,000s digit is most meaningful, the 1,000s and 100s digits of a test result are defined as the stem and leaf of the data, respectively: 0 to 2 for stems and 0 to 9 for leaves. However, the number of the stems is only three, therefore, the stems are further divided into

Of,Os,Oe, lz, It, If, Is, le,2z,and2t where 'z' is for the leaves 0 and 1, 't' is for 2 and 3, 'f' is for 4 and 5, 's' is for 6 and 7, and 'e' is for 8 and 9.

Step 2:

0s 7 1 Oe 8 8 8 8 9 9 9 7 l z 0 0 0 0 0 0 1 1 1 1 1 1 1 13 I t 2 2 2 3 3 3 3 3 3 3 3 11 1 f 4 4 5 5 5 5 5 5 5 5 5 11 Is 6 6 6 6 6 6 6 7 7 9 1 e 8 8 8 8 8 8 8 9 9 9 9 9 12 22 0 1 1 3

2 3 *B*vse.e*.,*",.a.-*b,+B.***,* '* %--

I Step 3: Summarize the frequency of leaves for each stem as shown above.


Exercise 6.4 (Percentile, Quartile, Interquartile Range, Median, and Mode)

(1) Since n = 70 is even, r = (n + l)k = 71 x 0.9 = 63.9, which is not an integer. Therefore,

PO.^ = X(Lr]) + (~([rl) - ~ ( l r i - L'D - - x(163,9) + (x([a.sl) - ~ ( 1 6 3 . 9 ~ I(63.9 - l63.9D

= x ( ~ ~ ) + ( x ( ~ ~ ) - ~ ( ~ ~ ) ) ( 6 3 . 9 - 63) = 1,890 + (1,910 - 1,890) X 0.9

= 1,908 (2) Since r = (n + l)k = 7 1x 0.75 = 53.25 is not an integer,

43 = P0.75 = x(1.h + (~(r rh -xILr1, - LrD - - ~ ( ~ 5 3 . 2 5 ~ + (~(153.25) - ~ ( 1 5 3 . 2 5 ) l(53.25 - 153.258

= x ( ~ ~ ) + ( x ( ~ ~ ) - ~ ( ~ ~ ) ) ( 5 3 . 2 5 - 53) = 1,730 + (1,750 - 1,730) x 0.25

= 1,735

(3) IQR = q3 - 4, = - = 1,735 - 1,097.75 = 637.25

(4) Since r = (n+l)k = 71x0.5 = 35.5 is not an integer,

~ 0 . 5 = ~(1.j) + ( ~ ( l r i - ~ ( j r j ) )(' - 1'4 9

- - X ( ~ a . 5 ) + (~([3i.il) - ~([3i , i~)(35-5 - 135.5D

= x ( ~ ~ ) + ( x ( ~ ~ ) - ~ ( ~ ~ ) ) ( 3 5 . 5 - 35) = 1,421 + (1,452 - 1,421) x 0.5

= 1,436.5 (5) Modes = 1,102, 1,3 15, and 1,750 (trimodal)

Exercise 6.5 1 (Frequency Measures)

Exercise 6.6 1 (Bar GraphIHistogram) Step 1 : Determine the number of bins (k).

Since & = f i = 8.4 , k = 9.

I Step 2: Determine the width of a bin (w). R max- min 2,265 -375

Since - = - - =210,usew=220. k k 9

I Step 3: Determine the limits of the histogram. . k w - R

1 G min- --- 9 x 220 - 1890

= 375 - = 330 2 2

u = l + k w = 3 3 0 + 9 ~ 2 2 0 = 2 , 3 1 0

6-22 Chapter 6 Random Sampling and Data Description i 1


Exercise 6.7

Step 4: Prepare a frequency table based on the bins determined.

Relative Cumulative Class IntervaI (X ) Frequency Relative

Frequency Frequency

Step 5: Develop a histogram. -

I T 0 2 g

Test Results

(Box Plot)

Since the minimum = 375 is greater than ql - 1.5 x IQR = 14 1.9 and the maximum = 2,265 is less than 93 + 1.5 x IQR = 2,690.9, the left and right whiskers extend to the minimum and maximum, respectively. Thus, there are no outliers in the test results which exceed 1.5 IQR from the edges of the box.

I Test Results (unit: hr)

7- 1 Introduction 7-4 Sampling Distributions

7-2 General Concepts of Point Estimation 7-5 Sampling Distributions of Means

7-3 Methods of Point Estimation Answers to Exercises

7-1 Introduction

* 8 s m - a &%s ee* -a a" W w M - n - m W # " W - s -wmm-#"w wa -- = m w w a w- e -* wmvw#*a.w w- us***-

O Describe the terms parameter, point estimator, and point estimate. O Identify two major areas of statistical inference. O Distinguish between point estimation and interval estimation. O Determine the point estimate of a parameter.

Parameter A parameter (denoted by 0) represents a characteristic of the vovulation under ( 0 ) study. It is constant but unknown in most cases.

(e.g.) Parameters Mean (p), variance (02), proportion (p), correlation coefficient (p), and regression coefficient (p)

Statistical Statistical inference refers to making decisions or drawing conclusions about a Inference population by analyzing a sample from the population. Two major areas of

statistical inference can be defined: 1. Parameter estimation: Estimates the value of 0. (e.g.) p. = 150 2. Hypothesis testing: Tests an assertion of 0. (e.g.) Ho: p = 150

Parameter Parameter estimation is further divided into two areas: Estimation 1. Point estimation: Estimates the exact location of 0. (e.g.) p = 150

2. Interval estimation: Establishes an interval that includes the true value of 0 with a designated probability. (e.g.) 145 < p < 155

7-2 Chapter 7 Point Estimation of Parameters

Point A point estimator (denoted by 6 ) is a statistic used to estimate 0. It is a random Estima?r variable because a statistic is a random variable.

( 0 ) n (e.g.) Point estimator of p: sample mean X = X i l n

A single numerical value of 6 determined by a particular random sample is

called a point estimate of 8 (denoted by 0).

Example 7.1

7-2 General Concepts of Point Estimation

(Point Estimate) Suppose that the life length of an INFINITY light bulb (3 has a normal distribution with mean p and variance 02. A random sample of size n = 25 light bulbs was examined and the sum of the life lengths was 14,900 hours. Estimate the mean life length (F) of an INFINITY light bulb.

m

w 'Bw* --- 0 Explain the terms unbiased estimator, minimum variance unbiased estimator (MVUE), standard

error, and mean square error ( M S E ) ofan estimator. 0 Select the appropriate point estimator of 8 in terms of unbiasedness, minimum variance, and

minimum mean square error each. - w*m" I* JXnXr a_nlsje*( -_ w _ ( _ a V * . _ ^ _ - - P m w _ ' * & * ~ - - B ~ P m - - - - e - -------

Exercise 7.1

Unbiased An unbiased estimator is a point estimator ( 6 ) whose expected value is equal to Estimator the true value of 8, i.e.,

~ ( 6 ) = 8 .

Note that several unbiased estimators can be defined for a single parameter 8.

The line width (X) of a tool used for semiconductor manufacturing is assumed to be normally distributed with mean p and variance 02. A random sample of size n = 20 tools was measured and the sum of the line width measurements was 13.2 micrometers (pm). Estimate the mean line width (p) of the tool.

As shown in Figure 7-1, the bias of 6 is the difference of the expected value of

6 from the true value of 8:

bias = ~ ( 6 ) - 8

7-2 General Concepts of Point Estimation 7-3

Unbiased Estimator

(cont.)

Figure 7-1 Bias of a point estimator 6 .

(Unbiased Estimator) Let X I , X2,. . . , Xn denote a random sample of size n from a probability distribution with E(X) = p and V(X) = 02. Show if the sample mean

1 is an unbiased estimator of the population mean p.

I Since E(X) = p , X is an unbiased estimator of p.

Exercise 7.2 Let X I , X2,. . . , X7 denote a random sample from a population having mean y and (MR 7-21 I variance d. Show if the following estimator

I is an unbiased estimator of p.

Minimum A minimum variance unbiased estimator (MVUE) of 0 is the unbiased 6 with Variance the smallest variance (see Figure 7-2). By using the MVUE of 0, the unknown Unbiased parameter 0 can be estimated accuratelv and precisely.

Estimator (MVUE)

Figure 7-2 Variances of unbiased estimators of 0. -

Example 7.3 (Minimum Variance Unbiased Estimator) Let XI, X2,. . . , Xn denote a random sample of size n (>I) from a population with E(X) = p and V(X) = 0'. Both X ,

and X are unbiased estimators of p because their expected values are equal to p. Of the two estimators, which is preferred to estimate p and why?

7-4 Chapter 7 Point Estimation of Parameters 1 1

I I i

Example 7.3 (cont.)

c$ Exercise 7.3

Standard Error

( 0 6 1

t' Example 7.4

B Exercise 7.4

o-rr The variances of X, and X are

V(X, ) = o2

Since V(X,) = o2 > ~ ( 2 ) = o2 1 n for n > 1, X is preferred to estimate p with higher accuracy and precision.

Let X I , X2,. . . , Xn denote a random sample of size n (>3) from a population with E(X) = p and V(X) = 02. Both (X, + X2 + X,) / 3 and X are unbiased estimators

of p because their expected values are equal to p. Of the two estimators, which is preferred to estimate p and why?

The standard error of a point estimator (denoted by o6 ) is the standard deviation

of a point estimator 6 . It can be used as a measure to indicate the precision of parameter estimation.

If o6 includes unknown parameters that can be estimated, use of the estimates of

the parameters in calculating o6 produces an estimated standard error

(denoted by s6 or se(6)).

In Example 7-1, X has a normal distribution with mean p and variance o2 and a random sample of size n = 25 was examined.

1. (Standard Error) Assuming o2 = 402, determine the standard error of the n

sample mean ( X ). Note that X = Xi I n - N(p, o2 1 n) . i = l

0-rr o 40 o- =-=-=8.0 hrs. "& , I5

2. (Estimated Standard Error) Suppose that o2 is unknown and the sample variance sX2 = 352. Calculate the estimated standard error of the sample mean

( X I . h 6 s, 35

S-=-=-=-- - 7.0 hrs. X & & J 2 5

In Exercise 7-1, X i s normally distributed with mean p and variance o2 and a random sample of size n = 20 was measured.

1. Assuming o2 = 0.05~, determine the standard error of the sample mean (2 ). 2. Suppose that o2 is unknown and the sample variance s$ = 0.04~. Calculate the

estimated standard error of the sample mean ( 2 ).

7-2 General Concepts of Point Estimation 7-5

Mean The mean square error (MSE) of a point estimator 6 is the expected squared Square Error

(MSE) o f difference between 6 and 0:

Estimator M S E ( ~ ) = E(B - 0)' = E[G - E ( B ) ] ~ + [e - E ( B ) ~

= v (6) + (bias) ' = ~ ( 6 ) for an unbiased 6 because bias = 0.

Note that a biased estimator of 0 with the smallest MSE is sometimes used for precise estimation of 0 while sacrificing the accuracy of estimation on 0 (see Figure 7-3).

r Example 7.5

Figure 7-3 A biased 6 , with a smaller mean square error than that of the

unbiased 6 , .

(Mean Square Error o f Estimator) Suppose that the means and variances of

6 , and 6 , are ~ ( 6 , ) = 0 , ~ ( 6 ~ ) = 0.90, ~ ( 6 , ) = 5 , and ~ ( 6 ' ) = 4 ,

respectively. Which estimator is preferred to estimate 0 and why?

F O ~ 6 , , biasgl = E(O,) - 0 = 0 - 0 = 0

M S E ( ~ , ) = ~ ( 6 , ) + (biasgl )' = 5 - 0 = 5

For 62, biasg, = E ( 0 2 ) - 0 = 0.90 - 0 = -0.10

M S E ( ~ , ) = ~ ( 6 ~ ) + (bias6, )2 = 4 + 0.010'

By subtracting M S E ( ~ , ) from MS.(@ ) ,

M S E ( ~ , ) - M S E ( ~ , ) = 5 - (4 + 0 .010~) = 1 - 0.010'

The preferred estimator of 0 for precise estimation depends on the range of 0 as follows:

if 0 2 10 because M S E ( ~ , ) I M S E ( ~ , ) and 6, is unbiased

62, if 0 < 10 because M S E ( ~ , ) > M S E ( ~ , )


Exercise 7.5 Let XI, X2,. . . , X5 denote a random sample of size n = 5 from a population with I E(X) = 70 and V(X) = 52. Two estimators of p are proposed:

I Of thc two estimators, which is preferred to estimate p and why?

7-3 Methods of Point Estimation

0 Explain the utility of the maximum likelihood method. 0 Find a point estimator of 0 by using the maximum likelihood method.

Maximum Likelihood

Method

The method of maximum likelihood is used to derive a point estimator of 0. This method finds a maximum likelihood estimator of 9 which maximizes the likelihood function of a random sample Xl, X2, . . . , Xn

L(9)= f(xl ;0)f(x2;9) . . . f (xn ;9 ) , whereXl,Xz, ..., Xn-i.i.d. f(x;0)

(Maximum Likelihood Estimator) Let X I , X2,. . . , X,, denote a random sample of size n from an exponential distribution with the parameter h. Find the maximum likelihood estimator of h.

U-W The probability density function of an exponential distribution is

f (x; h) = he-h"

Thus, the likelihood function of X I , X2, . . . , Xn is

Then, the log likelihood function is n

l n ~ ( h ) = n i n h - h C x i

I Now, the derivative of in L(h) is

By equating this derivative of In L(h) to zero, the point estimator of h which

maximizes L(h) is

7-4 Sampling Distribution 7-7

Exercise 7.6 Let XI, X2,. . . , Xn represent a random sample of size n from a geometric (MR 7-21) distribution with parameterp. Find the maximum likelihood estimators ofp. I

7-4 Sampling Distribution

Explain the term sampling distribution.

Sampling A sampling distribution is the probabilitv distribution of a statistic (a function of Distribution random variables such as sample mean and sample variance). The sampling

distribution of a statistic depends on the following: (1) The distribution of the population (2) The size of the sample (3) The method of sample selection

(e.g.) Sampling distribution of a sample mean

7-5 Sampling Distributions of Means

Explain the central limit theorem (CLT). ~e termine the distribution of a sample mean by applying the central limit theorem.

IX-e_^el *- -rwX(" IIIII(*XCL- - -*(i-%i wXIIYI(I LX-B*.(XI*l(~eYX I#*w-ma I-l. *-#,. .* ni* *esX--Le* */ l-X*P.IXI * P %" * - I d X P

Sampling Suppose that a random sample of size n is taken from a normal distribution with Distribution of mean p and variance 02. Then, the sampling distribution of the sample mean is

n 0

(Derivation) X = Xi 1 n - N(p,-) , where X - N(p, 02) i=l

n

Since XI, X2,. . . , X, are independent and normally distributed with the same

E(X) = p and V ( X ) = 0 2 , the distribution of X is normal with mean and variance


Sampling (x, + x 2 n + . . . + X n Distribution of v(X) = V =-[v(x,)+ V(X,)+ . - .+ v(x,)] -

T r

A (cont.)

Central Let XI, X2,. . . , Xn denote a random sample of size n taken from a population (X ) Limit with mean y and variance 02. Then, the limiting form of the distribution of the -

sample mean X is (CLT) n o2 x - p

x = z X i / n - ~ ( p , - ) a Z=-- n o / & i'J(0,l)

i=l

as n approaches M. This normal approximation of X is called the central limit theorem (CLT).

As displayed in Figure 7-4, the distributions of the sample means from uniform, binomial, and exponential distributions become normal distributions as their sample sizes n become large. In most cases, if n 2 30, normal approximation of X will be satisfactory regardless of the distribution of X. In case that n < 30 but the distribution of X is close to the normal, normal approximation of X will be still held.

(a) (b) (c)

Figure 7-4 Sampling distribution of X : (a) X - uniform; (b) X - binomial with p = 0.2; (c) X - exponential with h = 1.

Based on the central limit theorem, the sampling distribution of 2 where X is normal or non-normal is as follows:

1. Case 1: Normal population, X - N(p, o2 )

7-5 Sampling Distributions of Means 7-9 1

Central 2. Case 2: Non-normal population with p and o2 Limit (1) Case 2.1: Normal approximation applicable

Theorem (CLT) (cant.)

n o 2 = Xi I n - N(p,-) , if n 2 30 or the distribution of X is close to i=l n

the normal (2) Case 2.2: Normal approximation inapplicable

It would be difficult to find the distribution of 2 if n < 30 and the distribution of X is significantly deviated from the normal. In this case, use non-parametric statistics for statistical inference (see Chapter 15).

I o- Since n 2 30, normal approximation is applicable to X even i f X is exponentially distributed. Thus, the sampling distribution of X is

Example 7.7 (Central Limit Theorem; Sampling Distribution of Sample Mean) Suppose that the waiting time (X; unit: min.) of a customer to pick up hislher prescription at a drug store follows an exponential distribution with E(X) = 20 min. and V(X) = 52. A random sample of size n = 40 customers is observed. What is the distribution of the sample mean?

Exercise 7.7 Suppose that the number of messages (X) arriving to a computer server follows a Poisson distribution with E(X) = 10 messages/hour and V(X) = 10. The number of messages per hour was recorded for n = 48 hours. What is the distribution of the sample mean?



Exercise 7.1 1 1. (Point Estimate)

Exercise 7.2 I (Unbiased Estimator)

1 Since ~ ( 6 ) = p, 6 is an unbiased estimator of p.

Exercise 7.3

Exercise 7.4

Exercise 7.5

(Minimum Variance Unbiased Estimator)

The variances of (XI + X 2 + X , ) I3 and X are

1 Since v o - > v(X) = - for n > 3, X is preferred as 1 )-3- n 1 the point estimator of p for more accurate and precise estimation on p.

, 1 . (Standard Error)

2. (Estimated Standard Error)

(Mean Square Error of Estimator)

For 6,,

bias. = E ( @ 1 ) - p = 7 0 - 7 0 = 0 01


Exercise 7.6

Exercise 7.7

follows a Poison distribution. Therefore, the approximate distribution of X I is

Since M S E ( ~ , ) = 5.0 < MSE(~ , ) = 9.4, 6, is selected for more accurate

and precise estimation on p.

(Maximum Likelihood Estimator) The probability mass function of a geometric distribution is

f (x; P) = (1 - PIX-' P

Thus, the likelihood fhnction of XI, X2,. . . , Xn is

Then, the log likelihood function is

Now, the derivative of In L(p) is

By equating the derivative of In L(p) to zero, the point estimator o fp is

The maximum likelihood estimator o f p is the reciprocal of X

(Central Limit Theorem; Sampling Distribution of Sample Mean)

Since n = 48 2 30, normal approximation is applicable to X even if X

8 Statistical Intervals for a Single Sample

8- 1 Introduction

8-2 Confidence Interval on the Mean of a

Normal Distribution, Variance Known

8-3 Confidence Interval on the Mean of a

Normal Distribution, Variance Unknown

8-4 Confidence Interval on the Variance and

Standard Deviation of a Normal

Population

8-5 A Large-Sample Confidence Interval for

a Population Proportion

8-6 A Prediction Interval for a Future

Observation

8-7 Tolerance Intervals for a Normal

Distribution

Summary of Statistical Intervals for a Single

Sample



8-1 Introduction

0 Distinguish between confidence, prediction, and tolerance intervals. __& -eBXn_jmPM^(-jWms* -k#M%wm- smm-# wa*mmm-d wmaea-ms - ~ - M M e m m ~ z - e e . w M **" "' '' -" - - x nc-iw* em- xn a x nn *

Statistical While point estimation estimates the exact location of a parameter (8), interval Intervals estimation establishes bounds of plausible values for 8. Three types of statistical

intervals are defined: 1. Confidence Interval (CI): Bounds a parameter of the population distribution.

(e.g.) A 90% CI on p where X- N(p, 02) indicates that the CI contains the value of p with 90% confidence (see Section 8.2).

2. Prediction Interval (PI): Bounds a fbture observation. (e.g.) A 90% PI on a new observation where X- N(y, 02) indicates that the PI contains a new observation with 90% confidence (see Section 8.6).

3. Tolerance Interval (TI): Bounds a selected proportion of the population distribution. (e.g.) A 95% TI on X with 90% confidence indicates that the TI contains 95% ofxvalues with 90% confidence (see Section 8.7).

8-2 Chapter 8 Statistical Intervals for a Single Sample

8-2 Confidence Interval on the Mean of a Normal Distribution, Variance Known

0 Interpret a 100(1 - a)% confidence interval (CI). Cl Explain the relationship between the length of a CI and the precision of estimation. Ll Determine a sample size to satisfy a predetermined level of error (E) in estimating p. 0 Identify the point estimator of y when o2 is known and the sampling distribution of the point

estimator. Ll Establish a lOO(1 - a)% CI on p when o2 is known.

_fVIwa_U_#m-w-w-.w*- X( %- -im--e -A X- *tw- -.*n(l*UIX9(X llAnnm#mw=eu*8 iarun*a-emmmn-e a-* * awiinaaaw "- -s*a*e an nn x * x a * n t B M m S W rxnlirrmu m a r - - M m n i 4 a r u i r

Confidence Interval

A 100(1 - a)% confidence interval on a parameter (8 ) has lower and/or upper bounds ( I I 0 I u), which are computed by using a sample from the population. Since different samples will produce different values of I and u, the lower- and upper-confidence limits are considered values of random variables L and U which satisfy the following:

P ( L l O I U ) = l - a , O l a l l

Since L and U are random variables, a CI is a random interval. A 100(1 - a)% CI indicates that, if CIS are established from an infinite number of random samples, 100(1 - a)% of the CIS will contain the true value of 8.

The types of confidence intervals include: 1 . Two-Sided CI: Specifies both the lower- and upper-confidence limits of 8,

such as 1 I 0 I u. (e.g.) 730 I k 5 750 hrs, where X = life length of an INFINITY light bulb

2. One-Sided CI: Defines either the lower- or upper-confidence limit of 8. (e.g.) 730 I p~ or pxI 750 hrs, whereX= life length of an INFINITY light bulb

Length of CI The length of a C I refers to the distance between the upper and lower limits (u - and I ) . The wider the CI, the more confident we are that the interval actually contains

Precision of the true value of 8 (see Figure 8-I), but less informed we are about the true value Estimation of 0.

X

710 730 750 770 (life length; hrs.)

I (f95%CI--+I I

Figure 8-1 Confidence intervals on the mean life length (8 = k) of an INFINITY light bulb with selected levels of confidence.

The length of a CI on 8 is inversely related to the precision of estimation on 8: the wider the CI, the less precise the estimation of 8.

8-2 Confidence Interval on the Mean of a Normal Distribution, Variance Known 8-3

Sample Size The distance of a point estimate of 0 from the true value of 0 represents an error Selection (denoted by E) in parameter estimation:

(cont.) ~ = 1 0 - 0 1

=I T - p 1 in estimating p

To establish a 100(1 - a)% CI on p which does not exceed a predefined level of E, the sample size is determined by the formula

2

(in case n is not an integer, round up the value)

Inference Parameter of interest: y Context (3

Point estimator of y: X - N(p,-) , o2 known n

x - p z z=-- N(071) "I&

Confidence A 100(1 - a)% CI on p when 0, is known is Interval - (J - CF

Formula X - Z , , ~ - < ~ I X + Z , , ~ - for two-sided CI &- & 0

x - z , - < p for lower-confidence bound h -

- (3 p I X + z , - for upper-confidence bound

&

(Derivation) Two-sided - confidence interval on p, o2 known x -

By using Z = - - N(0,l) , o / &

P ( L < ~ I u ) = ~ - a z P(- z,,, 5 z I z,,,)= I - a

Therefore, - 0 - 0

L = X - z a I 2 - and U = X + z a 1 2 - & h

Note that, for a one-sided CI, use z, instead of z,,, to derive the corresponding limit.

4 Chapter 8 Statistical Intervals for a Single Sample

Example 8.1 Suppose that the life length of an INFINITY light bulb (X; unit: hour) follows the normal distribution N(y, 402). A random sample of n = 30 bulbs is tested as shown below, and the sample mean is found to be ? = 780 hours.

1 No. Life Length No. Life Length No. Life Length

7 814 17 804 27 829 8 820 18 754 2 8 82 1 9 753 19 715 29 8 16 10

,awma-***a,-rB*B#m.~

1. (Confidence Interval on p, o2 Known; Two-Sided CI) Construct a 95% two- sided confidence interval on the mean life length (p) of an INFINITY light bulb.

I 1 - a = 0 . 9 5 3 a = 0 . 0 5 95% two-sided CI on p:

2. (Sample Size Selection) Find a sample size n to construct a two-sided confidence interval on p with an error = 20 hours from the true mean life length. Use a = 0.05.

Exercise 8.1 The diameter of a hole (X; unit: in.) for a cable harness is normal with 02= 0.012. (MR 8-10) A random sample of n = 10 yields an average diameter (2 ) of 1 SO45 in. I

I 1. Construct a 99% upper-confidence bound on the mean diameter (p) of the hole.

I 2. Find a sample size n to construct a two-sided confidence interval on y with an error = 0.003 inch from the true mean hole diameter. Use a = 0.05.

8-3 Confidence Interval on the Mean of a Normal Distribution, Variance Unknown 8-5

8-3 Confidence Interval on the Mean of a Normal Distribution, Variance Unknown

O Explain the characteristics of a t distribution and read the t table. O Identify the point estimator of p when o2 is unknown and the sampling distribution of the point

estimator. Establish a 100(1 - a)% CI on p when o2 is unknown.

t Distribution The probability density function of a t distribution with v degrees of freedom is

The mean and variance of the t distribution are E(X) = 0 and V(X) = v /(v - 2) for v > 2

A t distribution is symmetric about zero, unimodal, and bell-shaped like the standard normal, but has thicker tails than the standard normal (see Figure 8-4 in MR). As v approaches w, the t distribution is close to the standard normal.

t Table The t table (see Appendix Table IV in MR) provides 100a upper-tail percentage points ( t , , ) o f t distributions with various degrees of freedom (v):

P(Tv > t , , = a Since a t distribution is symmetric about zero,

- tl-a," - -ta,v

(e.g.) Reading the t table (I) P(T12 > t) = 0.05 , to,5,1, = 1.782

(2) P(T12 < t) = 0.05 P(T1, > t) = 0.95

Since t0.95,,2 = tl-0.05,12 = -t0.05,12 9 t0.95,12 =

Inference Parameter of interest: p Context

(5 Point estimator of p: X - N(p, -) , 02 unknown

n

Confidence A 100(1 - a)% CI on w when o2 is known is Interval

Formula S X - tai2,, - < p I X + ta/2,v - &-

for two-sided CI &

S X - t,,, - < p 6-

for lower-confidence bound

- S p 5 X + t,,, - for upper-confidence bound

&


CI (Derivation) Two-sided confidence interval on u, 0, unknown Formula F - p

By using T = - - s 1 J;; ,

I t , , , , = 1 -a "-"

Therefore, S L = X - ~ ~ , , , ~ - and U = X + - S J;; J;;

Note that, for a one-sided CI, use t, instead of t,,, to derive the corresponding limit.

Large Sample If the sample size is large (n 2 30), the t-based CI formula on p is similar to the 2 CI based CI formula on p below, regardless of whether the underlying population is

normal or non-normal according to the central limit theorem (presented in Section 7-5).

- S - S X - z a t 2 - < p 5 X + z a , , - for two-sided CI

6- J;; r Example 8.2 (Confidence Interval on y, o2 Unknown; Two-Sided CI) For the light bulb life length data in Example 8.1, the following quantities have been obtained:

n = 30, T = 780, and s 2 = 40,

Assume that the variance o2 is unknown. Construct a 95% two-sided confidence interval on the mean life length (p) of an INFINITY light bulb.

h 1 - a = 0 . 9 5 a=0.05; v = n - 1 = 3 0 - 1 = 2 9 95% two-sided CI on p:

Note that this t-based CI is wider than the corresponding z-based CI 765.7 I p 5794.3 in Example 8.1.

8-4 Confidence Interval on the Variance and Standard Deviation of a Normal Population 8-7

8-4 Confidence Interval on the Variance and Standard Deviation of a Normal Population


CI Explain the characteristics of a distribution. 0 Read the table. CI Identify the point estimator of o2 and the sampling distribution of the point estimator. Cl Establish a 100(1 - a)% CI on 02.

An Izod impact test was performed on n = 20 specimens of a PVC pipe product. The sample average and sample standard deviation were x = 1.25 and s = 0.25, respectively. Assume that Izod impact strength (X) is normally distributed. Construct a 95% upper-confidence bound on the mean Izod impact strength (p).

X2 Distribution The probability density function of a X2 distribution with v degrees of freedom is

The mean and variance of the distribution are E(X) = v and V(X) = 2v

A distribution is unimodal and skewed to the right (see Figure 8-8 in MR). As v approaches w, the distribution is close to a normal distribution.

x2 Table The table (see Appendix Table I11 in MR) provides 100a upper-tail percentage points (xi, , ) of distributions with various degrees of freedom (v), i.e.,

p(x: >&")=a

(e.g.) Reading the Table 2

(1) P(x;, > x )= 0.1, =XE.1,14 Z21.06

(2) P ( X ~ < x 2 ) = 0 . 1 2 ~ - P ( x : ~ > x 2 ) = 0 . 9 , =Xi,9,14 =7.79

Inference Parameter of interest: G~

Context

Point estimator of 02: s2 = ,X,,X2 ,..., Xn-i.i.d. ~ ( p , o ~ ) n - 1


Confidence (, - 1)s2 (n - 1)s2 Interval l o l , for two-sided CI

Formula X a i 2 , v X 1-a i 2 , ~

(n - 1)s 2

I o2 for lower-confidence bound X a i 2 , v

(n-1)s2 o 1 2 for upper-confidence bound

X I-a/ 2,v

(Derivation) Two-sided confidence interval on o2 (n - 1)s2

By using the test statistic x2 = - X2 (v) ,

r Example 8.3

Therefore,

L = Xa12,v

Note that, for a one-sided CI, use instead of x i l2 to derive the corresponding limit.

(Confidence Interval on oZ; Two-Sided CI) Suppose that the life length of an INFINITY light bulb (X) follows a normal distribution with mean p and variance 02. A random sample of n = 16 bulbs is tested, and the sample variance is found to be s2 = 44'. Construct a 95% two-sided confidence interval on the variance of the life length of an INFINITY light bulb (0').

OT. 1 - a = 0 . 9 5 : a=0 .05 ; v = n - 1 = 16-1 =15 95% two-sided CI on 02:

(n-1)s' (n-1)s' l o I

2 Xai2 ,v XI-a12,v

8-5 A Large-Sample Confidence Interval for a Population Proportion 8-9

8-5 A Large-Sample Confidence Interval for a Population Proportion


Identify the point estimator o f p and the sampling distribution of the point estimator

A rivet is to be inserted into a hole. A random sample of n = 15 holes is selected, resulting in the sample standard deviation of the hold diameter = 0.008 mm. Assume that the hole diameter is normally distributed. Construct a 99% upper- confidence bound on the variance of the hole diameter (02).

Establish a 100(1 - a)% CI onp.

Inference Parameter of interest: p Context X

Point estimator ofp: p = - , where X - B(n, p ) n

- N(0,l) if np and n(1- p ) > 5

Sampling The mean and variance of a binomial random variable X - B(n, p ) are Distribution of E(X) = np and V(X) = np(1- p )

i, Thus,

Recall that in Section 4-7 a binomial distribution B(n,p) approximates to a normal distribution i f p is not close to either zero or one and n is relatively large so that both np and n(1-p) are greater than five.

Therefore, if np and n(1- p ) > 5, the approximate distribution of P is


Confidence A

Interval P - z,, , I p < P + z , , , for two-sided CI Formula

I ^

p - za,, I p for lower-confidence bound

p I P + z a I 2 '(I - ')

for upper-confidence bound

(Derivation) Two-sided confidence interval* on p

By using the test statistic Z =

P ( L S ~ I U ) = I - ~

3 P(-z,,, < z < z , , , ) = I - ~

By using k in place o f p because the upper and lower limits of the C1 include the unknown parameter p,

= 1 - a

Therefore,

L = P - z a i 2 /? and U = ~ + Z ~ ! ~

* For a one-sided CI, use z , instead of z,,, to derive the corresponding limit.

Sample Size In estimating p, the following formulas are used to select a sample size n for a Selection predefined level of error:

2

n = ( ) p ( b p) if information o f p is available

2

= (2) x 0.25 if information o fp is unavailable

Example 8.4 A sample of n = 40 bridges in HAPPY County is tested for metal corrosion, and x = 28 bridges are found corroded.

I . (Confidence Interval onp; Two-Sided CI) Construct a 95% two-sided confidence interval on the proportion of corroded bridges (p) in the county.

8-6 Prediction Interval for a Future Observation 8-1 I

Example 8.4 I x 28 0 . ~ r 1 - a = 0 . 9 5 s a=0 .05 ; j= -=-=0 .7 (cont.) n 40

Since both nj? = 40 x 0.7 = 28 and n(1- j3) = 40 x 0.3 = 12 are greater than

five, the sampling distribution of k is approximately normal.

1 95% two-sided CI on a:

I 2. (Sample Size Selection) Determine a sample size n to establish a 95% confidence interval o n p with an error = 0.05 from the true proportion.

8-6 A Prediction Interval for a Future Observation


- O Identify the distribution of the prediction error X,,, - X . Cl Establish a 100(1 - a)% prediction interval (PI) for a new observation.

---mm%m s wse w w w **# m.-&w ** * -* #+ w* * * a sr> w e *- *-ass VS* we=* e wmaeax a & -a** sse- e - a % wm * W " a

The proportion of defective integrated circuits produced in a photolithography process is being studied. A random sample of n = 300 circuits is tested, revealing x = 13 defectives.

1. Construct a 95% lower-confidence bound on the proportion of defective integrated circuits @).

2. Determine a sample size n to establish a 95% confidence interval o n p with an error = 0.02 from the true proportion. Use j = 0.05 as an initial estimate o f p for sample size calculation.

Sampling Suppose that XI, X2,. . . , X, is a random sample from a normal population with Distribution Of mean p and variance o2 and we wish to predict a new observation X,+, . If X is , - E = X,,, - X used as a point estimator of X,+1, then the distribution of the corresponding

b prediction error E is

1 X,,, and X are independent C

--


Sampling Thus, -

Distribution of - Xn+l - x

E = X,+, - X - ~ ( 0 , 1 ~ ) , if 02 is known

(cont.) o 1 + - -

T = Xn+l - x l--T

- t(n - I), if o2 is unknown

Prediction Interval F - Z , / ~ ~ 1 + - for two-sided CI, o2 known

Formula -

r Example 8.5


- d : d : 2 ~ - t , ~ ~ , , - , s 1+-IX,+lSF+t,12, , - ls 1 + - fortwo-sidedC1,o unknown

2 (Prediction Interval on X,+I, o Unknown; Two-Sided CI) For the light bulb life length data in Example 8.2, the following quantities have been obtained:

Construct a 95% two-sided prediction interval on the life length of the next light bulb tested (X31).

h 1 - a = 0 . 9 5 a a = 0 . 0 5 95% two-sided PI on X31:

I Note that this t-based PI is wider than the corresponding t-based CI on p 765.1 I p 5794.9 in Example 8.2.

For the Izod impact test in Exercise 8.2, the following quantities have been obtained:

n = 20, ?=1.25, and s2 = 0 . 2 5 ~

I Construct a 95% two-sided prediction interval on the impact strength of the next specimen of PVC pipe tested (X2,).

8-7 Tolerance Interval for a Normal Distribution 8-1 3

8-7 Tolerance Intervals for a Normal Distribution

~ w e M ~ M m " - ~ ~ m ~ ~ ~ - m e - - - - w s e ~ ~ . ~ w . m ~ - w ~ ~ ~ 8 ~ - ~ A

0 Establish a y tolerance interval with 100(1 - a)% confidence for a normal population.

Tolerance Suppose that the life length (X ) of an INFINITY light bulb follows a normal Interval distribution with mean p and variance 0'. Then, the interval ( p - 1.960,

p + 1.960), called tolerance interval, includes 95% (y, coverage percent) of the

light bulb population in terms of life length.

When p and 0' are unknown, the 95% tolerance interval may be established as ( X - 1.96s , X + 1.96s ) by using X and s2. However, due to sampling variability in Z and s2, the estimated tolerance interval includes less than 95% of the population.

Thus, when p and o2 are unknown, a y tolerance interval with confidence level 100(1 - a)% is established by ( i? - ks , X + ks ), where k is found in Appendix Table XI of MR. Note that as the sample size n approaches w, the value of k is close to the z-value associated with y. r Example 8.6 (Tolerance Interval on X; Two-Sided CI) For the light bulb life length data in Example 8.2, the following quantities have been obtained:

n = 3 0 , X=780,and s2 =402

Construct a 95% two-sided tolerance interval with 95% confidence for the life length measurements of light bulbs.

h k = 2.529 for n = 30, y = 0.95, and confidence level = 0.95 95% two-sided TI is

(X-ks, X + k s ) s (780 - 2.529 x 40 , 780 + 2.529 x 40 )

(678.8, 881.2)

I Note that this TI is wider than the corresponding t-based CI on p 765.1 I p I 794.9 in Example 8.2.

6 I

<

1 Exercise 8.6 For the Izod impact test in Exercise 8.2, the following quantities have been

L (MR 8-61) obtained: I

n = 2 0 , 2=1.25,and s2 =0.25'

Construct a 95% two-sided tolerance interval on the impact strength of PVC pipe with 90% confidence.

8-1 4 Chapter 8 Statistical Intervals for a Single Sample

Summary of Statistical Intervals for a Single Sample

A. Confidence lntervals

p, o2 known

p, o2 unknown

02 of a normal distribution

p of a binomial distribution

B. Prediction lntervals

- Xn+l, o2 known X 2-z,,,o I+-IX, , , <X+za, ,o J :

8-6 - X Xntl, o2 unknown

C. Tolerance lntervals

X - N(p, 0 2 ) , o2 known

X - N(p, o2 ) , o2 unknown

* y: coverage percent ** k: tolerance level factor found in Appendix Table XI of MR



Example 8.1 (Inference on p, c2 Known)

(1) Choose File * New, click Minitab Project, and click OK.

(2) ITY light bulbs on the worksheet.

INFINITY

(3) Choose Stat * Basic Statistics % l-Sample Z. In Variables, select INFINITY. Click Confidence Interval and enter the level of confidence in Level. In Sigma, enter the population standard deviation assumed. Then click OK.

I The assumed slgma = 40.0 I

I

I

:""" ....... '." ..................... Variable N Mean StDev SE Mean j 95.0 'r C I INFINITY 3 0 780.00 40.02 7 . 3 0 ( 765.68, 794.32) .......................................

Z Confidence Intervals I

8-1 6 Chapter 8

Example 8.2

Statistical Intervals for a Single Sample

(Inference on p, o2 Unknown) (1) Choose File , New, click Minitab Project, and click OK.

1 (2) Enter the life length data of INFINITY light bulbs on the worksheet.

I (3) Choose Stat , Basic Statistics , 1-Sample t. In Variables, select INFINITY. Click Confidence Interval and enter the level of confidence in

I T Confidence Intervals I Varlable N Hean StDev SE Bean INFINITY 30 780.00 40.02 7 . 3 1 i ( 765.06, 794.94) i .......................................

V 1 Example 8.4 (Inference on p)

(1) Choose Stat , Basic Statistics , 1 Proportion. Click Summarized data and enter the number of bridges surveyed in Number of trials and the number of bridges corroded in Number of successes. Then click

Example 8.4 (cont.)


(2) Enter the level of confidence in Confidence level and the hypothesized proportion in Test proportion. Check Use test and interval based on

I Test and Confidence Interval for One Proportion I Test of p = 0.5 vs p nor = 0.5 .......... "*.."..." ............. *.. Sample X N S a m p l e p i I 95.0 5 C I ; 2-Value P-Value

28 40 0.700000 { (0.557987, 0.842013) I 2.53 0.011 .............., " ,.................

8-1 8 Chapter 8 Statistical Intervals for a Single Sample


Exercise 8.1

Exercise 8.2

Exercise 8.3

1. (Confidence Interval on p, o2 Known; Upper-Confidence Bound)

1 - a = 0.99 3 a = 0.01 99% upper-confidence bound on p:

1 2. (Sample Size Selection)

(Confidence Interval on p, o2 Unknown; Upper-Confidence Bound)

1 - a = 0 . 9 5 a = 0 . 0 5 ; v = n - 1 = 2 0 - 1 = 1 9 95% upper-confidence bound on p:

S p I x + t,,, -

&

(Confidence Interval on 02; Upper-Confidence Bound)

1 - a = 0 . 9 9 3 a = 0 . 0 1 ; v = n - 1 = 1 5 - 1 = 14 99% two-sided CI on 02:

(n- l ) s2 (31 ,

XI-a,"

(15-l)x0.008' s o 2

x ?-0.01,14

Exercise 8.4

Exercise 8.5

Exercise 8.6


I 1 . (Confidence Interval onp; Lower-Confidence Bound)

Since both n j = 300 x 0.043 = 13 and n(1- F ) = 300 x 0.957 = 287 are

greater than five, the sampling distribution of is approximately normal.

1 95% lower-confidence bound onp:

1 2. (Sample Size Selection)

(Prediction Interval on Xn+l, o2 Unknown; Two-Sided CI)

~ - r r I - a = 0.95 3 a = 0.05 95% two-sided PI on X21:

1 X n + , 1X+ta12,n-1S

(Tolerance Interval on X; Two-Sided CI) ~ - r r k = 2.564 for n = 20, y = 0.95, and confidence level = 0.90

1 95% two-sided TI is (X-ks, X+ks)

3 (1.25-2.564~0.25, 1.25+2.564~0.25) (0.61, 1.89)

9 Tests of Hypotheses for a Single Sample

9-1 Hypothesis Testing 9-5 Tests on a Population Proportion

9-2 Tests on the Mean of a Normal 9-7 Testing for Goodness of Fit

Distribution, Variance Known 9-8 Contingency Table Tests

9-3 Tests on the Mean of a Normal MINITAB Applications

Distribution, Variance Unknown Answers to Exercises

9-4 Hypothesis Tests on the Variance and

Standard Deviation of a Normal

Population

9-1 Hypothesis Testing

Cl Explain the terms null hypothesis, alternative hypothesis, test statistic, acceptance region, rejection region, critical value, type I error probability (a), type 11 error probability (P), and power of a test (1-P).

Cl Establish the acceptance and rejection regions of a hypothesis test at a . O Determine the type I1 error probability and power of a test.

Explain the relationships between a and P. O Explain the difference between strong conclusion and weak conclusion.

Identify the procedure of hypothesis testing.

Hypothesis A hypothesis is an assertion about the parameters (0) of one or more populations under study. There are largely two kinds of hypotheses: 1. Null hypothesis (Ho): States the presumed condition of 0 (based on

experience, theory, design specification, regulation, or contractual obligation) that will be held unless there is strong evidence against it. Note that Ho should always specify an value of 0.

(e.g.) Ho: px= 750 hrs, where X is the life length of an INFINITY light bulb

9-2 Chapter 9 Tests of Hypotheses for a Single Sample

Hypothesis 2. Alternative hypothesis (HI): States the condition of 0 that would be (cont.) concluded if Ho is rejected. In parallel to the types of confidence intervals (see

Section 8-2), the following types of Hl are defined depending on the directionality of 0 assumed in HI . (1) Two-sided HI: Indicates directionality of 0.

(e.g.) HI: k+ 750 hrs (2) One-sided HI: Indicates the directionality of 0.

(a) Lower-sided HI: Includes the lower-side inequality sign. (e.g.) Hl: px< 750 hrs

(b) Upper-sided HI: Includes the upper-side inequality sign. (e.g.) HI: px > 750 hrs

Test Statistic A test statistic refers to a statistic used for statistical inference about 0. (e.g.) Test statistic for inference on k w h e r e X - ~ ( p , a ~ ) and a2 is known:

Test Regions Two regions of a test statistic (see Figure 9-1) are established for testing Ho against H I : 1. Acceptance region: The region of a test statistic that will lead to failure to

E&CJ Ho. 2. Rejection (critical) region: The region of a test statistic that will lead to

reiection of Ho. Note that the boundaries between the acceptance and rejection regions are called critical values.

For example, in Figure 9-1, in testing Ho: px= 750 vs. HI: px# 750 where X denotes the life length of an INFINITY light bulb, the test statistic of p is the sample mean and the critical values of the sample mean are 730 and 770 hours. Therefore, we will conclude the following:

(1) If 730 I T 5 770 (acceptance region), fail to reject Ho. (2) if Z < 730 or 2 > 770 (rejection region), reject Ho.

Rejection region

730 770 b

Mean life length of light bulb ( X ) Figure 9-1 Acceptance and rejection rejections for Ho: px = 750 vs. HI: px + 750, where X (unit: hours) is the life length of an INFINITY light bulb.

Test Errors The truth or falsity of a hypothesis can never be known with certainty unless the entire population is examined accurately and thoroughly. Therefore, a hypothesis test based on a random sample may lead to one of the two types of wrong conclusions (see Table 9-1):

(1) Type I error: Reject Ho when Ho is true. (2) Type I1 error: Fail to reject Ho when Ho is false.

~~ ~- m............ ......_ ............... - ... . . . . . . . . . . . . .__.... . . . . . . . . .

( ~ v r r r . ) U ~ ~ j J g I v n i

Fail to reject HO Reject Ho Ho is true Correct decision Type I error (HI is false) (Correct acceptance) (False rejection)

Truth

The probabilib of type Ierror (denotedas a) and the probabi/ity of type 11 %\\%\\L%\%*~\\\\%~\<<\%~L\\L\LL<\~\\\<%\--s.

a = P(type1 enor) = P(xe1ect Ho\ HQ-IS tme) p = P(type I1 error) = P(fai1 to reject HO I HO is false)

Note that the type I error probability (a) is also called the level of significance of a test. Small values of a such as 0.05 and 0.01 are typically used to make it difficult to reject Ho if it is true.

Power of Test The power of a statistical test indicates the probability of rejecting Ho when Ho is false, indicating the ability (sensitivity) of the test to detect evidence against Ho:

power = P(reject Ho I HO is false) = 1 - P(fai1 to reject Ho I Ho is false) = 1 - P

Test Regions, The test regions, hypotheses, a , P, and power of a test are related to each other.

The acceptance and reiection regions of a test statistic 6 are determined based CG P' and on a and the hvuothesized value of 0 (denoted as 00) in Ho (note that a and 00 are

Power G i f i e d by the analyst). If L and U denote the lower and upper limits of an acceptance region, respectively, and we are testing Ho: 0 = 80, the acceptance

region of 6 is determined as follows: 1 - a = P(fai1 to reject Ho I Ho is true) = P(fai1 to reject Ho 1 0 = 80)

[P(L S 6 I U 1 0 = go), for two- sided H,

= { ~ ( ~ ~ 6 ~ 0 = 0 , ) , for lower- sided H,

k ( 6 5 U I 8 = go), for upper- sided H ,

On the other hand, the P and power of the test are determined based on the acceptance region of the test and the true value of 0 as follows:

f3 = P(fai1 to reject Ho I Ho is false) = P(fai1 to reject Ho ( 0 # 80)

[P(L 5 6 I u 1 0 f go), for two- sided H,

= / ~ ( ~ < 6 ~ 0 # 0 , ) , for lower- sided H,

[ ~ ( 6 S U I 0 + go), for upper- sided H,

and power = 1 - p

9-1 Hypothesis Testing 9-3

Test Errors (cont.)

is truck -Mw- ( ~ a l s e mm-w-eM acceptancei -w-e -m%ms - -* (Correct * *- -d rejection) a - s m - The probability of type I error (denoted as a ) and the probability of type I1 error (denoted as P) are conditional probabilities as follows:

a = P(type I error) = P(reject Ho I Ho is true) p = P(type I1 error) = P(fai1 to reject Ho I Ho is false)

Note that the type I error probability (a) is also called the level of significance of a test. Small values of a such as 0.05 and 0.01 are typically used to make it difficult to reject Ho if it is true.

Power of Test The power of a statistical test indicates the probability of rejecting Ho when Ho is false, indicating the ability (sensitivity) of the test to detect evidence against No:

power = P(reject Ho I Ho is false) = 1 - P(fai1 to reject Ho I Ho is false) = 1 - P

Test Regions, The test regions, hypotheses, a , P, and power of a test are related to each other.

The acceptance and reiection regions of a test statistic 6 are determined based P9 and on a and the hvpothesized value of 8 (denoted as 4) in Ho (note that a and Bo are Power specified by the analyst). If L and U denote the lower and upper limits of an

acceptance region, respectively, and we are testing Ho: 8 = €lo, the acceptance

region of 6 is determined as follows: 1 - a = P(fai1 to reject Ho I Ho is true) = P(fail to reject HO 1 8 = 80)

[P(L i 6 5 u I 8 = go), for two- sided HI

= { p ( ~ i 6 ~ 8 = e , ) , for lower- sided HI

b(6 5 U I 8 = 8,), for upper- sided H,

On the other hand, the P and vower of the test are determined based on the acceptance repion of the test and the true value of 8 as follows:

p = P(fail to reject Ho 1 Ho is false) = P(fai1 to reject Ho 1 8 # 00)

I P ( L i 6 5 ~ l 8 # 8 , ) , fortwo-sidedH,

= ~ ( ~ < 6 1 8 + 8 , ) , for lower- sided HI

I P ( ~ 5 U ( 8 3 e,), for upper- sided H,

and power = 1 - p

9-4 Chapter 9 r Example 9.1

Tests of Hypotheses for a Single Sample

Suppose that the life length of an INFINITY light bulb (X; unit: hour) is normally distributed with o2 = 402. We wish to test Ho: px= 750 versus HI: p x f 750 with a random sample of size n = 30 light bulbs.

1 . (Acceptance/Rejection Regions) Construct the acceptance and rejection regions of the test on p at a = 0.05.

t + ~ The test statistic of p is the sample mean with the following sampling distribution:

The acceptance region I I X 1 u for Ho: px= 750 vs. HI: px# 750 satisfies the following:

1 - a = 1 - 0.05 = 0.95 = P(fai1 to reject Ho 1 p = 750) = ~ ( 1 1 X I u ( p = 7 5 0 )

1 Accordingly, the critical values are

Therefore, acceptance region: 735.7 5 2 I 764.3 rejection region: i? < 735.7 and 2 > 764.3

I 2. (p and Power of Test) Assume that the true value of px= 730 hrs. Find the P and power of the test if the acceptance region is 735.7 I 2 I 764.3.

I h 6 = P(fai1 to reject Ho 1 Ho is false)

= P(735.7 I X 5764.3 ( p = 730)

I power = 1 - p = 0.782

9-1 Hypothesis Testing 9-5


The heat evolved from a cement mixture (X;, unit: callg) is approximately normally distributed with o2 = 22. We wish to test Ho: b = 100 versus HI: b# 100 with a sample of n = 9 specimens.

1. Find the acceptance region of the test at a = 0.01.

2. Suppose that the acceptance region is defined as 98.3 I T 5 101.7. Find the P and power of the test if the true mean heat evolved is 103.

Relationship The relationships between a and P are explained with illustrations in Figure 9-2. between a and p under H , under H ,

Hypothetical distributions

of X - N ( p , o2 1 n) for

Ho: 8 = 80andH1: 8 # 80

under H , under H,

1. As the acceptance region widens, a decreases but p increases, provided that the sample size n does not change.

-- -

under H, under H ,

2. As n increases, both a and p decrease, provided that the critical values are held constant.

under H , under H,

3. When Ho: 0 = 8 0 is false, p increases as the true value of 8 approaches to 8 0 and vice versa.

Figure 9-2 Relationships between a and P.


Strong vs. The analyst can directly control both a (probability of rejecting Ho when HO is Weak true) and (hypothetical value of €9, but not P (probability of accepting Ho when

Conclusion HO is false; p depends on the true value of 8). Thus, it is considered that "rejection of Ho" is a strong conclusion and "acceptance of Ho" is a weak conclusion. Consequently, rather than saying we "accept HO," we prefer saying "fail to reject Ho," which indicates that we have not found sufficient evidence to reject HO.

Hypothesis The formal procedure to make a statistical decision about hypotheses consists of Testing the following four steps throughout the rest of the chapters:

Procedure Step 1: State Ho and HI. Step 2: Determine a test statistic and its value. Step 3: Determine a critical value(s) for a. Step 4: Make a conclusion, such as we reject or fail to reject HO at a.

(Note) The four-step procedure of hypothesis testing is basically same with the eight-step procedure of hypothesis testing in MR.

9-2 Tests on the Mean of a Normal Distribution, Variance Known

R Test a hypothesis on p when o2 is known (z-test). R Calculate the P-value of a z-test. O Compare the a-value approach with the P-value approach in reporting hypothesis test results. i

! R Explain the relationship between confidence interval estimation and hypothesis testing. O Determine the sample size of a z-test for statistical inference on y by applying an appropriate

sample size formula and operating characteristic (OC) curve. O Explain the effect of the sample size n on the statistical significance and power of the test.

Inference Parameter of interest: y Context o

Point estimator of p: X - N(y,-) , 02 known n

- p - N(0,l) Test statistic of p: Z = - o/,L

Test Step 1 : State Ho and HI. Procedure Ho: ~ = p l j

(2-test) H I : y f po for two-sided test

y > plj or p < po for one-sided test

Step 2: Determine a test statistic and its value.

? - P O 2, =- of&

9-2 Tests on the Mean of a Normal Distribution, Variance Known 9-7

z-test (cont.)

Significance Probability

(P-value)

a-value vs. P-value

Approach

Interval Estimation

and Hypothesis

Test

Sample Size Formula

Step 3: Determine a critical value(s) for a. z,,, for two-sided test; z, for one-sided test

Step 4: Make a conclusion. Reject HO if

Izo I > za, for two-sided test

zo > z, for upper-sided test

zo < -z, for lower-sided test

The significabce probability (denoted as P) of a test statistic refers to the smallest level of significance (a) that would lead to rejection of Ho. Thus, the smaller the P-value, the higher the statistical significance (an inverse relationship between P- value and statistical significance).

For example, the P-value of a test statistic zo can be computed by using the following formulas:

1 1 for two-sided test

for upper-sided test

for lower-sided test

(Note) @(z) = P(Z 5 z) : cumulative distribution function of Z

Two approaches are available to use in reporting the result of a hypothesis test: 1. a-value approach: States the test result at the value of a preselected. This

reporting method lacks providing the weight of evidence against Ho. 2. P-value approach: Specifies how far the test statistic is from the critical

value(s). Once the P-value is known, the decision maker can draw a conclusion at any specified level of significance (a) as follows:

Reject Ho at a , i f P I a Fail to reject Ho at a , otherwise

Note that the P-value approach is more flexible and informative than the a-value approach.

A close relationship exists between a two-sided 100(1 - a)% confidence interval (CI) on 0 (1 5 0 I u) and the test of a null hypothesis about 0 (Ho: 8 = 80) at a as follows:

Reject Ho at a , if the two- sided CI does not include 8,

Fail to reject Ho at a , otherwise

For example, suppose that a 95% CI on p is 75 1 I p I 779 and we are testing Ho: p = 750 vs. HI: p # 750 at a = 0.05. Since the CI of p does not include the hypothesized value p = 750, we will fail to reject Ho at a = 0.05.

Formulas are used to determine the sample size of a particular test for particular levels of p (or power = 1 - p), a , and 6 (= 10 - 001, difference between the true value of 0 and its hypothesized value). For a z- test on a single sample, the following formulas are applied:


Sample Size Formula n =

( ~ a i 2 + z p ) 2 0 2 , where 6 = p - p o for two-sided test (cant.) 6

Note that the sample size requirement increases as a , P, and 6 decrease and o increases.

Operating Operating characteristic (OC) curves in Appendix A of MR plot P against d (for Characteristic Z- and t-tests) or 3L (for X2- and F-tests) for various sample sizes n at a = 0.01 and

(OC) 0.05, i.e., Curve P =An, d or h, a )

By using an appropriate OC chart, an adequate sample size can be determined for a particular test to satisfy a predefined test condition.

Table 9-2 displays a list of OC charts in MR and a formula of the OC parameter d for a z-test on p. By using the table, the appropriate OC chart for a particular z- test is chosen (e.g., for a one-sided z-test at a = 0.05, chart VIc is selected).

Table 9-2 Operating Characteristic Charts for z-test - Single Sample

Test a Chart VI (Appendix A in MR) OC parameter

0.05 Two-sided

0.01 z-test -.

0.05 (c) (7 (7 One-sided

Effect of As the sample size n increases, both the statistical significance (inverse to P- Sample Size value) and power (= 1 - P) of the test increase. For example, Table 9-3 presents

P-values and powers of testing on p for the following conditions:

X - N p,- and x = 5 0 . 5 - [:I

Ho: px= 50 VS. H I : px+ 50 a = 0.05 and the truevalue o f p = 50.5

The P-value column indicates that, for the same value of i? = 50.5, Ho is rejected at a = 0.05 when n = 100 because P I a , while fro i s ' h t rejected at a = 0.05 when n 2 50 because P $ a.

9-2 Tests on the Mean of a Normal Distribution, Variance Known 9-9

Statistical vs. Practical

Significance

tr Example 9.2

The statistical significance of a test does not necessarily indicate its practical significance. Sincie the power of a test increases as the sample size increases, any small departure of 9 from the hypothesized value O0 will be detected (in other words, Ho: 0 = O0 will be rejected) for a large sample, even when the departure is of little practical significance. Therefore, the analyst should check if the statistical test result has also practical significance.

For the light bulb life length data in Example 8.1, the following results have been obtained:

n = 3 0 , T = 7 8 0 , 0' =402

95% two-sided CI on p: 765.7 I p I 794.3

1. (Hypothesis Test on k, o2 Known; Two-Sided Test) Test Ho: p = 765 hrs vs. HI : p # 765 hrs at a = 0.05.

Step 1 : State Ho and HI. Ho: p = 765 HI: ~ $ 7 6 5

Step 2: Determine a test statistic and its value. X - po - 780 - 765

zo =-- = 2.05 0 / & 40/@

Step 3: Determine a critical value(s) for a. za12 = z0.0512 = '0.025 =

Step 4: Make a conclusion. Since lzol = 2.05 > z,,,, = 1.96, reject Ho at a = 0.05.

2. (P-value Approach) Find the P-value for this two-sided z-test.

h P = 2[1- @(I z0 I)] = 2[1- @(I 2.05 I)] = 2[1- 0.9801 = 0.04

Since P = 0.04 I a = 0.05, reject Ho at a = 0.05.

3. (Relationship Between CI and Hypothesis Test) Test Ho: p = 765 hrs vs. HI: y # 765 hrs at a = 0.05 based on the 95% two-sided CI on p.

h Since the 95% two-sided CI on y, 765.7 I p 2794.3, does not include the

hypothesized value 765 hrs, reject Ho at a = 0.05.

4. (Sample Size Determination) Determine the sample size n required for this two-sided z-test to detect the true mean as high as 785 hours with power of 0.9. Apply an appropriate sample size formula and OC curve.

h (1) Sample Size Formula power = P(reject HO I Ho is false) = 1 - p = 0.9 P = 0.1 6 = y - y o ~ 7 8 5 - 7 6 5 ~ 2 0


Example 9.2 (cont.)

(2) OC Curve To design a two-sided z-test at a = 0.05 for a single sample, OC Chart VIa is applicable with the parameter

By using d = 0.5 and P = 0.1, the required sample size is determined as n = 44 as displayed below. Note that this sample size is similar with the sample size n = 42 determined by using a sample size formula.

(a) O.C. curves fur different values of n for the two-sided nomu4 test for a level of significance a = 0.05 I

Exercise 9.2 For the hole diameter data in Exercise 8.1, the following results have been (MR 9-28) obtained: I

n = 10, ?=1.5045,and o2 =0.012

1. Test the hypothesis that the true mean hole diameter is greater than 1.50 in. at a = 0.01.

2. Find the P-value for this one-sided z-test.

3. Determine the sample size n required for this one-sided z-test to detect the true mean as high as 1.505 inches with power of 0.9. Apply an appropriate sample size formula and OC curve.

9-3 Tests on the Mean of a Normal Distribution, Variance Unknown /

Test a hypothesis on y when o2 is unknown (t-test). O Determine the sample size of a t-test for statistical inference on p by using an appropriate

operating characteristic (OC) curve.

9-3 Tests on the Mean of a Normal Distribution, Variance Unknown 9-1 1

Inference Context

Test Procedure

(t-test)

Operating Characteristic

(OC) Curve

r Example 9.3

Parameter of interest: y 0

Point estimator of p: X - N(p,-) , o2 unknown n

-, X - p Test statistic of p: T = ----- - t ( v ) , v = n - 1

s /&

Step 1 : State Ho and HI. Ho: y = p o HI: p # pg for two-sided test,

p > po or y < yo for one-sided test.


2 - P O To = ------ - t (n - 1) s / &

Step 3: Determine a critical value(s) for a. t,,,,,-, for two-sided test; t u n , for one-sided test

Step 4: Make a conclusion. Reject Ho if

t o 1 > t n 1 for two-sided test

to > t,,,-, for upper-sided test

to < -t,,,-, for lower-sided test

Table 9-4 displays a list of OC charts in MR and a formula of the OC parameter d for a t-test on p. By using the table, the appropriate OC chart for a particular t-test is selected (e.g., for a two-sided t-test at a = 0.01, chart VIfis selected).

Table 9-4 Operating Characteristic Charts for 1-test - Single Sample

Test a Chart VI (Appendix A in b4R)

OC parmeter - , ..

Two-sided

t-test

One-sided * - - -"-"---- -" * -------* * - * - - - * - w - - w w - - - - ----

(Note) For 6, use a sample standard deviation or subjective estimate.

For the life length data in Example 8.2, the following results have been obtained:

n=30 , Z=780,and s 2 =402

1. (Hypothesis Test on p, 0' Unknown; Two-Sided Test) Test Ho: p = 765 hrs vs. HI: 1-1 # 765 hrs at a = 0.05.

9-1 2 Chapter 9 Tests of Hypotheses for a Single Sample

Example 9.3 (cont.)

* Step 1 : State HO and H I . Ho: P = 765 HI: ~ $ 7 6 5


Step 3: Determine a critical value(s) for a. -

f a / 2,n-1 - '0.05 /2,30-1 = t0.025.29 = 2.045

Step 4: Make a conclusion. Since ltol = 2.05 > = 2.045 , reject Ho at a = 0.05.

2. (Sample Size Determination) Determine the sample size n required for this two-sided t-test to detect the true mean as high as 785 hours with power of 0.9. Apply an appropriate OC curve.

hr power = P(reject Ho / Ho is false) = 1 - p = 0.9 3 p = 0.1 6=p-j.10 =785-765=20

To design a two-sided t-test at a = 0.05 for a single sample, OC Chart VIe is applicable with the parameter

By using d = 0.5 and P = 0.1, the required sample size is determined as n =

45 as displayed below.

d

(e) O.C. curves for different values of n for the two-sided t-test for a level of siBnif~cance a = 0.05.

9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Population 9-1 3 7

/

m Exercise 9.3 For the Izod impact test data of a PVC pipe product in Exercise 8.2, the following (MR 9-35) results have been obtained: I

I The ASTM standard requires that Izod impact strength must be greater than 1.0 ft-lblin.

I 1. Test if the Izod impact test results satisfy the ASTM standard at a = 0.05.

I 2. Determine the sample size n required for this one-sided t-test to detect the true mean as high as 1.10 with power of 0.8. Apply an appropriate OC curve.

9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Population

Test a hypothesis on o2 (x2-test). O Determine the sample size of a %'-test for statistical inference on o2 by using an appropriate

operating characteristic (OC) curve.

Inference Parameter of interest: o2 Context

Point estimator of 02: s2 = i=l ,Xl,X2 ,..., Xnwi.i.d. N ( ~ , O ~ ) n-1

(n-1)s' Point estimator of 02: x = - X ( v ) , v = n - 1

o

Test Step 1: State Ho and HI. Procedure Ho: 0 2 =oo 2

a2-test) HI: o2 # 0; for two-sided test,

o2 > 0; for upper-sided test

o2 < (3; for lower-sided test.


Step 3: Determine a critical value(s) for a. 2 2 xa; 2,n-1 and x ~ - ~ , 2,n-l for two-sided test 2 for upper-sided test 2 x ~ - ~ , ~ - ~ for lower-sided test


X2-test Step 4: Make a conclusion. Reject Ho if (cont.) xo 2 > x ~ ~ ~ , ~ - ~ 2 or xo 2 < x ~ - ~ ~ ~ , ~ - ~ 2 for two-sided test

2 2 xo > x,,~-, for upper-sided test 2 2 xo < x ~ - ~ , ~ - ~ for lower-sided test

Operating Table 9-5 displays a list of OC charts in MR and a formula of the OC parameter h Characteristic for a X2-test on 2. By using the table, the appropriate OC chart for a particular

(OC) x2-test is selected (e.g., for an upper-sided x2-test at a = 0.05, chart VIk is Curve selected).

Table 9-5 Operating Characteristic Charts for r2-test " ,*

Test a Chart VI (Appendix A in MR) OC parameter

0.05 Two-sided

0.01

r Example 9.4 For the light bulb life length data in Example 8.3, the following results have been obtained:

n = 16, s2 = 442

95% two-sided CI on 02: 32S2 l o2 5 68. l2

1 . (Hypothesis Test on 02; Two-Sided Test) Test Ho: o2 = 402 vs. HI: o2 f 402 at a = 0.05.

~l-rr Step 1 : State Ho and H I .

Ho: o2 =402

HI: o2 +402


Step 3: Determine a critical value(s) for a. 2 2 2 2

X0.0512,16-1 = X0.025,15 = 28.65 and XI-0.05/2,16-1 = X0.975,15 = 6.91

Step 4: Make a conclusion. Since xi = 18.15 P ~ i , ~ ~ ~ , ~ ~ = 28.65 and

= 18.15 rr xi,,, = 6.91, fail to reject Ho at a = 0.05.

I 2. (Relationship Between F/I and Hypothesis Test) Test Ho: o2 = 402 vs. HI: o2 # 402 at a = 0.05 based pn the 95% two-sided CI on 02.

9-5 Tests on a Population Proportion 9-1 5

Example 9.4 (cont.)

n- Since the 95% two-sided CI on 02, %2S2 I o' 5 68. 12 , includes the hypothesized value 402, fail to reject Ho dt a = 0.05.

3. (Sample Size Determination) Determine the sample size n required for this two-sided x2-test to detect the true standard deviation as high as 50 with power of 0.8. Apply an appropriate OC curve.

h power = P(reject Ho I Ho is false) = 1 - B = 0.8 s /3 = 0.2

To design a two-sided X2-test at a = 0.05, OC Chart VIi is applicable with the parameter

By using h = 1.25 and /3 = 0.2, the required sample size is determined as n =



1.25 A

( I ) O.C. curves for d i i kml values of n fa the tvm-dddd chi-square test fa a lml of significance a = 0.05

For the hole diameter data in Exercise 8.3, the following results have been obtained:

n = 15 and s2 = 0.008~

I . Test if there is strong evidence to indicate that the standard deviation of the hole diameter exceeds 0.01 mm. Use a = 0.01.

2. Determine the sample size n required for this upper-sided x2-test to detect the true standard deviation mean which exceeds the hypothesized value by 50% with power of 0.9. Apply an appropriate OC curve.

9-5 Tests on a Population Proportion

0 Test a hypothesis o n p (2-test) for a large sample. 0 Determine the sample size for statistical inference o n p by using an appropriate sample size

formula.


Inference Parameter of interest: p Context - X

Point estimator ofp: P = - , where X - B(n, p ) n

Test statistic ofp: - - N(0,l) if np and n(1- p) > 5

Test Step 1 : State Ho and H I . Procedure Ho: p 'PO

(2-test) HI: p stpo for two-sided test, p > po or p < p~ for one-sided test.


Step 3: Determine a critical value@) for a. z,,, for two-sided test; z , for one-sided test


)zO 1 > za,, for two-sided test

zo > z , for upper-sided test

z, < -z, for lower-sided test

Sample Size For a hypothesis test onp, the following formulas are applied to determine the Formula sample size:

Y Example 9.5

L

for two-sided test

L

for one-sided test

For the corroded bridge data in Example 8.4, the following results have been obtained:

x 28 n = 4 0 and j= -=-=0 .7

n 40

1. (Hypothesis Test onp; Two-Sided Test) Test Ho: p = 0.5 vs. H I : p # 0.5 at a = 0.05.

h Step 1: State Ho and HI. Ho: p = 0.5 H I : ~ $ 0 . 5

Example 9.5 (cont.)

9-7 Testing for Goodness of Fit 9-1 7

I Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. za12 = z0.05/2 = '0.025 =

Step 4: Make a conclusion. Since lz,l= 2.53 > z,,,~, = 1.96, reject Ho at a = 0.05.

2. (Sample Size Determination) Determine the sample size n required for this two-sided z-test to detect the true proportion as high as 70% with power of 0.9. Apply an appropriate sample size formula.

b power = P(reject Ho I Ho is false) = 1 - p = 0.9 p = 0.1

1 . Test if the fraction of defective units produced is than 0.05. Use a = 0.05. 2. Determine the sample size n required for this one-sided z-test to detect the true

proportion as low as 2% with power of 0.8. Apply an appropriate sample size formula.


9-7 Testing for Goodness of Fit

For the circuit test data in Exercise 8.4, the following results have been obtained:

A x 13 n=300 and p=-=-=0.043

n 300

0 Distinguish between nominal and ordinal variables. Ll Explain why the expected frequency of each class interval should be at least three in the goodness-

of-fit test. bution.

X I ( * * i i ~ _ n ~ q X I M & * l m ~ ~ X U T Ch.mlii--e


Categorical A categorical variable is used to represent a set of categories. Two types of Variable categorical variables are defined depending on the significance of the order of the

category listing: 1. Nominal variable: The order of listing of categories is meaningful.

(e.g.) gender (male and female) hand dominance (left-handed, right-handed, and ambidextrous)

2. Ordinal variable: The order of listing of categories is meaningful. (e.g.) education (< 9 years, 9 - 12 years, and > 12 years)

symptom severity (none, mild, moderate, and severe)

Inference The underlying probabilitv distribution of the population is unknown. Thus, we Context wish to test if a particular distribution fits the population.

(e.g.) Ho: X - Poisson distribution with 3L (discrete distribution)

Ho: X - ~ ( p , 02 ) (continuous distribution)

Test Statistic k X2 = c (oi -'i12 - ~ ~ ( k - ~ - l )

i=l Ei where: k = Number of class intervals (bins)

0, = Observed frequency of class interval i E, = Expected frequency of class interval i p = Number of parameters of the hypothesized distribution that are

estimated by sample statistics

As the observed frequencies are close to the corresponding expected frequencies, the statistic x2 becomes small. Thus, x2 is used to test the null hypothesis Ho: X follows a articular distribution. Table 9-6 can be used to calculate the test statistic d

Table 9-6 Goodness-of-Fit Test Table class 0be;erved Expact4

btewals Frequency Probability Frequency Bi(=npi) Oi-Ei i Qi Pi

(Caution) Minimum expected frequenc Y If an expected frequency is too small, X can be improperly large by a small departure of the corresponding observed frequency from the expected frequency. Although there is no consensus on the minimum value of an expected frequency, the value 3,4, or 5 is widely used as the minimum.

To avoid this undesirable case, when an expected frequency is very small (say, < 3), the corresponding class interval should be combined with an adjacent class interval and the number of class intervals k is reduced by one.

9-7 Testing for Goodness of Fit 9-19

Test Step 1 : State Ho and HI. Procedure Ho: X - A particular distribution.

&'-test) HI : X -/- The hypothesized distribution.


1. Estimate the parameteds) of the hypothesized distribution if their values are not provided.

2. Define class intervals and summarize observed frequencies (Oi's) accordingly.

3. Estimate the probabilities (pi's) of the class intervals.

4. Calculate the expected frequencies (Ei = npi) of the class intervals. If an expected frequency of a class interval is too small (< 3), combine it to an adjacent class interval. Then, repeat steps 2.2 to 2.4.

k (Oi - E , ) ~ 5. Calculate the test statistic X; =

i=l Ei

Step 3: Determine a critical value for a. 2

Xa,k-p-1


Example 9.6

(Caution) Upper-sided critical region Since the test statistic xo2 becomes smaller as the hypothesized distribution fits better, no lower limit is set as a critical value in the goodness-of-fit test.

(Goodness-of-Fit Test; Discrete Distribution) The number of e-mails per hour (3 coming to Ms. Young's e-mail account is assumed to follow a Poisson distribution. The following hourly e-mail arrival data are obtained during n = 100 hours:

2 7 3 5

**-- s-e-D e*--w*w*ws- ---me ~ ~ ~ w B M ~ ~ - - = m M m w - ~ - ~ - - # a m - -~wm"wa- , " -mm~em"

Conduct a goodness-of-fit test at a = 0.05. Step 1: State Ho and HI.

Ho: X - Poisson distribution with h (Note) E(X) = V(X) = 3L Hl: X -/- Poisson distribution

9-20 Chapter 9

Example 9.6 (cont.)



1. Estimate the parameter of the hypothesized distribution. O x 6 0 + l x 2 8 + 2 x 7 + 3 x 5

= E(X) = = 0.57 100

The number of parameters estimated is p = 1.

2. Define class intervals and summarize observed frequencies (0;'s)

3. Estimate the probabilities (pj's) of the class intervals.

4. Calculate the expected frequencies (El = np;) of the class intervals. If an expected frequency is too small (< 3), adjust the class intervals.

Since the expected frequency of the last class interval in the above table is less than three, combine the last two cells as follows:

Observed Expected

# e-mails Frequency Probability Frequency oi - Ei (Oi - E ; )

per hour Oi Fi Ei (= n j i )

3 (0 ; - E ; ) ~ 5. Calculate the test statistic: = = 3.14

i=l Ei

Step 3: Determine a critical value for a. 2 - 2 2

X a , k - p l - X0.05,3-1-1 = X0.05,l = 3.84

Step 4: Make a conclusion.

Since = 3.14 2 Xi,05,, = 3.84, fail to reject Ho at a = 0.05. In

other words, the number of e-mail arrivals per hour follows a Poisson distribution at a = 0.05.

9-7 Testing for Goodness of Fit 9-21

4 Exercise 9.6 1 Consider the following frequency table (n = 100) of a random variable X (MR 9-59) 1 ,y 0 1 2 3 4

I Bins in X Frequency

I

The mean and variance of the scores are 82.2 and 13.3~, respectively. Test if a normal distribution fits the test scores at a = 0.05. - Step 1 : State Ho and H I .

Ho: X - ~ ( 8 2 . 2 J 3 . 3 ~ )

HI: X +- ~ ( 8 2 . 2 J 3 . 3 ~ )

Example 9.7


(Goodness-of-Fit Test; Continuous Distribution) The final scores (X) of n = 40 students in a statistics class are summarized as follows:

1. Estimate the parameter of the hypothesized distribution. Since y and o2 are known, skip this step and the number of parameters estimated is p = 0.

I 2. Define class intervals and summarize observed frequencies (Oi's) accordingly.

Bins in Z Observed Expected

Xi - p Frequency Probability Frequency Bins in X d Oi Pi Ei (= nPi

3. Estimate the probabilities (pi's) of the class intervals. pl = P(X < 60) = P(Z < -1.67) = 0.05

p, =P(6OIX<70)=P(-1.67IZ<-O.92)=0.13

p, = P(70 I X < 80) = P(-0.92 I Z < -0.17) = 0.25

p, = P(80 I X < 90) = P(-0.17 I Z < 0.59) = 0.29

p , = P(90 5 X) = P(0.59 I Z) = 1 - P(Z < 0.59) = 0.28


Example 9.7 (cont.)

4. Calculate the expected frequencies (Ei = npi) of the class intervals. If an expected frequency is too small (< 3), adjust the class intervals.

Since the expected frequency of the first class interval in the previous table is less than three, combine the first two class intervals as follows:

Bins in Z Observed Expected

Xi - Y Frequency Probability Frequency Bins in X (T 'i Pi Ei (= nPi

4 (Oi - E , ) ~ 5. Calculate the test statistic: = = 1.49

i=l Ei . .

Observed Expected Frequency Frequency

Ei (= np, ) Oi - Ei (oi - E~ )x,

Bins in X oi


X a , k - p - l - X0.05,4-0-1 = X0.05,3 = 7.8

Step 4: Make a conclusion. Since = 1.49 2 X i , 0 5 , 3 = 7.81 , fail to reject Ho at a = 0.05.

Exercise 9.7 A random number generator that uniformlv produces numbers (X) between zero and one is tested. The distribution of n = 100 numbers from the generator is summarized as follows:

I Bins in X Frequency

9-8 Contingency Table Tests

Describe a contingency table. Conduct a contingency table test for independencelhomogeneity of categorical variables.

9-8 Contingency Table Tests 9-23

r x c An r x c contingency table (see Table 9-7) refers to the table having r rows (for Contingency the categories o f 4 and c columns (for the categories of Y ) where each cell i j

Table (category i of X and category j of Y) contains the corresponding observed frequency Oii.

Table 9-7 An r x c Contingencv Table

1 011 01 2 ... 0, . . . 01, 2 1 0 2 1 0 2 2 ... 0 2 , ... . $ . 0 2 c

... ... X ; i Oil

Oi2 . . . 0, . . . Oic ... ...

r Or1 Or2 ... 0 . ... Oi-c * ~ a - - ~ ~ s - ~ - w . " . * ~ x ~ m --,a w--,,.,-,,.,waa,..,, ' ,,m,,.m,vw, "-"" "-Bw "*,XRRRR ,ms.w-~ww-~s~-~,,,,,- ' ,,

Inference We wish to test the association between two categorical variables (X and Y ) by Context using an r x c contingency table for independence or homogeneity as follows:

1. Independence: To examine if X and Y are independent, a representative sample is selected from a single population and then each element in the sample is classified into one of r categories in Xand one of c categories in Y.

(e.g.) Classifying a sample of residents in HAPPY County in terms of gender (X) and occupation (Y).

2. Homogeneity: To examine if r populations (X) are homogeneous in terms of Y, representative samples are selected from the r populations and then the elements of each sample are classified into c categories in Y.

(e.g.) Classifying five samples of residents from different counties (A') in terms of occupation (Y).

Recall that P(A I B) = P(A) , P(B I A) = P(B) , and P(A n B) = P(A)P(B) for two independent events A and B (see Section 2-6). Likewise, the relationship of two categorical variables is considered independent/homogeneous if

(1) P ( X = xi I Y) = P ( X = x i ) = pi, ,

(2) P ( Y = y j I X ) = P ( Y = y j ) = p , , , o r

I where: P ( X = x, I Y) and P(Y = y , I X ) are conditional probabilities, t L I P ( X = x, ) and P(Y = y j ) are marginal probabilities, and

P ( X = x, , Y = y , ) is the joint probability of X and Y.

Test Statistic (0, - E , ) ~ x 2 = t z - x 2 ( v ) , v=( r - l ) ( c - l )

j=1 i=l E, where: 0, = observed frequency of cell ij

n . n . j ni.n., E, = expected frequency of cell i j = np, = npi ,p . j = n"- = -

n n n


Test Statistic (Derivation) Determination of degrees of freedom I

(cont.) v = k - p - 1 = r e - [ ( r - l ) + ( c - I)]-1 = r c - r - c + l = ( r - l ) ( c - 1 ) since: k = # cells = rc

p = # marginal probabilities that should be estimated by sample statistics to determine E,'s = (r - 1) + (c - 1)

As the observed frequencies are close to the corresponding expected frequencies, the value o f 2 becomes small. Thus, the statistic x2 is used to test the null hypothesis Ho: X and Yare independent for independence or Ho: X's are homogeneous in terms of Y for homogeneity. Table 9-8 can be used to calculate the test statistic 2 .

Totals 1 n., ... . . .

(Caution) Minimum expected frequency Like the minimum expected frequency for the goodness-of-fit test (see Section 9- 2), if an expected frequency is too small (say, < 3), x2 can be improperly large for a small departure of the observed frequency from the expected one. Thus, any category whose expected frequency is small (< 3) should be combined with an adjacent category.

Test Step 1 : State Ho and HI. Procedure

a2-test) (1) Case 1: Testing for Independence. Ho: X and Yare independent. HI: X and Yare not independent.

(2) Case 2: Testing for Homogeneity. Ho: &'s (r populations) are homogenous in terms of Y. HI: Xi's (r populations) are not homogenous in terms of Y.


(o,-E,)~ 2 ni.n.j - x ((r - 1)(c - 1)) , where EV = - j=1 i=1 E, n


xL-test (cont.)

r Example 9.8

Step 3: Determine a critical value for a. 2

Xa,(r - l ) (c - l )

Step 4: Make a conclusion. Reject Ho if 2 2

X O ' X a , ( r - I ) ( ~ - I )

(Caution) Upper-sided critical region Since the test statistic xo2 becomes small as the null hypothesis of independence1 homogeneity is true, no lower limit is set as a critical value in the independence1 homogeneity test.

(Contingency Table Test; Independence) Grades in ergonomics (X) and grades in statistics (Y) of n = 100 students are summarized as follows:

Statistics Grade (Y) A B Others

Test if grades in ergonomics (X) and grades in statistics (Y) are independent at a = 0.05. - Step 1 : State Ho and HI.

Ho: Grades in ergonomics (X) and grades in statistics (Y) are independent.

H I : Grades in ergonomics (X) and grades in statistics (Y) are not independent.

Step 2: Determine a test statistic and its value. Statistics Grade ( Y )

A B Others Totals

Ergonomics Grade (3

/ 4 8 2 1 i

Others 4 8.6 10.6

"3 i 13.9 !

*.b ~ , d ~ " *~ ,~~,"~,~~, .~ ~a~%.~m.e.#'m, wv,m" w m ~ , ~ " . m ~ m . , ~ . * # A " # ~ ~ ~ m ~ 8 , " ~ # . u . v ~ > " ~ w*~*-.aB,w.w.d,2b*~ a.w.~,d*-.-s- * -~M~.d*easw, a " a ~ ~ - . ~ , ~ . * " ~ a - * ~ . ~ . e . e . ~ , " , m *- n sw" ,m,mwmvm*A.#

Totals 26 3 2 42 !" ' 1'00 ,....m..., "," ,-.-. x*Aw ".*"" v .," ... . .,.,,.-,,-.,-.,Ia ,I,I,",I,",I,I,I"',I,I,I,I.,I",I,I,I,I,I,I,I,I,I,I,I,I,I",I,I,I,I,I,I,I"..,I,I,I,I,I,I,I",I,I ,I *"," ... ", .,.,.., ">" ",1,1",1,1,11" l.llll. "11"/11 ," ", "l". " .,.,,-..

Step 3: Determine a critical value for a. 2 2 2

Xa,(r-l)(c-1) = X0.05,(3-1)(3-1) = X0.05,4 = 9'49


Since = 19.5 > Xi,05,4 = 9.49 , reject HO at a = 0.05. ~t is

concluded that grades in ergonomics (X) and grades in statistics (Y) are not independent at a = 0.05.

9-26 Chapter 9 Tests of Hypotheses for a Single Sample 4 4 :

I Failure Type (Y) A 3 C D

m 1

I Mounting Front 22 46 18 9


I Positionlq Back 4 17 6 12 -ae*-"* ese ~ ~ ~ w ~ m ~ ~ ~ ~ a ~ ~ - s . ~ w ~ v ~ m 8 ~ . ~ ~ A ~ ~ ~ ~ ~ ~ ~ ~ ~ B . ~ ~ w ~ ~ ~ ~ w . ~ - ~ m m ~ - ~ * w . m ~ " m ~ m ~ . s ~ ~ ~ - ~ . ~ . ~ ~ ~ ~ - ~ - ~ m ~ ~ - ~

The failures of an electronic component are under study. There are four types of failures (Y) and two mounting positions (X) at the device. A summary of n = 134 failures are as follows:

Test if the type of failure (Y) is independent of the mounting position (X) at a = 1 0.05.

(Contingency Table Test; Homogeneity) A random sample of n = 300 adults with different hand sizes (2') evaluates two

I mouse designs (Y). The evaluation results are summarized as follows:

I -

Mouse Designs (Y) Conventional New

Small Hand Size Medium 3 5 65

20 80 Groups (X) La%e 3 0 70

a eu -- w-s--a-*- mmm.w a-wa flmvam ----- ----- %** m"-mw--asmmwmmm-.. -- --s--*-aw--"m- -"-

Test if users in different hand-size groups (X) have homogeneous opinions on the mouse designs at a = 0.05.

m- Step 1: State Ho and HI. Ho: Users in different hand-size groups are homogeneous in terms

of opinions on the mouse designs. HI: Users in different hand-size groups are not homogeneous in

terms of opinions on the mouse designs.

Step 2: Determine a test statistic and its value. Mouse Designs (Y)

Conventional j Totals New

Hand Size Medium Groups (X)

Large 30 70

B 28.3 71.7 1 100 .w.vm...-a~w-~ws..~.*a-~ ~ e # w w ~ ~ - ~ ~ m - * e e - - ~ " b*m- Iggggggggggggggg w,smsm.-.*a*wB, gggg.gggggggggg.".gg" ," ,.,,-,,,,-*,, ., ..,i ,, ,%*,m-.,**

Totals 1 85 2 15 1 300 " ,.,,,.,, ,,a--,,,,.,,,,-,B e wmm,e-mm--a mm- .,,,, a, ,-,,-,,,, w. 8A8*wfl ~ ~ w*d,a*s*%m--,7 ".~" B*B.s,w, ..,,,.,," ,.,,, . ,,.*..,--,,,,.,.w.a,.", ,...


Xa,(r-l)(c-1) = X0.05,(3-1)(2-1) = X0.05,2 = 5.99

Step 4: Make a conclusion. Since Xi = 5.7 > ~ i , ~ ~ , ~ = 5.99 , fail to reject Ho at a = 0.05. It is

concluded that users in different hand-size groups do not have significantly different opinions on the mouse designs at a = 0.05.



A random sample of n = 630 students in different class standings (X) is asked their opinions (Y) on a proposed change in core curriculum. The survey results are as follows:

Class Sophomore 70 130 Standing (X ) Junior 60 70

Senior 40 60 e--s-"ms--- -mm-.-m'a-----~m'm wwmswm-- ---me--=-m * - ----- sw -w*e * a mw -**+ a"B B a wa

Test if students in different class standings (X) have homogeneous opinions on the curriculum change at a = 0.05.



Example 9.1 (Power

(1)

- of Test)

Choose Stat ) Power and Sample Size > 1-Sample Z. Click Calculate power for each sample size. Enter the sample size selected in Sample sizes, the difference between the true mean and hypothesized mean to be detected in Difference, and the population standard deviation in Sigma. Then click O~tions.

Jnder Alternative Hypothesis select the type of alternative hypothesis, and in Significance level enter the probability of type I error (a).Then click OK twice.

I I Testing mean = null (versus n o t = null) Calculating power f o r mean = null + 20 Alpha = 0.05 Sigma = 40

Sample , ..................

30; 0.7819 i ....................

MINITAB Applications 9-29

Example 9.2 (Inference on p, c2 Known) I (1) Choose File % New, click Minitab Project, and click OK.

(2) ITY light bulbs on the worksheet.

r---A--------m - r INFINITY 1

(3) Choose Stat % Basic Statistics % 1-Sample Z. Click Test mean, enter the hypothesized mean life length, and select not equal (type of the alternative hypothesis) in Alternative. Enter the assumed population

2-Test

Test of mu = 765.00 vs mu not = 765.00 The assumed sigma = 40.0

................................... Varlable N Mean StDev SE Mean i Z P /- INFINITY 30 780.00 40.02 7.30 2.05 0.040 i -5. , .................. l.II ...........

"y"

,5, .ii.. 3:. ::

9.2.1,2

-30 Chapter 9 Tests of Hypotheses for a Single Sample

Example 9.2 (5) Choose Stat > Power and Sample Size % 1-Sample Z. Click Calculate (cont.) sample size for each power value. Enter the power of the test predefined

in Power values, the difference between the true mean and hypothesized mean to be detected in Difference, and the population standard deviation

(6) Under Alternative Hypothesis select the type of alternative hypothesis, and in Significance level enter the probability of type I error (a).Then click OK twice.

1 (7) Obtain the s a m ~ l e size reauired to satisfv the medefined test condition.

I Power and Sample Size ?I I I 1-Sample Z Test I

Testing mean = null (versus not = null) Calculating power for mean = null + 20 Alpha = 0.05 Sigma = 40 .................. isample i Target Actual

i Slze -z ?p i 43 i 0.9000 0.9064

--

................. .. Y


Example 9.3 (Inference on p, cr2 Unknown)

( I ) Choose File * New, click Minitab Project, and click OK.

(2) Enter the life length data of INFINITY light bulbs on the worksheet.

2hoose Stat * Basic Statistics * 1-Sample t. Click Test mean, ente hypothesized mean life length, and select not equal in Alternative. click OK.

:r the Then

T-Test of the Mean

I T e s t of mu = 765.00 vs mu not = 765.00 I ................................. Variable N Mean StDev SE Mean i T INFINITY 30 780.00 40.02 7.31 I2.05 0.049 9.3.1 , ................................

lu: Qs ~d l a

(5) Choose Stat * Power and Sample Size k 1-Sample t. Click Calculate sample size for each power value. Enter the power of the test predefined in Power values, the difference between the true mean and hypothesized mean to be detected in Difference, and the sample standard deviation in


Example 9.3 (cont.)

Example 9.5

(6) Under Alternative Hypothesis select the type of alternative hypothesis, and in Significance level enter the probability of type I error (a).Then

Cancel

(7) Ohtain the sample size reauired to satisfv the redefined test condition.

I Power and Sample Size *I I l-Sample t T e s t I

T e s t i n g mean = n u l l ( v e r s u s n o t = n u l l ) C a l c u l a t i n g power f o r mean = n u l l + 20 Alpha = 0.05 Sigma = 40 .................. I.

!Sample i T a r g e t A c t u a l i Slze rower ruwm 9.3.2 i 44 i 0.9000 0.9000 : .................

Y

(Inference on p)

(1) Choose Stat * Basic Statistics ) 1 Proportion. Click Summarized data and enter the number of bridges surveyed in Number of trials and the number of bridges corroded in Number of successes. Then click

Example 9.5 (cont.)


(2) Enter the level of confidence in Confidence level, the hypothesized proportion in Test proportion, and not equal in Alternative. Check Use

twice.

I Test and Confidence Interval for One Proportion I Tesc of p = 0.5 vs p not = 0.5 :.. ............................ S a m p l e X N S a m p l e p I 95.0 % C I j 2 - V a l u e P - V a l u e

2 8 40 0.700000 (0.557987, 0.842013); 2.53 0.011 9.5.1 ................................ --

(4) Choose Stat > Power and Sample Size > 1 Proportion. Click Calculate sample size for each power value. Enter the power of the test predefined in Power values, the true proportion to be detected in Alternative p, and

9-34 Chapter 9

Example 9.5 (cont.)

Example 9.8



(61 Ohtain the samnle size reauired to satisfv the medefined test condition.

I Power and Sample Size 1 I Test f o r One Proport ion I

Test ing proportion = 0 .5 (versus not = 0.5) Ca lcu la t ing power f o r proport ion = 0.7 Alpha = 0.05 Difference = 0.2 ................. Sample i Target Actual

Size ruwrr ruwrr 9.5.2 62 i 0.9000 0.9029 -

(Goodness-of-Fit Test; Independence)

(1) Choose File + New, click Minitab Project, and click OK.

CI-T j ~2 - --J-

I A Others 3

Others r--T--d---- 21 M I

Example 9.8 (cont.)

Example 9.9

-


(3) Choose Stat , Tables , Chi-Square Test. In Columns containing the table, select A, B, and Others which contain the frequencies. Then click OK.

Chi-square Test 1

I Expected counts are p n n t e d be lor observed counts

A B Others Total I 12 5 4 2 1 5.46 6.72 8.82

Total 2 6 3 2 42 100 ................................................................................................... !Chi-Sq = 7.834 + 0.440 + 2.634 +

0 .321 + 1.244 + 0.279 + 2.445 + 0 . 6 2 1 + 3.678 = 19.496

~ D F = 4 , P-Value = 0.0011 i

I .................................................................................................. *!

(Goodness-of-Fit Test; Homogeneity)

(1) Choose File > New, click Minitab Project, and click OK.

' (2) 1 on the mouse designs.

9-36 Chapter 9

Example 9.9 (cont.)


(3) Choose Stat % Tables * Chi-Square Test. In Columns containing the table, select Conventional and New which contain the frequencies. Then click OK.

I 1 Chi-Square Test

I I Expected counts a r e p r i n t e d below observed counts 1 Conventi New Tota l

1 35 65 100 28.33 71.67

Total 85 2 15 300 ....................................................... !

IChl-~q = 1.569 + 0.620 + 2.451 + 0.969 + 0.098 + 0.039 = 5.746

~ D F = 2 , P-Value = 0.057 - ......................................................... v: $t I B ~ , ~ $9

31



Exercise 9.1 1. (AcceptanceICritical Regions)

The test statistic of p is the sample mean with the following sampling distribution:

The acceptance region I I 2 I u for Ho: px = 100 vs. HI : px # 100 satisfies the following:

1 - a = 1 - 0.01 = 0.99 = P(fai1 to reject Ho I p = 100) = ~ ( 1 1 X 1 u 1 p = 1 0 0 )

Accordingly, the critical values are

Thus, acceptance region: 98.3 1 F 1 1 01.7 rejection region: F < 98.3 and F > 101.7

2. (p and Power of Test)

a = P(fai1 to reject HO 1 HO is false) = P(98.3 I 2 1101.7 1 p = 103)

power = 1 - = 0.973

9-38 Chapter 9

Exercise 9.2


1. (Hypothesis Test on y, o2 Known; Upper-Sided Test)

Step 1 : State Ho and H I . Ho: p = 1.50 HI: p > 1.50

Step 2: Determine a test statistic and its value. X-po - 1.5045-1.50

zo =-- = 1.42 CJ/& o .o1 /4G

Step 3: Determine a critical value(s) for a. z, = z,,,, = 2.33

Step 4: Make a conclusion. Since z, = 1.42 2 z,,,, = 2.33, fail to reject Ho at a = 0.01.

2. (P-value Approach)

P = 1 - Q(z,) = 1 - Q(1.42) = 1 - 0.922 = 0.078

Since P = 0.078 5 a = 0.01, fail to reject Ho at a = 0.01.

3. (Sample Size Determination)

(1) Sample Size Formula power = P(reject Ho I Ho is false) = 1 - p = 0.9 P = 0.1 6 = p - po = 1.50 - 1.505 = 0.005

(z, + ~ ~ ) ~ o ~ ( z ~ . ~ ~ + z ~ , ~ ) ~ o . o ~ ~ (2.33 + 1.28)~ ~ 0 . 0 1 ~ n = - - - - E 53

6 0.005~ 0.005~ (2) OC Curve

To design a one-sided z-test at a = 0.01 for a single sample, OC Chart VId is applicable with the parameter

d = I P - P O I - 161 - 10.0051 - - - ------ = 0.5 CJ 0 0.01

By using d = 0.5 and P = 0.1, the required sample size is determined as n = 55 as displayed below. Note that this sample size is similar with the sample size n = 53 determined by using a sample size formula.

d

(4 0 C. c w for &fiereat values of n for the one-sided n d test for a lml of s ~ l p u a = 0 01


Exercise 9.3 1. (Hypothesis Test on p, o2 Unknown; Upper-Sided Test) Step 1 : State Ho and HI.

Ho: y = 1.0 H,: p > 1 . 0

Step 2: Determine a test statistic and its value. x-Po - to =-- 1.25 -1.0

= 4.47 s / & 0 . 2 5 / f i


tct,n-l - t0.05,20-~ = t0.05,19 = .73

Step 4: Make a conclusion. Since to = 4.47 > to., ,,,, = 1.73, reject Ho at a = 0.05. We conclude

that the PVC pipe product meets the ASTM standard at a = 0.05.


power = P(reject Ho I Ho is false) = 1 - p = 0.8 a P = 0.2 6 ~ p - p ~ =1.10-l.O=O.lO

I To design a one-sided t-test at a = 0.05 for a single sample, OC Chart VIg is applicable with the parameter

I By using d = 0.4 and P = 0.2, the required sample size is determined as n =



Exercise 9.4 1. (Hypothesis Test on 02; Upper-Sided Test) Step 1 : State Ho and H I .

Ho: o2 =0.012

HI: 02>0 .012


Step 3: Determine a critical value@) for a. 2 2 2

Xa ,n- I = X0.0l,l5-1 = X0.01,14 = 29.14

Step 4: Make a conclusion. Since = 8.96 2 0,,14 = 29.14 , fail to reject Ho at a = 0.0 1.

There is no significant evidence which indicates that the standard deviation of the hole diameter is greater than 0.01 mm at a = 0.01.


power = P(reject Ho 1 HO is false) = 1 - j3 = 0.9 p = 0.1

To design an upper-sided X2-test at a = 0.01, OC Chart VII is applicable with the parameter

0 h=-=1.5

By using h = 1.5 and p = 0.1, the required sample size is determined as n =


(1) 0.C curves for different values of n for the one-sided (upper tail) chi-square test for a level of significance a = 0.01.

Exercise 9.5

Exercise 9.6


1. (Hypothesis Test onp; Lower-Sided Test)

Step 1 : State Ho and HI. Ha: p = 0.05 H I : p < 0.05


Step 3: Determine a critical value(s) for a. z , = zo,05 = 1.64

Step 4: Make a conclusion. Since z0 = -0.53 4: -z,,, = -1.64, fail to reject Ho at a = 0.05.


power = P(reject Ho I Ho is false) = 1 - p = 0.8 3 P = 0.2

(Goodness-of-Fit Test; Discrete Distribution)

Step 1 : State Ho and HI. Ho: X - Poisson distribution with h = 1.2 HI : X -/- Poisson distribution


1. Estimate the parameter of the hypothesized distribution. Since the value of h is given, skip this step. The number of parameters estimated is p = 0.

2. Define class intervals and summarize observed frequencies (Oi's) accordingly.



4 or more 4 0.03 3 1 0.3 - -*---- B - M - * m m -e*-m----- ------ -------*.aa*--.% -m, "**-=-

3. Estimate the probabilities (pi's) of the class intervals.


All the expected frequencies are three or above.

5 (Oi - E , ) ~ 5. Calculate the test statistic: X: = = 6.6

i=l Ei

I Step 3: Determine a critical value for a.

Exercise 9.7



Since = 6.6 2 Xi.05,4 = 9.49 , fail to reject Ho at a = 0.05.

(Goodness-of-Fit Test; Continuous Distribution)

Step 1 : State HO and HI. Ho: X - Uniform distribution f (x) = 1 , 0 I x I 1 HI: X + Uniform distribution

1. Estimate the parameter of the hypothesized distribution. Since the uniform distribution does not have any parameter, skip this step. The number of parameters estimated is p = 0.

I 2. Define class intervals and summarize observed frequencies (Oi's) accordingly.



3. Estimate the probabilities (pi's) of the class intervals. .2

p , = P(X < 0.2) = fidx = XI: = 0.2

Since the sizes of the bins are the same, the corresponding probabilities are equal to each other as 0.2.


None of the expected frequencies are less than three.

(0 , - E , ) ~ 5. Calculate the test statistic: X t = =1.70

i=l Ei


Xc~,k-~-l - X0.05,5-0-1 = X0.05,4 = 9.49


Since = 1.49 P = 7.81 , fail to reject Ho at a = 0.05. The

data does not indicate any abnormal functioning of the random number generator at a = 0.05.

Step 2: Determine a test statistic and its value: Failure Type (Y) i

P

A B C

Exercise 9.8

I Totals 1 26 63 24 21 L 134 *--- ----m-mx-w- ~ - ~ m ~ ~ - ~ - - - - m - ~ ~ - - ~ - m - a . w ~ m - - ~ m s ~ " ~ -,---

(Contingency Table Test; Independence) Step 1: State Ho and HI.

Ho: The type of failure is independent of the mounting position. HI: The type of failure is not independent of the mounting position.



Exercise 9.9


Xa,(r-l)(c-1) = XO.O5,(2-1)(4-1) = X0.05,3 = 7.8

Step 4: Make a conclusion. Since = 1 0 . 8 > ~ i , , ~ , ~ = 7.81, reject Ho at a = 0.05.

(Contingency Table Test; Homogeneity) Step 1: State Ho and HI.

Ho: Students in different class standings are homogeneous in terms of opinions on the curriculum change.

H I : Students in different class standings are not homogeneous in terms of opinions on the curriculum change.

I S t e ~ 2: Determine a test statistic and its value.

I Opinion (Y) Favoring opposing Totals

Totals


Xa,(r-l)(c-1) - X0.05,(4-1)(2-1) = X0.05,4 = 9.49

Step 4: Make a conclusion. Since = 27.0 > X i .,,, = 9.49, reject Ho at a = 0.05. It is

concluded that students in different class standings have significantly different opinions on the curriculum change at a = 0.05.

10 Statistical Inference for Two Samples

10-2 Inference for a Difference in Means of 10-4 Paired t-Test

Two Normal Distributions, Variances 10-5 Inference on the Variances of Two

Known Normal Populations

10-3 Inference for a Difference in Means of 10-6 Inference on Two Population

Two Normal Distributions, Variances Proportions

Unknown MINITAB Applications I c Answers to Exercises

10-2 Inference for a Difference in Means of Two Normal Distributions, Variances Known

Cl Test a hypothesis on p~ - p2 when o12 and 022 are known (z-test). Cl Determine the sample size of a z-test for statistical inference on pl - pz by using an appropriate

sample size formula and operating characteristic (OC) curve. O Establish a 100(1 - a)% confidence interval (CI) on p1 - p2 when o12 and 022 are known. 0 Determine the sample size of a z-test to satisfy a preselected level of error (I$) in estimating pl - p2.

Inference Parameter of interest: p~ - p2 Context 0: o;

Point estimator of pl - p2: X1 - X2 - N(pl - p 2 , - + -) "'1 n2

0 0; 2 2 (Note) Xl - N(pl ,') and X2 - N(p2, -) ,o1 and 0 2 known;

n1 n2 XI and X2 are independent

Test statistic of pl - p2: Z = ('1 - '2 ) - (PI - 112 ) _ N(O,l) I -

Isal p a p y s - l a ~ o l ~ o j "z- > Oz

1sal paprs-laddn ~ o j "z < OZ

1sao pap!s-OMI lroj ""2 < lozJ J! OH 13afax .uo!snpuo3 e ayeK :p days

L = 0 2

Og-('x- - '_XI 'anIan s)! pua q+s!+a+s )sa+ e au!uualaa :Z dais

z T i - 'ri=(Zx)y-('x)g=(Zx- - - - Ix)y -

a ~ u e ! . ~ e ~ pue ueaur q y ! ~ ~euuou S! z~ - I X - -

'Lia~!l~adsal ' 1 E D = ( Z ~ ) ~ pue ' [U 1 ;D = ( 'x)/i ' ' d = ( Z ~ ) g ' [rl = (Ix)g - - - - sa3uepeA put? sueaur y o ! ~ leuuou PUB ouapuadapu! ale Z~ pue I X a3u!S - -

.ouapuadapu! ale zx pup I X - -

10-2 Inference for a Difference in Means of Two Normal Distributions, Variances Known 10-3

Sample Size Formula

(cont.)


(OC) Curve

( z a + ~ ~ ) ~ ( o : + o i ) n = for one-sided test

(A - A d L

where: A = p, - p2 (true mean difference),

A + A, (hypothesized mean difference), and

Table 10-1 displays a list of OC charts in MR and a formula of the OC parameter d for a z-test on p1 - p2. By using the table, the appropriate OC chart for a particular z-test is chosen (e.g., for a one-sided z-test at a = 0.05, chart VIc is selected).

Table 10-1 Operating Characteristic Charts for z-test - Two Samples

Test a Chart VI OC p m e t e r - - G ' e A in %)

Confidence Interval

Formula

A 100(1 - a)% CI on p, - p2 when o12 and ~2~ are known is

- - 5 p, -)I, I X, - X, + z,,, for two-sided CI

I p l - p2 for lower-confidence bound

pl - p2 I Xl - X2 + Z, - + A for upper-confidence bound 1: :: (Derivation) Two-sided confidence interval* on p1 - p2, o12 and 022 known

By using the test statistic Z =

P ( L I ~ I U ) = I - ~

3 P(- z,,, I z I z,,,)= 1 - a

Therefore,

- - - - L=X,-X, -z , , , and U = X l - X 2 + z a 1 2

10-4 Chapter 10 Statistical Inference for Two Samples

Sample Size As an extension of the sample size formula for estimation on p with a preselected Formula level of error (a (described in Section 8-I), the following formula is used for

for estimation on pl - p2: Predefined 2

Error n = (y) (0; + 0; ) . where a , = n2 = n

Example 10.1 The life lengths of INFINITY (XI; unit: hour) and FOREVER (X2; unit: hour) light bulbs are under study. Suppose that XI and X2 are normally distributed with oI2 =

402 and 022 = 302, respectively. A random sample of INFINITY light bulbs is presented in Example 8.1 and a random sample of FOREVER light bulbs is shown below:

No Life Length No Life Length No Life Length 7 89 I I 755 2 1 837

7 769 17 787 8 836 18 788 9 847 19 794

The two random samples are summarized as follows:

I Brand of Sample Size Sample Mean Variance Light Bulb

- nl = 30 x, = 780 hrs o12 = 402

mean life length of an INFINITY light bulb is different from that of a FOREVER light bulb at a = 0.05.

h Step 1: State Ho and HI. Ho: -p2 = 0 HI: p1-p2+0

Step 2: Determine a test statistic and its value. (X, - X2) - 6O (780 - 800) - 0

z,, = -

-+- + --

I Step 3: Determine a critical value(s) for a. za12 = z0.0512 = z0.025 =

Step 4: Make a conclusion. Since lz,, 1 = 2.12 > zo,02j = 1.96, reject Ho at a = 0.05.

10-2 Inference for a Difference in Means of Two Normal Distributions, Variances Known 10-5


2. (Sample Size Determination for Predefined Power of Test) Determine the sample size n (= nl = n2) required for this two-sided z-test to detect the true difference in mean life length as high as 20 hours with 0.8 of power. Apply an appropriate sample size formula and OC curve.

w (1) Sample Size Formula power = P(reject HO I Ho is false) = 1 - p = 0.8 3 P = 0.2

(2) OC Curve To design a two-sided z-test at a = 0.05 for two samples, OC Chart Vla is applicable with the parameter

A - 120-0) d=dT=dy=0.4

o + o 402 + 30

By using d = 0.4 and P = 0.2, the required sample size is determined as n = 50 as displayed below.

3. (Contidence Interval; Two-Sided CI) Construct a 95% two-sided confidence interval on the mean difference in life length (pl - p2). Based on this 95% two- sided CI on p~ - p2, test Ho: p1- p2 = 0 VS. HI: p1- p2 # 0 at a = 0.05.

h 1 -a=0.95 3 a=0.05 95% two-sided CI on pl - p2:




Since this 95% two-sided CI on yl - y2 does not include the hypothesized value zero, reject Ho at a = 0.05.

4. (Sample Size Determination for Predefined Error) Find the sample size n (= nl = n2) to construct a two-sided confidence interval on yl - y2 within 20 hours of error at a = 0.05.

Two types of plastic (say, plastic 1 and plastic 2) are used for an electronics component. It is known that the breaking strengths (unit: psi) of plastic 1 (XI) and plastic 2 (X2) are normal with 0,' = 0 2 ~ = 1 .O. From two random samples of n, =

10 and n2 = 12, we obtain = 162.5 and = 155.0.

1. The company will not adopt plastic 1 unless its breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample test results, should they use plastic l? Use a = 0.05 in reaching a decision.

2. Determine the sample size n (= nl = n2) required for this one-sided z-test to detect the true difference in mean breaking strength as high as 11.5 psi with 0.9 of power. Apply an appropriate sample size formula and OC curve.

3. Construct a 95% upper-sided confidence interval on the mean difference in breaking strength (y1- y2).

10-3 Inference for a Difference in Means of Two Normal Distributions, Variances Unknown

O Test a hypothesis on yl - y2 when o12 and 0 2 ~ are unknown (t-test). 0 Determine the sample size of a z-test for statistical inference on yl - y2 by using an appropriate

operating characteristic (OC) curve. R Establish a 100(1 - a)% confidence interval (CI) on yl - y2 when 0 1 2 and 0; are unknown.

*---*-.#.aw--m -*----- w M m w m - - - m - u m m ~ - ~ s m - m ~ ~ m - - ~ ~ - ~ ~ * ~ - wa**m m-ms-s .s-s-dm -"* B--% ---- Inference Parameter of interest: P I - y2

Context 0: 0;

Point estimator of y1 - y2: X1 - X2 - N(yl - y 2 ,- + -) "'1 n2

0: 0 (Note) XI - N(pl ,-) and X2 - N(y2 ,A), o12 and oZ2 unknown; n1 n2

and X2 are independent


Inference Context

(cont.)

Test Procedure

(t-test)

Test statistic of pl - p2: Different test statistics of p1 - ~2 are used depending on the equality of o12 and 022 as follows:

(1) Case 1: Equal variances (oI2 = 022 = 02)

, (n, - 1)s: + (n2 - 1)s; where: S, = (pooled estimator of 02) n, + n 2 - 2

(2) Case 2: Unequal variances (oI2 # 022)

(Note) The equality of two variances (o12 = 022) can be checked by using an F test described in Section 10-5.

Step 1: State Ho and HI. Ho: PI - P2 = 60

HI: PI - p2 # 60 for two-sided test PI - p2 > 60 or pl - p2 < 60 for one-sided test


(1) Case 1: Equal Variances (oI2 = 022 = 02)

where: S, = n, + n2 - 2

(2) Case 2: Unequal Variances (oI2 # 022)

Step 3: Determine a critical value(s) for a. t,,,, for two-sided test; t , , for one-sided test


It,, 1 > t 2 for two-sided test

to > t , , for upper-sided test

to < - t , , for lower-sided test


Operating Table 10-2 displays a list of OC charts in MR and a formula of the OC parameter 2 2 2 Characteristic d for a t-test on p1 - p2 where ol = o2 = o and nl = n2 = n. Note that OC curves

(OC) are unavailable for a t-test when oI2 # 0 z 2 because the corresponding t fhrve distribution is unknown. By using the table, the appropriate OC chart for a

particular t-test is chosen (e.g., for a two-sided t-test at a = 0.01, chart VIfis selected). The sample size n* obtained from an OC curve is used to determine the sample size n (= nl = n2) as follows:

n * + l n=- , where n* from an OC curve

2

Table 10-2 Operating Characteristic Charts for t-test - Two Samples

Test a Chart VI (Appendix A in MR)

OC parameter

Two-sided

t-test

One-sided

subjective estimate.

Confidence A 100(1 - a)% CI on p1 - p2 when oI2 and 022 are unknown depends on the Interval equality of o12 and 022 as follows:

l70rmula (1) Case 1: Equal variances (oI2 = 022 = 02) I - -

I p1 - p2 I XI - X2 + ta, 2,vSp for two-sided CI

1 3, - 1, - t 1 2 , v S + - 5 pl - p2 for lower-confidence bound

'"1 n2 - -

pl -& I Xl -X2 +ta,2,vSp for upper-confidence bound

(2) Case 2: Unequal variances (o12 # 0 2 ~ )

- - I p, -p2 I XI - X , +ta12,, for two-sided CI

5 p, - p2 for lower-confidence bound

- - pl -& I Xl - X2 +ta,2,v for upper-confidence bound

CI and If the sample sizes are large (nl and n2 L 30), the z-based C1 formulas and test Hypothesis procedure in Section 10-2 can be applied to inference on pl - p2 regardless of

Test for whether the underlying populations are normal or non-normal according to the Large Sample central limit theorem (described in Section 7-5).


I For the light bulb life length data in Example 10.1, the following results have been obtained:

I Brand of Light Bulb Sample Size Sample Mean Sample Variance

- I INFINITY (1) nl = 30 x, = 780 hrs S: = 40' - I FOREVER (2) n2 = 25 x2 = 800 hrs

Sided Test) Assuming 01' # 0z2 , test if the mean life length of an INFINITY light bulb is different from that of a FOREVER light bulb at a = 0.05.

Step 1: State Ho and HI. Ho: ~1 - ~2 = 0 HI : PI - P2+ 0

Step 2: Determine a test statistic and its value. (XI - X2) - 60 (780 - 800) - 0

to = -

-+- + --

(s t In , +s i In,), v = - 2 =

(40, I 30 + 302 1 2512 - 2 ~ 5 4

(s: In , ( ~ i l n d 2 (40, 1 3 0 ) ~ L (302 1251,

Step 3: Determine a critical value(s) for a.

t a ~ ~ , ~ = t0.025,54 = 2.00

Step 4: Make a conclusion. Since Ito/ = 2.12 > to,o,5,5, = 2.00, reject Ho at a = 0.05.

2. (Sample Size Determination) Assuming 0 1 , = 0z2 , determine the sample size n (= nl = n2) required for this two-sided t-test to detect the true difference in mean life length as high as 20 hours with 0.8 of power. Apply an appropriate OC curve.

I h To design a two-sided t-test at a = 0.05 for two samples, OC Chart VIe is applicable with the parameter

I (n, - 1)s; + (n, - 1)s; (Note) s: =

n, +n2 -2

10-1 0 Chapter 10 Statistical Inference for Two Samples


For d = 0.28 and P = 0.2, n* = 100 as displayed below. 1 .o j - - - - - - - r - - - - - - 7 - - ~ - - - - - 3

(e) O.C. curves for different v d w of n for the two-sided t-test for a level of significance a = 0.05.

Thus, the required sample size n is

3. (Confidence Interval on p1- p2, o12 and o; Unknown but Equal; Two- Sided CI) Assuming 0 1 2 = 0z2 , construct a 95% two-sided confidence interval on the difference in mean life length (pl - p2).

95% two-sided CI on p1 - p2:

(Note) This t-based CI is wider than the corresponding z-based CI -38.5 I p l -p , 5 - 1 5 in Example 10-1.

4. (Confidence Interval on pl - p2, o12 and o; Unknown and Unequal; Two- Sided CI) Assuming 0 1 2 # 0 2 2 , construct a 95% two-sided confidence interval on the difference in mean life length (pl - p2). Based on this 95% two-sided C I o n p ~ - p 2 , t e ~ t H ~ : p ~ - p ~ = O ~ ~ . H ~ : p ~ - p ~ # O a t a = 0 . 0 5

10-4 Paired t-Test 10-1 1

I Since this 95% two-sided CI on p~ - p2 when 0 1 2 and 02' are unknown and unequal does not include the hypothesized value zero, reject Ho at a = 0.05.

I while another random sample of n2 = 16 gears from supplier 2 results in Z2 =

32 1 and s2 = 22. Assume that XI and X2 are normally distributed.


I 1. Assuming o12 = 022, test if supplier 2 provides gears with higher mean strength at a = 0.05.

Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measured in foot-pounds is an important characteristic. A random sample of n, = 10 gears from supplier 1 results in Fl = 290 and sl = 12,

2. Assuming oI2 = oZ2, determine the sample size n (= nl = n2) required for this one-sided t-test to detect the true mean difference as high as 25 foot-pounds with 0.9 of power. Apply an appropriate OC curve.

3. Assuming oI2 = 022, construct a 95% upper-confidence bound on pl - p2.

4. Assuming oI2 # 022, construct a 95% upper-confidence bound on - p2.

10-4 Paired t-Test

O Explain a paired experiment and its purpose. O Test a hypothesis on AD for paired observations when 0 D 2 is unknown (paired t-test). O Establish a 100(1 - a)% confidence interval (CI) on p~ for paired observations when 0~~ is

unknown.

Paired A paired experiment collects a pair of observations (XI and X2) for each specimen Experiment (experimental unit) and analyzes their differences (instead of the original data).

This paired experiment is used when heterogeneity exists between specimens and this heterogeneity can significantly affect XI and X2; in other words, XI and X2 are not independent.

For instance, in Table 10-3, the effect of a diet program on the change in weight is under study with n participants, who are heterogeneous. This heterogeneity is likely to confound the effect of the diet program on the weight change; in other words, the weight before (XI) and the weight after (X2) the diet program are not independent. To block the effect of the heterogeneity on the weight change, it is proper to analyze the differences (D) of the paired observations.


Paired Experiment

(cont.)

Inference Context

Test Procedure

(paired t-test)

Confidence Interval

Formula

CI and Hypothesis

Test for Large Sample

Parameter of interest: ~ 1 0 -

0; Point estimator of pD: D = X I - X , - N(pD, T), oD2 unknown;

X I and X , are independent. - D - P D Test statistic of PD: T = ------ - t ( v ) , v = n - 1 s, I&

Step 1: State Ho and HI. Hot pD = 60 H I : p~ ;t GO for two-sided test,

p~ > 60 or p~ < 60 for one-sided test.


Step 3: Determine a critical value(s) for a. t,,,,,-, for two-sided test; t , for one-sided test


t o 1 > t , 2 , n 1 for two-sided test

to > t,,,-, for upper-sided test

to < -t,,,-, for lower-sided test

S D -

D - t,, , , - < p D 5 D + t,, , , for two-sided CI &- &

for lower-confidence bound

SD PD 5 D + t , , - for upper-confidence bound

&

If the sample size is large (n 2 30), the z-based CI formulas and test procedure in Section 9-2 can be applied to inference on pD according to the central limit theorem.

10-4 Paired t-Test 10-1 3

r Example 10.3 The weights (unit: lbs) before and after a diet program for 30 participants are measured below.

7 198 182 17 184 176 27 215 206 8 165 160 18 173 156 28 206 20 1 9 180 178 19 179 167 29 165 156 10 172 171 20 168 152 30 170 154

w - w w A m - 9 - a w w w - w e A - The summary of the weight data is as follows:

Sample Size Sample Mean (no. participants) (weight loss) Sample Variance

n = 30 d = 10 lbs s i = 52

1. (Hypothesis Test on PD, OD' Unknown; Two-Sided CI) Test if there is a significant effect of the diet program on weight loss. Use a = 0.05.

h Step 1 : State Ho and HI. Ho: pD=o HI: b # O


Step 3: Determine a critical value(s) for a. t a /2 ,n- l = t0.05/2,30-1 = t0.025,29 = 2.045

Step 4: Make a conclusion. Since It, 1 = 10.95 > t0,025,24 = 2.045 , reject Ho at a = 0.05.

2. (Confidence Interval on b, OD' Unknown; Two-Sided CI) Construct a 95% two-sided confidence interval on the mean weight loss (pD) due to the diet program. Based on this 95% two-sided CI on PD, test Ho: p . ~ = 0 VS. HI: PD z 0 at a = 0.05.

rn 1 - a = 0 . 9 5 3 a=0.05; v = n - 1 = 3 0 - 1 =29 95% two-sided CI on

10-1 4 Chapter 1



Statistical Inference for Two Samples

Since this 95% two-sided CI on does not include the hypothesized value zero, reject Ha at a = 0.05.

A computer scientist is investigating the usefulness of two different design languages in improving programming tasks. Twelve expert programmers, familiar with both languages, are asked to code a standard function in both languages, and their coding times in minutes are recorded as follows:

7 16 10 6 8 14 13 1 9 2 1 19 2

11 13 15 -2 12 18 20 -2

*- d d - w - w w . - --m--w--- smm-s---- * - e e - - m s m-ma-mm - - *-=--- =--- -ms- -- w-mm"

The summary quantities of the coding time data are

n = 12, d = 0.67, and si = 2.962

1. Does the data suggest that there is no difference in mean coding time for the two languages? Use a = 0.05 in drawing a conclusion. 1

2. Construct a 95% two-sided confidence interval on the difference between the two languages in coding time (pD). Based on this 95% CI on b, test Ha: =

0 vs. H I : pD # 0 at a = 0.05.

10-5 Inference on the Variances of Two Normal Populations

Explain the characteristics of an F distribution. R Read the F table.

Test a hypothesis on the ratio of two variances (012/022) (F-test). O Determine the sample size of an F-test for statistical inference on 0 1 ~ 1 0 2 ~ by using an appropriate

operating characteristic (OC) curve. Establish a 100(1 - a)% confidence interval (CI) on 012/022.

10-5 Inference on the Variances of Two Normal Populations 10-1 5

F Distribution The probability density function of an F distribution with u and v degrees of freedom is

The mean and variance of the F distribution are E ( X ) = v l(v - 2) for v > 2

2v2(u + v - 2) V ( X ) = for v > 4

u(v - 2),(v- 4)

An F distribution is unimodal and skewed to the right (see Figure 10-4 in MR) like a X 2 distribution. However, an F distribution is more flexible in shape by having an additional parameter of degrees of freedom.

F Table The F table (see Appendix Table V in MR) provides 1 OOa upper-tail percentage points ( fa, , , , ) of F distributions with various values of u and v, i.e.,

P(Fu,v > fa,,,,

The lOOa lower-tail percentage points ( fl-a,,,v) can be found as follows:

1 fi-a,,,, =-

fa,,,,

(e.g.) Reading the F table

P(F7,15 > f ) = 0.05 : f = f0,05,7,15 = 2.71 (2) P(F7,,, > f)=0.95 : f = fO,,,,,,,, = 1 / f0,,,15,7 =l/3.51 = 0.28

Inference 0: Context Parameter of interest: - 0:

o2 s: Point estimator of + : - 0 2 s;

"1 "2 - (Note) S: =C(xli In, - 1 and S: =C(X,, -x,)' i n , - 1 ;

X 1 - N ( p l , o : ) and X , - N ( F ~ , ~ ; ) ;

X l and X , are independent

o2 s; l o 2 Test statistic of + : F = -----L - F(nl - 1, n, - 1 )

0 2 s; 10;

10-1 6 Chapter Statistical Inference for Two Samples

Test Procedure

(F-test)


(OC) Curve

Step I : State Ho and HI. 2 2

- 0 1 , o 0 1 -- Ho: 7

0: 2

HI : , t for two-sided test, 0 2 4 , o

2 2 g 0 1 , o >- or 5 < 0, for one-sided test. 2

0; 0 2 , o 0; 0 2 , o


Step 3: Determine a critical value(s) for a. 1

fa/2,n,- l ,n2-l and f i -a/2,nl-l ,n2-1 ) for two-sided test f a / 2 , n 2 - ~ , n 1 - 1

fa ,n , - I , ~ ~ - I for upper-sided test / \

1 for lower-sided test


f~ > fa /2 ,n l - l ,n2-1 0' f~ < fi-a/2,n,-l,n2-1 for two-sided test

f o > fa,nI-l ,n2-l for upper-sided test

f o < fi-a,n,-l,n,-l for lower-sided test

Table 10-4 displays a list of OC charts in MR and a formula of the OC parameter h for an F-test on 012/022 where n l = nz = n. By using the table, the appropriate OC chart for a particular F-test is chosen.

Table 10-4 O~eratinp. Characteristic Charts for F-test "

Test a Chart VI (~~pendix A in MR) OC parameter

10-5 Inference on the Variances of Two Normal Populations 1 0-1 7

Confidence s; o2 s2 < L < I f Interval fl-a/ 2,n2 -~ ,n , -I - 2 a/2,n2-l,n,-I for two-sided CI

Formula ' 2 0; S 2

s : - 0 <1 2 fi-a,n2-l,n,-l - for lower-confidence bound

S 2 0;

0: s; -7 5 fa,n2-~,n,-I for upper-confidence bound 0 2 S 2

(Derivation) Two-sided confidence interval* on 012/022

s; / 0; By using the test statistic F = ----- - F(n2 - 1, n, - 1) ,

s: 10;

Example 10.4

Therefore,

* For a one-sided CI, use fa instead of fa , , to derive the corresponding limit.

For the light bulb life length data in Example 10.1, the following results have been obtained:

Brand of Light Bulb Sample Size Sample Mean Sample Variance

INFINITY (1) nl = 30 Z, = 780 hrs S: = 402 -

FOREVER (2) n2 = 25 x, = 800 hrs s,2 = 302

1 . (Hypothesis Test on o12/022; Two-Sided Test) Test Ho: 0 1 ~ 1 0 2 ~ = 1 VS. HI: 012/022 # 1 at a = 0.05.

h Step 1: State Ho and HI.

Ho: o:/(T; = l

HI: of 10; $ 1


10-1 8 Chapter 10 Statistical Inference for Two Samples

Example 10.4 Step 3: Determine a critical value(s) for a. (cont.)

fa1 2,nl -l,n2 -1 = f0.05 12,30-1,25-1 = f0.025,29,24 = 2'22 and

fl-a12,nl-l,n2-1 = f0.975,29,24 = f0.025,24,29 2. I


Since, f0 = 1.78 2 f0025,29,2, = 2.22 and

fo = 1.78 g f0,975,29,24 = 0.46, fail to reject HO at a = 0.05.

2. (Sample Size Determination) Determine the sample size n (= nl = n2)

required for this two-sided F-test to detect the ratio of ' 3 1 to '32 as high as 1.5 with 0.8 of power. Apply an appropriate OC curve.

~hr TO design a two-sided F-test at a = 0.05, OC Chart Vlo is applicable with the parameter

~ = 0 ' = 1 . ' j '32

By using h = 1.5 and P = 0.2 (because power = 1 - P = 0.8), the sample size required is determined as n (= nl = n2) = 50 as displayed below.

1.50 A

(0) O.C. curves for diffnent values of n for the two-sided F-test for a level of significance a = 0.05,

3. (Confidence Interval on oI2/oz2; Two-Sided CI) Construct a 95% two-sided confidence interval on 01~1'32~. Based on this 95% CI on '31~/'322, test Ho: ~~1~1'322 = 1 VS. HI: '312/'322 # 1 at a = 0.05.

~hr 1 - a = 0.95 = a = 0.05 95% two-sided CI on '31'102~:

10-6 Inference on Two Population Proportions 10-1 9



Since this 95% two-sided CI on '312/'322 includes the hypothesized value unity, fail to reject Ho at a = 0.05.

In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etching rate is an important characteristic in this process. Two different etching solutions have been compared, using two random samples of 10 wafers for each solution. The observed etching rates (unit: milslmin) are as follows:

9.4 10.3 10.6 10.2 9.3 10.0 10.7 10.7 9.6 10.3 10.4 10.4

that the etching rates are normally distributed.

1. Test Ho: o12/'322 = 1 VS. HI: o12/'322 # 1 at a = 0.05. 2. Determine the sample size n (= nl = n2) required for this two-sided F-test to

detect the ratio of '3, to o2 as high as 2 with 0.9 of power. Apply an appropriate OC curve.

3. Construct a 95% two-sided confidence interval on '312/'322. Based on this 95% two-sided CI on 012/'322, test Ho: = 1 VS. HI: 0121'322 f 1 at a = 0.05.

10-6 Inference on Two Population Proportions

0 Determine the sample size of a z-test for statistical inference onpl -p2 by using an appropriate

Inference Parameter of interest: pl -p2


Inference (Note) X1 - B(n1, P I ) , X, - B(n,, p,) ; Context

n l p l , n,(l- p , ) , n2p,, and n2 (1 - p,) are greater than 5; (cont.) X, and X, are independent;

and P2 = - - n1 n2

Test statistic ofpl -p2: The test statistic ofpl -p2 depends on the equality ofpl and p2 as follows:

(1) Case 1: Unequal proportions (pl #p2)

(2) Case 2: Equal proportions @I = p2 =p)

Test Step 1 : State Ho and HI. Procedure Ho: PI -p2'60

(z-test) HI: PI -p2 f 60 for two-sided test,

pl - p2 > 60 or p1 - p 2 < 6o for one-sided test.

Step 2: Determine a test statistic and its value. (1) Case 1: Unequal proportions (pl # p2)

A A 8 -P2 -6 , 4-6-6, zO=~T=/-y ( 2 ) Case 2: Equal proportions (pl = pz =p)

Step 3: Determine a critical value(s) for a. z,,, for two-sided test; z , for one-sided test


Izol > zai2 for two-sided test

z, > z, for upper-sided test

2, < -2, for lower-sided test

10-6 Inference on Two Population Proportions 10-21

Sample Size For a hypothesis test onp l - p2, the following formulas are applied to determine Formula the sample size:

Confidence Like the test statistic onp , -p2, a 100(1 - a)% C I onpl -p2 depends on the Interval equality o fp l and p2 as follows:

Formula 1. Case 1: Unequal proportions (pl #p2) A A ,: - p2 - dv

a A

PI - p 2 -za/2 /- i p, - p2 for lower-confidence bound

A A

P I - P Z S P , -P2 + z a ~ for upper-confidence bound n2

2. Case 2: Equal proportions (pl =p2 = p )

A A

C pl - p i C P, - P2 + for two-sided CI

I pl - p2 for lower-confidence bound

for upper-confidence bound

Example 10.5 Random samples of bridges are tested for metal corrosion in the HAPPY and



1. (Hypothesis Test onpl -pz; Unequal Proportions; Upper-Sided Test) Assuming pl #p2, test if the proportion of corroded bridges of HAPPY County exceeds that of GREAT County by at least 0.1. Use a = 0.05.

h Since n , j l =40x0.7=28, n l ( l - j l ) = 4 0 x 0 . 3 = 1 2 , n 2 j 2 =30x0.5=15,

and n2 (1 - j2 ) = 30 x 0.5 = 15 are greater than five, the sampling

distributions of 4 and p2 are approximately normal.

Step 1 : State Ho and HI. Ho: pl -p2 = 0.1 Hi: pl -p2 > 0.1


Step 3: Determine a critical value@) for a. za = zo,05 = 1.64

Step 4: Make a conclusion. Since )zo 1 = 0.86 2 z,,,, = 1.64 , fail to reject Ho at a = 0.05.

2. (Sample Size Determination) Suppose that p, = 0.7 and p, = 0.5. Determine the sample size n (= nl = n2) required for this two-sided z-test to detect the difference of the two proportions with power of 0.9.

k power = P(reject Ho I Ho is false) = 1 - p = 0.9 = p = 0.1

3. (Confidence Interval onpl -pz; Unequal Proportions; Upper-Confidence Bound) Assumingpl +p2, construct a 95% upper-confidence bound on the difference of the two corroded bridge proportions (pl - p2).

95% two-sided CI onpl - p2:

PI -P* S j l - i ) 2 +Za



10-6 Inference on Two Population Proportions 10-23

A random sample of nl = 500 adult residents in MARICOPA County found that xl = 385 were in favor of increasing the highway speed limit to 70 mph, while another random sample of n2 = 400 adult residents in PIMA County found that x2 = 267 were in favor of the increased speed limit. 1. Does the survey data indicate that there is a difference between the residents of

the two counties in support of increasing the speed limit? Assume pl - p2 = 0 and use a = 0.05.

2. Suppose that p1 = 0.75 and p2 = 0.65. Determine the sample size n (= nl = n2) required for this two-sided z-test to detect the difference of the two proportions

I with power of 0.8. 3. Assuming pl = p2, construct a 95% two-sided confidence interval on the

difference between the favor proportions of the two counties for the speed , limit increase (pl - p2). Based on this 95% two-sided CI onpl - p2, test Ho: pl -

p2=Ovs. Hl :p l -p2+Oata=0 .05 .



Example 10.2 (Inference on 1 1 - pz, o12 and oz2 Unknown) I ( 1 ) Choose File * New, click Minitab Project, and click OK.

(2) Enter the life length data of the INFINITY and FOREVER light bulbs on the worksheet

c l - 1 g a & INFINITY /FOREVER, 3;

(3) Choose Stat * Basic Statistics * 2-Sample t. Click Samples in different columns and select INFINITY and FOREVER in First and Second, respectively. Select not equal in Alternative and type the level of confidence in Confidence level. Check Assume equal variances if o12 =



Two Sample T-Test and Confidence Interval I I TWO 3 m p ~ T f o r INFINITY vs FOREVER I I N M e a n S t D e v S E M e a n I

I N F I N I T Y 30 780.0 40.0 7.3 FOREVER 2 5 800.0 30.0 6.0

t I 1 10.2.4 1 ............ + ........... , 95t C I f o r m u I N F I N I T Y - m u FOREVER: i,(.:3?.;~-.:~-.ai .................... T - T e s t m u I N F I N I T Y = m u FOREVER ( v s n o t =) :;T..:.z?.:>?.*P.,y.j.;p.329i DF = 52

Two Sample T-Test and Confidence Interval

( TWO s m p l e T f o r INFINITY vs FOREVER 1 N M e a n S t D e v S E M e a n

I N F I N I T Y 30 780.0 40.0 7.3 FOREVER 2 5 800.0 30.0 6.0

95% C I f o r m u I N F I N I T Y - m u FOREVER:;.!..;::::, T - T e s t m u I N F I N I T Y = m u FOREVER ( v s Both u s e P o o l e d S t D e v = 35.8

(5) Choose Stat N Power and Sample Size N 2-Sample Z. Click Calculate sample size for each power value. Enter the power of the test predefined in Power values, the difference between the true mean and hypothesized mean to be detected in Difference, and the pooled estimate of standard




Example 10.3

fined test condition.

I Power and Sample Size

1 2-Sample t Tes t I T e s t i n g mean 1 = mean 2 ( v e r s u s no t =) C a l c u l a t i n g power f o r mean 1 = mean 2 + 20 Alpha = 0.05 Sigma = 35.8

(Inference on po, oDZ Unknown) (1) Choose File * New, click Minitab Project, and click OK.

(3) Choose Stat * Basic Statistics N Paired t. Select BEFORE and AFTER in


Example 10.3 I (4) Enter the level of confidence in Confidence level, the hypothesized (cont.) difference in mean in Test mean, and not equal in Alternative. Then

click OK twice.

( 5 ) Intemret the analvsis results.

Paired T-Test and Confidence Interval *I I Palred T f o r Before - After I

Example 10.5 (Inference on pl - p2)

(1) Choose Stat + Basic Statistics + 2 Proportions. Click Summarized data and enter the number of bridges surveyed in Trials and the number of

N Mean StDev SE Mean Before 30 181.00 2 2 . 7 6 4.15 After 30 171.00 2 3 . 6 3 4.31 Difference 30 10.000 S f 1 7 n

Ins.

1 10.3.2 1

10-28 Chapter 10


(2) Enter the level of confidence in Confidence level, the hypothesized difference in proportion in Test difference, and greater than in Alternative. Check Use pooled estimate of p for test if the proportions

/ Test and Confidence Interval for Two Proportions I

10.5.3

Test for ~ ( 1 ) - ~ ( 2 ) = 0.1 [vs > 0.1) :i~,~..~.,~:~~...~:~~.1.~~..:...9:,?~~ 5 5 5 5 5 5 ; .: -

(4) Choose Stat + Power and Sample Size + 2 Proportions. Click Calculate sample size for each power value. Enter the power of the test predefined in Power values and the true proportions of two populations



(5) Under Alternative Hypothesis select the type of alternative hypothesis, and in Significance level enter the probability of type I error (a).Then

I Power and Sample Size *I I T e s t f o r Two P r o p o r t i o n s I

T e s t i n g p r o p o r t l o n 1 = p r o p o r t l o n 2 ( v e r s u s >) C a l c u l a t i n g power f o r p r o p o r t l o n 1 = 0 . 7 a n d p r o p o r t l o n 2 = 0 . 5 Alpha = 0 . 0 5 D i f f e r e n c e = 0 . 2 1



Exercise 10.1 1. (Hypothesis Test on - p ~ , tsZ Known; Upper-Sided Test) Step 1 : State HO and H I .

Ho: PI - k 2 = 10 H I : ~ 1 - ~ 2 > 1 0


Step 3: Determine a critical value(s) for a. za = z ~ , ~ , = 1.65

Step 4: Make a conclusion. Since z, = -5.84 ;6 z,,, = 1.65, fail to reject Ho at a = 0.05. We do

not have significant evidence to support use of plastic 1 at a = 0.05.


(1) Sample Size Formula

n = i z , + ~ , 3 ) ~ ( 0 : + 0 ; ) - - (z,.05+zo.,)2(12+12)

(A - A,)2 (11.5-10)~

(2) OC Curve To design a one-sided z-test at a = 0.05 for two samples, OC Chart VIc is applicable with the parameter

A - A 11 1.5 - 101 = 1.06

By using d = 1.06 and P = 0.1, the required sample size is determined as n (= nl = n2) = 8 as displayed below.

1.06

(c) 0.C NMS for di&rent vahtea of n for ttv ~ r t ~ s i & d n d tcs for a lnrd of s@Sicanw a = 0.05. .


Exercise 10.2


3. (Confidence Interval; Upper-Confidence Bound) 1 - a = 0 . 9 5 3 a=0 .05 95% upper-confidence bound on p1- p2:

1. (Hypothesis Test on pl - pz, o12 and o; Unknown but Equal; Lower-Sided Test)

Step 1: State Ho and HI. Ho: p1- p2=0 H1: p1 - p 2 < 0


2 (n, - 1)s; + (n2 - l)s2 (Note) si =

n, + n2 - 2

Step 3: Determine a critical value@) for a. -

ta,v - t0.05,24 =

Step 4: Make a conclusion. Since to =-4.07<-to,05,24 =-1.711,rejectHoat a=0.05.

2. (Sample Size Determination) To design a one-sided t-test at a = 0.05 for two samples, OC Chart VIg is applicable with the parameter



When d = 0.66 and j3 = 0. I , n* = 20 as displayed below.

(g) O.C. curves for different values of n for the one-sided t-test for a level of significance a = 0.05.

Thus, the required sample size n is

I 3. (Confidence Interval on p1- pz, 01' and 02 Unknown but Equal; Upper- Confidence Bound)

1 95% upper-confidence bound on p1 - p2:

4. (Confidence Interval on p1 - p z , ~ ~ ' and 0 2 ' Unknown and Unequal; Upper-Confidence Bound)

1 - a = 0.95 3 a = 0.05

1 95% upper-confidence bound on pl - p2:


Exercise 10.3

Exercise 10.4

1 . (Hypothesis Test on y D, 0 D 2 Unknown; Two-Sided Test) Step 1 : State Ho and HI.

Ho: PD=O HI: P D # O



ta/2,n-1 - t0.05/2,12-1 = t0.025,1 I = 2.201

Step 4: Make a conclusion. Since I d o l = 0.783 P to,025,11 = 2.201, fail to reject Ho at a = 0.05.

Thus, we conclude that the two design languages are not significantly different in mean coding time at a = 0.05.

2. (Confidence Interval on yo, OD' Unknown; Two-Sided CI) 1 -a=0 .95 3 a=0.05; v = n - 1 = 1 2 - 1 = 1 1 95% two-sided CI on b:

Since this 95% two-sided CI on p . ~ includes the hypothesized value zero, fail to reject HO at a = 0.05.

1. (Hypothesis Test on o 1 2 / o z 2 ; Two-Sided Test) Step 1 : State Ho and HI.

2 2 Ho: o; =oi 3 0, l o 2 = l

HI : o ;#o; 3 o; /o; # l


Step 3: Determine a critical value(s) for a. f a , 2,nl -l,n2-l = f0.05 2,10-I,IO-I = f0.025,9,9 = 4.03 and


Exercise 10.4 Step 4: Make a conclusion. (cont.) Since fo = 3.33 > f0,,, ,,,,, = 4.03 and f o = 3.33 < f0,9,5,9,9 = 0.25 ,

fail to reject Ho at a = 0.05.

1 2. (Sample Size Determination)

To design a two-sided F-test at a = 0.05, OC Chart VIo is applicable with the parameter

When h = 2.0 and P = 0.1, the required sample size is determined as n (= nl =

n2) = 25 as displayed below.

I

(0) O.C. curves for di f fmt values of n for the two-sided F-test for a level of significance a = 0.05.

3. (Confidence Interval on O ~ ~ I O Z ~ ; Two-Sided CI)

1 - a = 0 . 9 5 3 a=0.05 95% two-sided CI on 0 1 ~ 1 0 2 ~ :

Since this 95% two-sided CI on 0 1 ~ 1 0 2 ~ includes the hypothesized value unity, fail to reject Ho at a = 0.05.

Exercise 10.5

7


1. (Hypothesis Test onpl -pz, Equal Proportions; Two-Sided Test) The survey data is summarized as follows:

X Sample County Sample Size Proportion

Since n, j, = 500 x 0.77 = 385 , n, (1 - j , ) = 500 x 0.23 = 115, n 2 j 2 =

400x 0.67 = 267, and n2 (1 - j 2 ) = 400x 0.33 = 133 are greater than five,

the sampling distributions of 4 and are approximately normal.

Step 1 : State Ho and HI. Ho: pi - p 2 = 0 Hi: pi - p 2 + 0


x, + x, - 385 + 267 (Note) j = - - = 0.72

n, + n2 500 + 400

Step 3: Determine a critical value(s) for a. 'a12 = z0.05!2 = '0.025 =

Step 4: Make a conclusion. Since lz, 1 = 3.32 > z,,,,, = 1.96 , reject Ho at a = 0.05.


power = P(reject Ho I Ho is false) = 1 - p = 0.8 a P = 0.2



3. (Confidence Interval onpl -pz; Equal Proportions; Two-Sided CI)

95% two-sided CI on pl - p2:

Since this 95% two-sided CI onpl -p2 does not include the hypothesized value zero, reject Ho at a = 0.05.

Simple Linear Regression and Correlation

1 1 - 1 Empirical Models 1 1-6 Confidence Intervals

1 1-2 Simple Linear Regression 11-6.1 Confidence Intervals on the

1 1-3 Properties of the Least Squares Slope and Intercept

Estimators 11-6.2 Confidence Interval on the

1 1-4 Some Comments on Uses of Regression Mean Response

1 1-5 Hypothesis Tests in Simple Linear 1 1-7 Prediction of New Observations

Regression 1 1-8 Adequacy of the Regression Model

1 1-5.1 Use of t-Tests 1 1-8.1 Residual Analysis

1 1-5.2 Analysis of Variance Approach 1 1-8.2 Coefficient of Determination

to Test Significance of 11-8.3 Lack-of-Fit Test

Regression 1 1-9 Transformations to a Straight Line 1 1 - 1 1 Correlation



11 -1 Empirical Models

independent variable, dependent variable, and random error. le linear model.

Regression Regression analysis is a statistical technique to model the relationship between Analysis two or more variables so that we can predict the response of a variable at a given

condition of the other variable(s).

11 -2 Chapter 11 Simple Linear Regression and Correlation

Regression For example, in Figure 1 1-1, the scatter diagram of price versus sales for a' Analysis plywood product indicates that the volume of sales linearly decreases as the price

(cont.) of the product increases. By modeling this linear relationship, we may want to know how many pieces of plywood would be sold if the price is set at $8.50. It may not be easy to answer the question because there is no simple, linear curve which passes through all the points in the diagram-the data randomly scatter along a straight line.

1 I I I I * 0 2 4 6 8 10

Price per Piece (x)

Figure 11-1 Scatter diagram of plywood price versus sales.

Simple A simple linear model which explains the linear relationship between two Linear variables (x and Y) is Model Y = p o + P l x + &

where: (1) Regression coefficients (Po and PI): The coefficients Po and PI denote the intercept and slope of the regression line, respectively. The slope PI measures the expected change in Y for a unit change in z.

(2) Independent variable (x; regressor, predictor): Since the independent variable x is under control with negligible error in an experiment, x represents controlled constants (not random outcomes).

(3) Random error (E): The variable E represents the random variation of Y around the straight regression line Po + Pix. It is assumed that E

is normally distributed with mean 0 and constant variance 02, i.e.,

E - N(o ,o~) (4) Dependent (response) variable (Y): Since outcomes in Y at the

same value of x can vary randomly, Y is a random variable with the following distribution:

y-N(P0 + P , x , ~ ~ )

In summary, four assumptions are made for the simple linear regression model: 1. Linear relationship between x and Y: The variables x and Yare linearly

related. 2. Randomness of error: The errors (spreads of Y along the regression line)

are random (not following any particular pattern).

1 1-2 Simple Linear Regression 1 1 -3

Simple 3. Constant oZ: The variance of the random errors is constant as 02. In other Linear words, the spreads of Y along the regression line are the same for different Model values of x (see Figure 1 1-2). (cont.) 4. Normality of error: The random errors are normally distributed.

0

0 XI x2 X

Figure 11-2 The variability of Y over x.

The analyst should check if the assumptions of the linear model are met for proper regression analysis, which is called assessment of model adequacy (presented in Section 11-8).

11 -2 Simple Linear Regression

O Explain the method of least squares. O Estimate regression coefficients by using the least squares method. 0 Calculate mean responses and residuals by using a fitted regression line. Cl Estimate error variance 02.

Least In the 1800s Karl Gauss proposed 'the method of least squares' to estimate the Squares regression coefficients of a linear model. Suppose that we have n pairs of Method observations (xi, yi), i = 1,2 , . . . , n, and the n observations are modeled by a

simple linear model

The least squares method finds the estimates of Do and (3, which minimize the sum of thesquares of the errors (denoted by SSE; called error sum of squares; sum of the squares of the vertical deviations of data from the regression line in Figure 1 1-3)

Chapter

Least Squares Method

(cont.)

(Derivation) Least square estimators po and Dl Recall that, in calculus, when we want to find the value of a variable which I minimizes (or maximizes) a certain function, the following procedures are used: !

(1) Take the derivative of the function with respect to the variable designated, (2) Set the derivative as equal to zero, and (3) Solve the equation.

Since the function SSE (or L) has two unknowns (Do and PI), we need to take a partial derivative with respect to each regression coefficient as follows:

1 1-2 Simple Linear Regression 1 1 -5

Least n

Squares By multiplying the first equation above by xi and the last equation by n each, Method i=l

(cont.)

By subtracting the last equation above from the first equation above,

Thus,

Lastly, from Equation (1 1-2),

2 Y i - S l t x i i=l

S o = '=' = y - p l x

n

Fitted By using the least squares estimates of Do and D l , the fitted (estimated) regression Regression line is determined as follows:

Model + S o +Six

By applying the fitted regression line, the mean response at xi is estimated. Note that an actual response yi can be different from the corresponding predicted response ji . The difference between the actual and predicted responses is called

the residual (denoted as e;), which is an estimate of the random error E~ :

e i = y i - j j i , i = 1 , 2 ,..., n

1-6 Chapter Simple Linear Regression and Correlation

Estimation of Error

Variance

( 0 2

Since the error variance o2 is unknown in most cases, estimation on o2 is needed. The point estimator of o2 is

n

(Derivation) SS, = y: - ny2 - bls, i=l

* A

SS, =f:ei2 =f:(yi - j i ) 2 = f:Li -(Bo + D , X ~ ) ] ~ because ji = P o + Pixi i=l i=l i=l

A - * -

= f: [vi - (y - pi? + p1xi ) r because Do = y - D l x i=l

i=l i=l

n

(Notes) SS, = S, = (yi - j?)2 (called total sum of squares)

n n

=SS, -2fils, +P:S, because S, = C y i ( ? - x i ) and C ( ? - x i ) = O i=l i=l

A >xy A Axy = SS, - 2 B 1 ~ v + PI -SY because P1 = - Sxx Sxx

n n n =cy,? - n j 2 -B1s, because SS, = C ( ~ , - J ) ~ = C y ' - n ~ '

1 1-2 Simple Linear Regression 11 -7 r Example 11.1 I The sales' volumes (Y) of a plywood product at various prices (x ) are surveyed as follows (sorted by price):

The summary quantities of the plywood price-sales data are

n = 3 0 , z y, = 1,930, z y,? = 138,656, z x i = 225, z xi = 1,775 , and

z xi y, = 13,360

Assume that x and Yare linearly related.

1 . (Estimation of Regression Coefficients) Calculate the least squares estimates of the intercept and slope of the linear model for x and Y.

The fitted regression line of the plywood price-sales data is A n

j = p , + p,x = 159.9 - 1 2 . 7 ~

I 2. (Calculation of Residual) Calculate the residual ofy = 84 at x = $6.

h The predicted sales at x = $6 is .. A

jX=, = P o +/3 ,~=159.9-12 .7~6=83.4

Thus, - ex=, - yX,, - jx=, = 84 - 83.4 = 0.6 (underestimate)

1 1-8 Chapter 1 1 Simple Linear Regression and Correlation

Example 11.1 3. (Estimation of 02) Estimate the error variance 0'.

(cont.) - 2

SS, = t y i -ny2 -&s, = 1 3 8 , 6 5 6 - 3 O x ( z ) -(-12.7)x(-1,115) i=l

Exercise 11.1 An article in the Journal of Sound and Vibration (Vol. 15 1, 199 1, pp. 383-394) (MR 11-9) described a study investigating the relationship between noise exposure (x) and

hypertension (Y). The following data (sorted by sound pressure level) are representative of those reported in the article:

I The summary quantities of the noise-hypertension data are

I Assume that x and Yare linearly related.

I 1. Calculate the least squares estimates of the intercept and slope of the linear model for x and Y.

1 2. Calculate the residual of y = 5 mmHg at x = 85 dB.

1 3. Estimate the error variance 4'

11-3 Properties of the Least Squares Estimators

" *

Identify the sampling distributions of the least squares slope and intercept estimators.

0 Check if the least square estimators fi, and Dl are unbiased estimators.

11-3 Properties of the Least Squares Estimators 11 -9

The sampling distribution of the least squares slope estimator 8, is Distribution of

Slope Estimator

(Dl) (Note) Estimated standard error of 8, =

(Derivation) Sampling distribution of 8, Recall that the least squares estimator of P1 is

n

Also recall that Y = P o + p 1 x + & - N(Po + P l x , o 2 ) , where E - N ( o , o ~ ) , andx

represents constants. Since 0, is a linear combination of the normal random

variables y i 's, Dl is normally distributed with the following mean and variance:

I - =-plCxi(xi-;) c because x ( x i - x ) = O

(Note) Since E(&) = pl , 8, is an unbiased estimator of P I .

11 -1 0 Chapter 1 1 Simple Linear Regression and Correlation

The sampling distribution of the least squares intercept estimator po is Distribution of

Estimator

(Derivation) Sampling distribution of Do Recall that the least squares estimator of Po is

Also recall that Y - N(Po + P,x, 0 2 ) , B, - N(Pl , 0 2 1 S , ) , and x denotes

constants. Since Do is a linear combination of the normal random variables y, 's i

and p, , So is normally distributed with the following mean and variance:

E(B,) = E(y - p , ~ ) = E(y) - ~ ~ ( f i , ) = E(Po + Dl?) -P I? = P o + p 1 x - p 1 x = p 0

v(po)=v(y-plx)=v(y)+x2v(pl)-2xcov(y,pl)=v(y)+x2v(pl) (Note) cov(j7,pl) = 0 because and p, are independent.

(Note) Since E(p0) = Po , Do is an unbiased estimator of Po .

11 -4 Some Comments on Uses of Regression

- * w I * I X I m w ~ ~ w a - -d -Wa--in * - - I x n - * a 1 X z m w m - ~ - - - m- --I. 1)",XU*XIcrr .-.FW(-= E1W --- *-*- - L l Explain common misuses of regression.

Common Care should be taken on the following to avoid technical misuse of regression: Misuses of 1 . Practical validity of regressor: A statistical significance can be found on the Regression relationship between independent (x) and dependent (Y) variables, although

they are not related from a practical sense. Therefore, the practical significance of the relationship among the variables should be checked separately.

1 1-5 Hypothesis Tests in Simple Linear Regression 11 -1 7 i

Common 2. Empiricavtheoretical validity of regressor: The direction and/or magnitude Misuses of of a regression coefficient in the model can contradict common understanding Regression of the relationship between the variables. Therefore, it should be checked if

(cont.) the sign and magnitude of each regression coefficient are reasonable as compared with previous findings.

3. Association of regressor with predictor: A significant relationship between the regressor and predictor in the model does not necessarily indicate a cause- effect relationship between the variables. Designed experiments are the only way to determine cause-effect relationships between variables.

4. Limited generalizability of the model: The regression relationship established between the variables may be valid only over the range(s) of the regressor(s) examined in the study. To apply the regression model beyond the original range(s) of the regressor(s), a new study should be designed for the extrapolation purpose.

11 -5 Hypothesis Tests in Simple Linear Regression

11 -5.1 Use of f-Tests

O Conduct hypothesis tests on PI and Po each. ~ U X * l w ~ X ( _ * J U w _ I ~ ~ - ~ m ~ M - ~ - L * ~ * ~ w ~ ~ ~ * * w M ~ - % - ~ w Mm -I %"-. - # "^* lil* XIX( *FX %"& *" I. iiX4aa X * e. XXlll', t_X I

Inference Parameter of interest: PI Context on

'x' [ o2 is unknown. Slope Point estimator of PI : PI = - - N P, ,- CP,) s,

Test statistic of PI : T - '1 - P ~ , o - t (n-2)

O - J i S ,

Hypothesis Step 1 : State HO and HI. Test on Ho: PI = PI,O

P1 HI: PI # P1,o (t-test)


Step 3: Determine a critical value(~) for a. t a t 2.n-2


It0 I > tat2.n-2

11 -1 2 Chapter 1 1

The Significance of

Regression

Inference Context on

Intercept ( P o )

Hypothesis Test on

Po (t-test)

r Example 11.2


It is typical to test if the regressor x is significant to explain the variability in Y by using Ho: = 0, which is called testing the significance of regression.

Note that failure to reject Ho: [3, = 0 may imply that (1) The regressor x is of little value in explaining the variability in Y, or (2) The true relationship between x and Y is not linear.

On the other hand, rejection of Ho: = 0 may imply that (1) The regressor x is of value in explaining the variability in Y, (2) The simple linear model is adequate, or (3) A better model could be obtained by adding a higher order term(s) of x

while a linear effect of x on Y remains significant.

Parameter of interest: P o

Test statistic of P o : To =

Step 1 : State Ho and H I . Ho: Po = Po,o H1: Po + P0,o

Step 2: Determine a test statistic and its value. 1

I



1'0 ( > tai 2,n-2

For the plywood price-sales data in Example 1 1-1, the following summary quantities have been calculated:

n = 30, b2 =3.2', E x i =225 , Sxx =87.5, 8 , =-12.7, and 8 , =159.9

1. (Hypothesis Test on PI) Test Ho: P1 = 0 at a = 0.05.

1 1-5 Hypothesis Tests in Simple Linear Regression 1 1-1 3




'u12,n-2 - t0.025,28 = 2.05

Step 1 : State Ho and HI. Ho: p1 = 0 HI: pl + O

I Step 4: Make a conclusion. Since It0\ = 37.4 > t,12,n-, = 2.05, reject Ho at a = 0.05.

2. (Hypothesis Test on Po) Test Ho: = 0 at a = 0.05. - Step 1: State Ho and HI. H0: Po = 0 HI: pozo


Step 3: Determine a critical value@) for a. 'a/2,n-2 = t0.025,28 2.05

I Step 4: Make a conclusion. Since (t,(=61.0>taI,,,-, =2.05,rejectHoata=0.05.

Exercise 11.2 For the noise-hypertension data in Exercise 1 1 - 1, the following summary (MR 11-25) quantities have been calculated:

n = 20, ir2 = 1.4', Ex, = 1,646, S, = 3,010.2, = 0.17, and a, = -9.81

1. Test HO: pl = 0 at a = 0.05.

2. Test HO: Po = 0 at a = 0.05.

I I -5.2 Analysis of Variance Approach to Test Significance of Regression

Explain the analysis of variance (ANOVA) technique. Describe the segmentation of the total variability of a response variable (SST) into the regression sum of squares (SSR) and error sum of squares (SSE).

Establish an ANOVA table and conduct a hypothesis test on P1. - - - l _ n P - X - - l j - - - M - - ~ ~ - - - - _ j ~ l l ~ w w = ~ - a ~ _ _ * - _ I * ~ ~ = P _ L - a a a X w - ~ --/tr,*-m---s s-

11 -14 Chapter 1 1 Simple Linear Regression and Correlation

Analysis of The analysis of variance (ANOVA) technique provides a statistical procedure to Variance evaluate the contribution of a variable (X,) to the variability in the response

(ANOVA) variable (Y). As illustrated in Figure 11-4, the ANOVA method divides the total variance of Y ( S T ) into the segments ( SS,, 's ) explained by variables Xi's and

the segment unexplained (SSE), implying that the larger the size of the segment SS,, , the higher the contribution of XI to the total variability in Y.

Figure 11-4 Segmentation of the total variability in Y (ST) .

ANOVA for The ANOVA method can be used to test the significance of regression (Ho: PI = Test on 0). For a simple linear model, as shown in Table 1 1 - 1, the total variability in Y 1

1 (SST or Syy, total sum of squares; degrees of freedom = n - 1) is partitioned into two components: i

(1) Variability explained by the regression model (SSR, regression sum of squares; degrees of freedom =I)

(2) Variability due to random error (SSE, error sum of squares; degrees of freedom = n - 2)

Table 11-1 Partitioning of the Total Variability in Y (SST) SST SSR ss.

(Total SS) (Regression SS) (Emr SS)

r= l r=l r = l

DF n - l 1 + n-2 - - m*&"w-am**m-mema -*B waa*s~-~-*--essB--b-~a* w ~ - m ~ ~ a - # - - - - ~ e ~ m . - - ~ w " * m . m ~ *sea%a--"--mm--e

(Note) SS: Sum of Squares; DF: Degrees of Freedom

A graphical interpretation of the ANOVA quantities is provided in Figure 1 1-5. The deviation of an observation yi from the corresponding mean y is divided into the yi-deviation explained by the regression line and the yi-deviation due to error. These deviation are squared and then added to those of the other observations to calculate SST, SSR, and SSE as follows:

ANOVA for Test on

PI (cont.)

1 1-5 Hypothesis Tests in Simple Linear Regression 11 -1 5

(variability ofy, due to regress~on)

Figure 11-5 Partitioning of the total variability ofy.

The ratio of SSR/l to S S E / ( ~ - 2) follows an F distribution:

F = - S S ~ l M S ~ F(l , n - 2) ---

SS, l (n -2 ) MS,

This statistic F is used to test Ho: P1 = 0, because F becomes as P1 = 0 is true. The quantities MSR (regression mean square) and MSE (error mean square) are adjusted SSR and SSE by their degrees of freedom, respectively. Note that

ANOVA The variation quantities from the ANOVA analysis are summarized in Table 11-2 Table to test the significance of regression.

Table 11-2 ANOVA Table for Testing the Significance of Regression Source of Sum of Degrees of Mean Variation Squares Freedom Square Fo

Regression SSR = fils, 1 MS, MS,IMS,

Error SS, = SS, - SS, n - 2 MSE n

Total SS, =zyi -ny2 n-1

Hypothesis Step 1 : State Ho and HI. Test on Ho: Pl = 0

P I HI: P l f 0 (F-test)


F, = - SSR I ' MSR F(l, - 2) --- SS, l(n - 2) MS,

I 11 -1 6 Chapter 1 1 Simple Linear Regression and Correlation

Hypothesis Step 3: Determine a critical value(s) for a. 1 Test on fa, l ,n-2 i

P1

(F-test) Step 4: Make a conclusion. Reject HO if (cont.)

f~ > fa,~,n-2

Relationship In testing the significance of regression, the ANOVA test procedure is equivalent between to the t-test procedure in Section 11-5.1, because the square of the t-test statistic

t- and becomes the F-test statistic when pl = 0. Therefore, use of either test procedure F-tests will lead to the same conclusion on 01.

- -- f i l S ~ Y because Sv,,

-- - M S ~ because fils, = SS, = MSR and e2 = MSE MSE

Note that the t-test is more flexible than the F-test in testing a hypothesis on PI , because the t-test can have either a one- or two-sided alternative hypothesis whereas the F-test can have only a two-sided alternative hypothesis. However, the F-test is used to evaluate the relative contributions of multiple regressors to the variability in Y (see Section 12-2).

Example 11.3 (ANOVA for Hypothesis Test on PI) For the plywood price-sales data in I Example 1 1 - 1, the following summary quantities have been calculated: n = 30, zyi =1930, y? =138,656, S , =-1,115, and p 1 =-12.7

Establish an ANOVA table and conduct a hypothesis test on PI at a = 0.05.

I- 2

SS, = Ay,? - njJ2 = 138,656 - 3Ox(%) = 14,492.7 i=l

I SS, = filsxy = -12.7 x (-1,115) = 14,208.3

SS, = SS, - S S P = 14,492.7 - 14,208.3 = 284.4

I "

Source of Sum of Degrees of Mean Variation Squares Freedom Square fo

1 Regression 14,208.3 1 14,208.3 1398.9

I Error 284.4 2 8 10.2

I Total 14,492.7 29

(Note) t: = (-37.41~ = 1398.9 = fo


d Exercise 11.3 (MR 11-25)

1 1-6 Confidence Intervals 11-17

Step 1: State Ho and H I . Ho: p1 = 0 H1: P I # 0

Step 2: Determine a test statistic and its value. SSR 11 MSR - 14,208.3

f o = - = 1,398.9

SS,l(n-2) MS, 10.2

Step 3: Determine a critical value(s) for a. for,l,n-2 = f0.05,1,30-2 = 4.20

Step 4: Make a conclusion. Since f , = 1,398.9 >fa,,,,-, = 4.20, reject Ho at a = 0.05.

For the noise-hypertension data in Exercise 1 1 - 1, the following summary quantities have been calculated:

Establish an ANOVA table and conduct a hypothesis test on P1 at a = 0.05.

I 1 6 Confidence Intervals

11-6.1 Confidence Intervals on the Slope and Intercept

Establish confidence intervals on P1 and PO each.

Inference Parameter of interest: p1 Context on

Slope Point estimator of P1 : ( P I ) -

(Note) Estimated standard error of P , = - A /;x:

Confidence A 100(1 - a)% confidence interval on is Interval on

P 1 I - 2 - 2 4 I 1 + 2 - 2 i-' 0 1 sn

Inference Parameter of interest: P , Context on

- fi - Intercept Point estimator of p , : D , = y - P l x - N ( P o 1

(Note) Estimated standard error of p , = ij2 - + -

1 1

11 -1 8 Chapter 1 1 Simple Linear Regression and Correlation

Confidence A 100(1 - a)% confidence interval on Po is Interval on

Po

1. (Confidence Interval on PI) Establish a 95% two-sided confidence interval on the slope. !

Example 11.4

h- 1 - a = 0 . 9 5 z a=0.05 95% two-sided CI on PI:

BI - t a / 2 , n - 2 J K 5 P l 5fi1 + t a / 2 , n - 2 J E

=. - 12.7 - t o , 0 5 , 2 , 3 , - 2 ~ ~ ~ ~ I 5-12.7 +to,05,2,30-2 d 3.22 187.5

I 3 -12.7 - 2.05 x 0.34 I PI 5 -12.7 + 2.05 x 0.34

1 a -13.4 I P , 5-12.0

For the plywood price-sales data in Example 1 1- 1, the following summary quantities have been calculated:

n=30, x x i =225, 6' =3.22, S , =87.5, a, =-12.7,and Po =159.9

I 2. (Confidence Interval on Po) Establish a 95% two-sided confidence interval on the intercept.

I 1 -a=0.95 a a=0 .05 95% two-sided CI on Po:

and Po =-9.81

1. Establish a 95% two-sided confidence interval on the slope. 2. Establish a 95% two-sided confidence interval on the intercept.



n = 20, z x i = 1,646, B2 =1.4', S, =3,010.2, S , =3,010.2, 0, = 0.17,

1 1-6 Confidence Intervals 1 1 -1 9

11-6.2 Confidence Interval on the Mean Response

0 Identify the sampling distribution of the mean response estimator at x = xo. Establish a confidence interval on the mean response at x = xo.

Sampling The mean response at x = x o is Distribution of Pyls = E ( Y XO)=E[(PO + P ~ ~ + E ) I ~ O ~ = P O + P P o

Mean Response

Estimator The point estimator of p ylxo is

The sampling distribution of P Ylxo is

(Derivation) Sampling distribution of fi Ylxo

Recall that

Do - N p0,o -+- and B1 - N PI,- ( 2 [ : ] ) 1:) Since fi ylxo is a linear combination of Po and p1 , fi ylxo has a normal distribution

with the following mean and variance:

E ( P y l x o ) = ~ ( B o + b l ~ o ) = ~ ( B o ) + x o ~ ( B i ) = ~ o + x o h =Prix0

(Note) Since E(P ) = p ylxo , fi ylxo is an unbiased estimator of ~ Y I , .

Inference Parameter of interest: p yl,o Context on

Mean Point estimator of p yIxo : P ylxo = b o + B ~ X O - N Response

a t x = x , o2 is unknown.

( P Y l X o

11-8 Adequacy of the Regression Model 11 -25



h The normal probability plot displays that the residuals lie closely along a straight line. In addition, the standardized residual plot shows more than 95% of the standardized residuals (29130 = 96.7%) fall in the interval (-2,2).

1 Therefore, it is concluded that the residuals are normally distributed. (Notice that there is no indication of an outlier in the plywood price-sales data because all the standardized residuals are within the interval (-3,3).)

Next, the plots ei vs. j j and ei vs. xi indicate that the residuals are random and have a constant variance. The variability of the residuals is slightly increased in the mid ranges of y and x; however, this deviation from the constant variance is not serious enough to reject the adequacy of the regression model.

In the noise-hypertension data in Exercise 11-1, the following residual plots are obtained for the fitted regression model j = -9.8 1 + 0 . 1 7 ~ :

-3 -2 - 1 0 I 2 3

Residuals (e,)

(a) normal probability plot (b) standardized residual plot

Observation Number

Sound Pressure

E Level 5 * 6 O (x: dB) ...............

0. .

a". ............ - - . . . . . . - - -

(c) e, vs. ji (d) e, vs. xi

1 Discuss the adequacy of the regression model by using the residual plots.

11 -8.2 Coefficient of Determination (I?)

O Explain the term coeflcient of determination. O calculate the coefficient of determination of a regression model.

-,-.- ..... B2ws,BMsB . -s*m* mw"w,d *-~sm,w,m ....ass.*W ............ "m%#*B.m.w.b ~%esem w*m8aM-ea- .m.sm*a 8acv,B* . . ~ . d ~ ~ a . d . . m ~ . m . ~ v ~ ~ - - ~ - ~ ~ . . ~ m . ~ ~ ~ w A ~ * ~ * ~ ~ . ~ ~ ~ ~ . ~ ~ ~ m . . w m # ~ a * ~ . ~ 2 d . ~ . e ~ . ~ ' ~ . M m . a . # a ~ w a . . ~ * . ~ a ~ . . ~ e#.mw.Bw#w~~d~..a.

Coefficient of The coefficient of determination (denoted as R2) indicates the amount of Determination variability of the data explained bv the regression model. In other words, R2 is the

( R ~ ) proportion of the total variability in Y which is accounted for by the regression model:


I

I 2 SSR 14,208.3 R =-- - = 98.0% SS, 14,492.7

1

Coefficient of 2 5's Determination R =R- - I -%, 0 < R 2 < 1

(RZ) SST SST

(cont.) (Caution) Use of R~ A large value of R~ does not necessarily indicate the adequacy of the regression model. By adding new regressors or high-order terms of the regressors to an existing model, R~ will always increase even if the new model is less desirable than the existing model in terms of simplicity, stability, and statistical significance. For instance, n data points of x and y can be 'perfectly' fit by a polynomial of degree n - 1 (e.g., four data points can be perfectly fit by a polynomial with a degree of three). Although the value of the perfect-fit model is one, the high-order polynomial is complex to use and often unstable when a different sample of data is used. Lastly, unless the error sum of the squares (SSE) of the new model is reduced by the error mean square of the existing model, the new model will have a larger error mean square (MSE), resulting in a decreased level of statistical significance.

I The fitted regression model explains 98.0% of the variability in the sales of the plywood product.

Example 11.8 (Coefficient of Determination) For the plywood price-sales data in Example 11- 1, the following summary quantities have been calculated:

SS, = 14,492.7, SS, = 14,208.3, and SS, = 284.4 Calculate the coefficient of determination ( R ~ ) of the fitted regression model 9 = 159.9 - 1 2 . 7 ~ .

11 -8.3 Lack-of-Fit Test


me e* * ? -n_wll_/l a & mw - a a Ull jlC-* U-P_neCl I * * -^* a -" - *- -n- . XFII

O Describe the purpose of the lack-of-fit test on a regression model. O Conduct a lack-fit-test on a fitted regression model.


SS, = 124.2, SS, = 88.5, and SS, = 35.7 Calculate the coefficient of determination ( R ~ ) of the fitted regression model j = - 9 . 8 1 + 0 . 1 7 ~ .

Lack-of-Fit The lack-of-fit (goodness-of-fit) test on a regression model identifies if the order Test of the model is appropriate. For a lack-of-fit test, repeated observations of Y are

needed for at least one level of x.

1 1-8 Adequacy of the Regression Model 1 1 -27

ANOVA for The ANOVA method underlies the lack-of-fit test on a regression model. Lack-of-Fit Suppose that the observations for m levels ofx are as follows:

Test X No. s%pateQ @ims

Y

By applying the ANOVA technique, as shown in Table 1 1-3, the error sum of squares (SSE; degrees of freedom = n - 2; see Section 11-5.2) is further partitioned into two components:

(1) Variability due to pure error (SSpE, pure error sum of squares; degrees of freedom = n - m)

(2) Variability due to the lack-of-fit of the model (SSLOF, lack-of-fit sum of squares; degrees of freedom = m - 2)

Table 11-3 Partitioning of the Error Sum of Squares (SSE) 4% ~ P R SSWF

(EfiorSS) (Pure m r SSj (Lacksf-& S$l

D F n-2 - n-m + m-2 - --*----* ---m-'#- v%a-rma-xmrrrxa-----

m

(Note) n = n, ; SS: Sum of Squares; DF: Degrees of Freedom

A graphical interpretation of SSE, SSPE, and SSLoF is provided in Figure 11-7. The deviation of an observation yij from the corresponding estimate ji at xi is divided

into the deviation ofyV from the corresponding mean Ti due to pure error and the

deviation of yi from the corresponding estimate ji due to the lack-of-fit of the

Figure 11-7 Partitioning of the variability of y due to error.

11 -28 Chapter 11 Simple Linear Regression and Correlation

ANOVA for model. These three deviations are squared and then added to those of the other Lack-of-Fit observations to calculate SSE, SSpE, and SSLOF. In Figure 1 1-7, the error due to the

Test lack-of-fit of the model could have been zero if a higher order model (dotted line) (cont.) was employed.

The ratio of SSLo,W/(m - 2) to SSp~l(n - m) follows an F distribution: 1

Since the statistic F becomes as the order of the model is appropriate, F is used to test Ho: The order of the regression model is correct. Note that the quantities MSLoF (lack-of-fit mean square) and MSPE (pure-error mean square) are adjusted SSLOF and SSPE by their degrees of freedom, respectively.

ANOVA The lack-of-fit and pure-error quantities from the ANOVA analysis are Table and summarized in Table 11-4 to test for the adequacy of model.

Lack-of-Fit Test Table 11-4 ANOVA Table for the Lack-of-Fit Test - - - ~ -. - - - ~ ~ - - - - -

Source of 1

Sum of Degrees of Mean Variation Squares Freedom Square Fo i

Lack-of-Fit S S ~ o ~ m - 2 Ms,m M s ~ ~ l M S ~ ~

Pure Error SSPE n - m M ~ P E

Error SSE n - 2 A mm-waw* ew*meww-mww wMwe- mwvs -ww * a

MSE am -w ma%- mms - dw - *-**A %- *- a * "V Yrn - *" A W

Step 1: State Ho and HI. Ho: The order of the model is correct. HI: The order of the model is not correct.




f~ ' fa,m-2,n-m Example 11.9 (Lack-of-Fit Test) For the plywood price-sales data in Example 1 1 - 1, the I following summary quantities have been calculated:

I m

n = 30, m = 6 , SS, = 284.4, SSLoF = C(yi - ji )* = 2.4, and

I Conduct a lack-of-fit test on the regression model jj = 159.9 - 1 2 . 7 ~ at a = 0.05.

1 1-9 Transformation to a Straight Line 1 1-29

I Pure Error 282.0 24 11.8

Example 11.9 (con?.)

I Error 284.4 2 8 nwmX( % n _ ~ ' w w * L n ~ f ~ j _ x " . ~ - s s m - - ~ ~ - - ~ ~ - w ~ ~ , ~ * w ~ " - w . - ' - z - - e m ~ ~ ~ * - w # * ~ i*eeanniea ,-,, a.x,", a.x.w.rx.naarar.""ir

An ANOVA table to test the lack-of-fit of the regression model is as follows: Source of Sum of Degrees of Mean Variation Squares Freedom Square Fo

Step 1 : State Ho and HI. Ho: The order of the model is correct. HI: The order of the model is not correct.

I Step 2: Determine a test statistic and its value. MSL, - 0.6 fo = - - - = 0.05 MS, 11.8

Step 3: Determine a critical value(s) for a. fa,m-2,n-m = f0.05,4,24 = 2.78

Step 4: Make a conclusion. Since fo = 0.05 > f a , , -,,,-, = 2.78, fail to reject HO at a = 0.05. It

is concluded that the order of the model is adequate.

Exercise 11.9 For the noise-hypertension data in Exercise 1 1-1, the following summary I quantities have been calculated:

I m

n=2O, m=10, SS, =35.7, SS,, = G ( j i -ii)' =4.8.and i=l

I Conduct a lack-of-fit test on the regression model j. = -9.8 1 + 0 . 1 7 ~ at a = 0.05.

I 1 9 Transformations to a Straight Line

O Explain the relationship between linear and non-linear models. -_(j. -ls ---Y_--s----w= ------- - - ~ - - * ~ w a - ~ ~ ~ ~ ~ " ~ w w -a #-4 *witm (CUYP- (i XIWIIXI PI - i -*- Transformation A non-linear relationship between independent (x) and dependent (Y) variables

of Nonlinear can be transformed into a linear model by mathematical operations. Examples are Model presented in Table 1 1-5.

P


Transformation Table 11-5 Transformation of Nonlinear Models (illustrated) of Nonlinear Nonlinear Model Linear Model

Model (cont.) Y = P ~ ~ ~ ~ ~ E InY=lnpo + P , x + E

11 -1 1 Correlation

Calculate the sample correlation coefficient (R) between random variables X and Y. LI Construct a confidence interval on the population correlation coefficient (p) between X and Y.

Conduct a hypothesis test on p.

Correlation vs. A close relationship exists between the correlation of random variables Xand Y Regression and the regression model of independent and dependent variables x and Y. Recall

that in the regression model the regressor x is assumed to be a mathematical variable, not a random variable (see Section 11-1).

Suppose that X and Yare random variables and the observations (X., Yi), i = 1,2, . . . , n, are obtained from a bivariate normal distribution. The population correlation coefficient p, (see Section 5-5) is

The conditional probability distribution of Y given X= x is a normal distribution with mean and variance (see Section 5-6)

CJ Y CJ Y = P o + P 1 x , where p o = p y -pxpm- and PI = p H - -

CJx '3,

Therefore, it is identified that the conditional probability distribution of Y given X = x is equivalent to the probability distribution of a simple linear regression model with mean DO + plx and variance 02.

1 1 - 1 1 Correlation 1 1 -31

Estimation of The estimator of p,, called sample correlation coefficient (denoted as R), is Correlation n

Coefficient z ( x i - X)(? - Y) (PXY)

R - 6 x & y 6 , - d s = / w ' Sm - 1 I R < 1

x ( x i - X ) 2 X ( ? -F12 i=l i=l

The square of R becomes the coefficient of determination R' of the regression model for Y and X having a bivariate normal distribution:

2 SSR (Derivation) R = - SST

2 sh - S R =-- because SS, = Syy S,S, S,SST

=- A S , 'ism because Dl = - SST S ,

-- - S S ~ because SSR = B 1 ~ , y j

SST

Inference Parameter of interest: p , ~ Context on s x

Pxr Point estimator of pxr: R = - S X S Y

Test statistic of p~ R-Po - t(n - 2) for testing Ho: p = 0

=

Zo = (arctanh R - arctanh po)- for testing Ho: p = p~ (i 0), n t 25

Confidence An approximate 100(1 - a)% confidence interval on p is Interval on

PXY

Hypothesis Step 1 : State Ho and HI. Test on H o : P ' PO

PI Hi: p*po (t-test)

Step 2: Determine a test statistic and its value. - - t n - 2) for testing Ho: p = 0

=

ZO = (arctanh R - arctanh pO)Jn - 3 for testing Ho: p = po (# 0), n 2 25

I 1-32 Chapter 11 Simple Linear Regression and Correlation

Hypothesis Step 3: Determine a critical value(s) for a. Test on t,,,,,-, for testing Ho: p = 0

P 1 (t-test) z,,, for testing H o : p = po (+ 0), n 2 25

(cont.) Step 4: Make a conclusion. Reject Ho if

It,l> t,12,,-2 for testing Ho: p = 0

IzOI > zaI2 for testing H o : p = po (# 0), n 2 25

Example 11.10 The relationship between ergonomics grade (X) and statistics grade (Y) is under study. A random sample of n = 50 students is selected who have taken both ergonomics and statistics courses. Their final scores in ergonomics and statistics are summarized as follows:

n = 50, S , = 6,652.5, S , = 3,999.0, and S , = 3,303.1

I Assume that ergonomics grade and statistics grade have a bivariate normal distribution.

I 1. (Sample Correlation Coefficient) Calculate the sample correlation coefficient I

4 1

(R) between ergonomics grade (X) and statistics grade (Y).

I 2. (Confidence Interval on pxy) Construct a 95% confidence interval on the population correlation coefficient (p) between X and Y. I " tanh(arctanh r - arctanh r + - Jn-3

arctanh 0.64 - - z0.05 1 2 zo,0512 ) < p 1 tanh(arctanh 0.64 + - JG' - , I s

3. (Hypothesis Test on pm) Test if p m # 0 at a = 0.05.

Step 1: State Ho and HI. Ho: p = O HI: p # O


I Step 3: Determine a critical value(s) for a. ta/2,n-2 = t0.0512,50-2 = t0.025,48 = 2.01



--

1 1- 1 1 Correlation I 1-33

Step 4: Make a conclusion. Since Itol = 5.78 > t , /2 ,n-2 = 2.01, reject Ho. It is concluded that

ergonomics grade (3 and statistics grade (Y) are significantly related at a = 0.05.

The measurements of weight (X) and systolic blood pressure (Y) for n = 26 randomly selected males in the age group 25 to 30 are collected, resulting in the following summary quantities:

n = 26, S , =4,502.2, S , =15,312.3, and S , =6,422.2

Assume that weight and systolic blood pressure are jointly normally distributed. 1. Calculate the sample correlation coefficient (R) between X and Y. 2. Construct a 95% confidence interval on the population correlation coefficient

(p) between X and Y.

3. Test if p~ # 0 at a = 0.05.



Examples

11.1-3,5,6,8

I (Simple Linear Regression)

I ( I ) Choose File * New, click Minitab Project, and click OK.

(2) eet.

& I Pr~ce 1 Sales Volume / I

1 I I

95 / I

1 (3) Choose Stat * Regression * Regression. In Res~0nSe select Sales - Volume and in predictors select Price.

(4) Click Graphs. For Residuals for Plots, select Regular or Standardized. Under Residual Plots, check Normal plot of residuals, Residuals versus fits, andlor Residuals versus order. In Residuals versus the

MINITAB Applications 11 -35

Examples 11.1-3,5,6,8

(cont.)

(5) Click Options. Check Fit intercept. In Prediction Intervals for new observations, enter the value of price of interest for the confidence interval on the mean response andor prediction interval of a future observation. In Confidence level, type the level of confidence. Then OK

click

1 (6) Click Results. Under Control the Display of Results, click the level of

I Regression Analysis

I The r e g r e s s ~ o n equa t ion 13 , , S a l e s Volume = 160 - 12.7 P r l c e

Source i DF MS P j Regression 1 ...A$;;, 14208 13g8.9: 0,000 1 Residual Er ro r ; 28 5 .,,.. ?,!4,! 10 11.3 T o t a l : 29 14493 : ............................................................................. *..

Obs P r l c e S a l e s Vo F i t StDev Flt Resldual S t Resld 1 5.0 95.000 96.190 1.032 -1.190 -0.39 2 5.0 98.000 96.190 1.032 1.810 0.60 3 5.0 96.000 96.190 1.032 -0.190 -0.06 4 5.0 92.000 96.190 1.032 -4.190 -1.39

5.0 99.000 96.190 1.032 2.810 0.93

. . . ........ . h.7.7. ........ -1.12 11.1.2 ,. .....8.8.... .. b'...b'....Hj. ...........BB. 0.774 y4 44............ 0.552 : 0.18

1.3y?'* 0.50 / - _ P - -

R denote8 en obse rva t lon wl th a l a r g e standardized r e s l d u a l

( P t e d l c t e d Values


Example 11.7 (Residual Analysis)

(1) Normal probability plot

Normal Probablll~ Plot of the Reslduals @*on.. 8. *I" la)

B ; E ;rl .. .. I '*

2 ,

2 .

Residual

I I Reslduals Versus Me Fnted Values mipas. 8. S.I.' W I

I Flned value

(2) Standardized residual plot

Reslduals Versus the Order of the Data '..Dm.. ts s.,.. W)

; 0 ........................ ............. .

6

16 m 30

Ob~ewstlon Order

(4) e, vs. x,

Answers to Exercises 11 -37


Exercise 11.1 ( 1. (Estimation of Regression Coefficients)

Exercise 11.2

The fitted regression line of the noise-hypertension data is

j=po +PIX=-9.81 +0.17x

2. (Calculation of Residual) A A

jx=*5 = P o + P1x = -9.81 + 0.17 x 85 = 4.76

ex,85 = YxES5 - jx=85 = 5 - 4.76 = 0.24

1 3. (Estimation of d )

I 1 . (Hypothesis Test on j31) Step 1 : State Ho and HI.

Ho: p1 = 0 H I : P l f O



tc(/z,n-2 - t0.025,18 = 2.10

Step 4: Make a conclusion. Since Ito(= 6.68 > t,,2,n-2 = 2.10, reject Ho at a = 0.05.

1 1-38 Chapter 1 1 Simple Linear Regression and Correlation

Exercise 11.2

(cont.) 2. (Hypothesis Test on Po)

Step 1 : State Ho and HI. Ho: po = 0 HI : po#o


Step 3: Determine a critical value(s) for a. ta/2,n-2 = t0.025,18 = 2.10

Step 4: Make a conclusion. Since Ito 1 = 4.6 > t,,, ,,-, = 2.10, reject Ho at a = 0.05.

Exercise 11.3 (ANOVA for Hypothesis Test on P I ) I

Regression 88.5 1 88.5 44.7

Error 35.7 18 2 .O

Total 124.2 19 n _ % ~ l x b _ _ e m ~ ~ ~ - ~ - m * B ~ - ~ W m ~ ~ m - - ~ u r r a a r r - * x l ~ m x w ~ x ~ r r r r a m M ~ r a l - a . r - ~ ~ - ~ - - m

(Note) t i = 6.7' = 44.7 = f o

Step 1 : State Ho and HI. Ho: PI= 0 HI : P I # 0

Step 2: Determine a test statistic and its value. SSR 11 MSR - 88.5 - 44.7 - ----- f0 =

S S ~ i(n - 2) M S ~ 2.0

Step 3: Determine a critical value(s) for a. fa , l ,n-~ = f0.05,1,20-2 = 4.41

Step 4: Make a conclusion. Since fo = 44.7 > f ,,,,,-, = 4.41, reject HO at a = 0.05.

Exercise 11.4

Exercise 11.5

. Answers to Exercises 1 1-39

1. (Confidence Interval on PI) 1 - a = 0 . 9 5 a a = 0 . 0 5 95% two-sided CI on P1:

8 1 - t a / i 7 n - 2 J G ~ ~ i '81 + t a / i . n - 2 J l S l r

=) 0.17 - to,05,2,20-2 I,/- I P1 I 0.17 + to.05,2,20-2 d- 3 0.17-2.10XO.O3IP, I 0 . 1 7 + 2 . 1 0 ~ 0 . 0 3

3 0.12 I P, 10.23

2. (Confidence Interval on Po) 1 - a = 0.95 a a = 0.05 95% two-sided CI on Po:

(Confidence Interval on y ylx, )

0 yls5 = Po + 8,xo = -9.81 + 0.17 x 85 = 4.7

1 - a = 0 . 9 5 a a = 0 . 0 5 95% two-sided CI on y ,Ig5 :

11 -40 Chapter

Exercise 11.6

Exercise 11.7


(Prediction Interval on yo)

jo =so + s l x o =-9,81+0.17~85=4.7

1 - a = 0.95 3 a = 0.05 The 95% two-sided prediction interval on yo at x = 85 dB is

(Note) The prediction interval 1.7 1 yo 17.8 is wider than the corresponding 1 1 confidence interval on the mean response 4.1 5 p,,,, 5 5.4 in Exercise 11-5. I

1 1

(Residual Analysis) The normal probability plot displays that the residuals lie closely along a straight line. In addition, the standardized residual plot shows that all the standardized residuals fall in the interval (-2,2). Therefore, it is concluded that the residuals are normally distributed.

Next, the plots ei vs. 5i and ei vs. xi indicate that the residuals are random and have a constant variance. The variability of the residuals is slightly reduced in the low ranges of y and x; however, this deviation from the constant variance is not serious enough to reject the adequacy of the regression model.

Exercise 11.8 1 (Coefficient of Determination)

2 SSR - 88.5 I R =- --- - 71.3% SS, 124.2

I The fitted regression model explains 71.3% of the variability in blood pressure rise.

Answers to Exercises 11 -41

Exercise 11.9

Exercise 11.10

(Lack-of-Fit Test) An ANOVA table to test the lack-of-fit of the regression model is as tbllows:

Source of Sum of Degrees of Mean Variation Squares Freedom Square Fo

Lack-of-Fit 4.8 8 0.6 0.20

Pure Error 30.8 10 3.1

Error 35.7 18

Step 1 : State Ho and HI. Ho: The order of the model is correct. HI: The order of the model is not correct.


MS,, - 0.6 - 0.20 fo = ------ - - - MS, 3.1

Step 3: Determine a critical value(s) for a. fa,m-2,n-m = f0.05,8,10 = 3.07

Step 4: Make a conclusion. Since fo = 0.20 P = 3.07, fail to reject Ho at a = 0.05. It

is concluded that the order of the model is adequate.

1 . (Sample Correlation Coefficient)

2. (Confidence Interval on pxy)

arctanh r - - arctanh r + -

arctanh 0.77 -- z0.051 2 ) < p < tanh(arctanh 0.77 + - JZT - J26-3

> 0.55 < p 10.89

3. (Hypothesis Test on pxy) Step 1: State Ho and H I .

Ho: p=O HI: p $ 0


12 Multiple Linear Regression

12- 1 Multiple Linear Regression Model

12-2 Hypothesis Tests in Multiple Linear

Regression

12-2.1 Test on Significance of

Regression

12-2.2 Test on Individual Regression

Coefficients and Subsets of

Coefficients

12-3 Confidence Intervals in Multiple Linear

Regression

12-3.1 Confidence Intervals on

Individual Regression

Coefficients

12-3.2 Confidence Interval on the

Mean Response

12-4 Prediction of New Observations

12-5 Model Adequacy Checking

12-5.1 Residual Analysis

12-5.2 Influential Observations

12-6 Aspects of Multiple Regression

Modeling

12-6.1 Polynomial Regression Models

12-6.2 Categorical Regressors and

Indicator Variables

12-6.3 Selection of Variables and

Model Building

12-6.4 Multicollinearity



12-1 Multiple Linear Regression Model

Explain the assumptions of a multiple linear regression model. R Transform a non-linear model into a linear model. O Calculate residuals. O Estimate the error variance 02.

Multiple As an extension of a simple linear model, a multiple linear model explains the Linear linear relationship between the response variable (Y) and k (> 1) multiple Model regressors (xi's):

Y = P O +Plxl + " ' + P k ~ k + &

where: (1) Regression coefficients (Po, PI , . . . , Pk): The coefficient Po denotes

12-2 Chapter 12 Multiple Linear Regression

Multiple the intercept of the model and the coefficients Pj's, j = 1 to k, denote Linear the slopes of the regressors. Each slope coefficient Pj measures the Model expected change in Y per unit change in xi when the other regressors (cont.) are held constant (this is why Pj's are sometimes called partial

regression coefficients). (2) Independent variables (XI, xz, . . . , xk; regressors, predictors):

Since the regressors xi's will be controlled with negligible error, xj's represent controlled constants (not random outcomes).

(3) Random error (E): The variable E represents the random variation of Y around the linear line Po + P,x, + .. . + P xk . It is assumed that

E is normally distributed with mean 0 and constant variance 02, i.e., 1 E - N ( o , ~ ~ )

(4) Dependent (response) variable (Y): Since outcomes in Y at the same values of xl, x2,. . . , xk can vary randomly, Y is a random variable with the normal distribution

Y - N(Po +P,x, +...+ P k x k , 0 2 )

Like the assumptions of a simple linear model (see Section 1 1-I), the multiple linear model has four assumptions:

4

(1) Linear relationship between x,'s and Y i

(2) Randomness of error (3) Constant o2 (4) Normality of error

The analyst should check if these four assumptions are met for proper regression analysis (see the assessment of model adequacy in Section 12-5).

Note that computers are mostly used in multiple regression analysis so that the detailed calculation process is not presented in this chapter.

Transformation Like the transformation of a non-linear model in simple linear regression analysis of Nonlinear (see Section 1 1-9), a non-linear relationship between xj7s and Y can be

Model transformed into a linear model by applying appropriate mathematical operations. Therefore, any regression models with regressors linearly combined can be treated as a linear model.

(e.g.) Transformation of a non-linear model

(1) Y=Po + P I X I + P 2 ~ 2 + P I Z X ~ X ~ + E

Y = P O + Plxl + P2x2 + P3x3 + E , where j3, = PI2 and x, = xlx2

1 3 3 Y = PO + Plxl + P,x, + P3x3 + P4x4 + E , where x, = - and x, = x,

XI

Matrix Matrix notations and operations are useful in multiple regression. Suppose that n Notation observations consisting of k regressors and the response variable, (xi,, xi2,. . . , xik,

yi), i = 1,2, . . . , n, are modeled by a multiple linear model

yl = P o +Plxi l + . . . + P k ~ i k + E ~ , i = 1 , 2 ,..., n

12-1 Multiple Linear Regression Model 12-3

Matrix The system of n equations is expressed in matrix notation as follows: Notation y = X $ + &

(cant.) where:

I!'] and & =I] P = .

P k E n

Model The least squares method finds the estimates of regression coefficients $ which Building minimize the sum of the squares of the errors

The least squares estimators of $ are

fi = (xfx)-' x'y

Thus, the fitted regression model is

i = xfi

Residual and

Error Variance

Standard Error of

Regression Coefficients

B j

The residuals of the fitted model are e = y - i

The estimate of the error variance G~ is

(J A =- S S ~ - - "' - "x" , where p = number of regression coefficients n - P n - P

= k + 1 (k = number of regressors)

The covariances of 6 are

COV($) = 02(x'x))-l = a2c , where C = (x'x)-'

This covariance matrix is a 07 x p ) symmetric matrix whosejjth component is the

variance of B and whose ij" component is the covariance of B, and fi , where i

and j = 0 to k, i # j.

Then, the estimated standard error of fi is

se(B, ) = &&

Thus, e = y - 9 = 301 - 291.8 =9.2 (underestimate)


Standard The standard error of a regression coefficient is a useful measure to represent the Error of precision of estimation for the regression coefficient: the smaller the standard

P, error, the better the estimation precision.

(cont.)

I 2. (Estimation of 02) Estimate the error variance 02.

U-F n = 12 and p (number of regression coefficients) = 5

Example 12.1

I s S ~ - - 242.7 = 15.6' 6 2 = - - -- n - p 12-5

The electric power (Y) consumed each month at a chemical plant is under study. It

By using a statistical software program, a fitted regression model of the power consumption data including all the four regressors (xl, x2, x3, and x4) and the corresponding error sum of squares (SSE) are obtained as follows:

i = -102.71 + 0 . 6 1 ~ ~ + 8 . 9 2 ~ ~ + 1 . 4 4 ~ ~ + 0 . 0 1 ~ ~ and SSE = 1,699.0

1. (Calculation of Residual) Calculate the residual ofy = 301 when XI = 65"F, x2 = 25 days, x3 = 91%, and x4 = 94 tons.

U-F The estimate of the power consumption given that xl = 65, x;, = 25, x3 = 91, and xq = 94 is

j = -102.71 + 0 . 6 1 ~ 6 5 + 8 . 9 2 ~ 2 5 + 1 . 4 4 ~ 9 1 + 0 . 0 1 ~ 9 4 = 291.8

12-2 Hypothesis Tests in Multiple Linear Regression 12-5


I The pull strength of a wire bond is under study. Information of pull strength ('y), die height (XI), post height (xz), loop height (xg), wire length (x4), bond width on

I the die (xs), and bond width on the post (xg) is collected as follows:

By using a statistical software program, a fitted regression model of the pull strength data including four regressors ( ~ 2 , x3, x4, and x ~ ) and the corresponding error sum of squares (SSE) are obtained as follows:

5 = 7.46 - 0 . 0 3 ~ ~ + 0 . 5 2 ~ ~ - 0. lox4 - 2 . 1 6 ~ ~ and SSE = 10.9 1 I. Calculate the residual of y = 9.0 when x2 = 18.6, x3 = 28.6, x4 = 86.5, and x5 =

2.0.

2. Estimate the error variance a2.

! 12-2 Hypothesis Tests in Multiple Linear Regression

12-2.1 Test on Significance of Regression

CI Test the significance of multiple regression.

12-6 Chapter

Significance of Regression

Test Procedure

(F-test)

Multiple Linear Regression

The test for significance of regression determines if there is at least one xj, out of k regressors, that has a significant linear relationship with the response variable Y, i.e.,

Ho: P I = P 2 = = P k = O (3 P = O )

HI: Pi # 0 for at least one j = 1 to k

Like simple regression, the ANOVA method is used to test the significance of regression. For the multiple linear model, in Table 12-1, the total variability in Y (SST or S,,, total sum of squares; degrees of freedom = n - 1) is partitioned into two components:

(1) Variability explained by the regression model (SSR, regression sum of squares; degrees of freedom = k)

(2) Variability due to random error (SSE, error sum of squares; degrees of freedom=n-k-1 = n - ( k + l ) = n - p )

Table 12-1 Partitioning. of the Total Variabilitv in Y (SST) Q - \ ,,

SST SSR SSE (Total SS) (Regression SS) (Error SS)

DF n w - w a w * - # a * . A 8 ~ " * m ~ ~ w w ~ - * v ~ . -

(Note) SS: Sum of

The ratio of S&/k to SSE/(n - p ) follows an F distribution: SSR l k

F = - --- M S ~ ~ ( k , n - p) SS, l(n - p) MS,

Since F becomes as Ho: Pl = P2 = ... = Pk = 0 is true, it is used to test the significance of regression. Note that MSR (regression mean square) and MSE (error mean square) are adjusted SSR and SSE by their corresponding degrees of freedom, respectively. Also note that MSE is an estimate of the error variance:

The variation quantities from the ANOVA analysis are summarized in Table 12-2 to test the significance of regression.

Table 12-2 ANOVA Table for Testing the Significance of Regression v - Source of Sum of Degrees of ~ e &

Fo P-value Variation Squares Freedom Square Regression SSR k MSR MSRIMSE

Error SSE n - P MSE

Total n - 1


Test Step 1 : State HO and HI. Procedure ff0: p1 = p 2 = ... = p k-0 -

(F-test) HI: p j + O foratleastonej,j= 1 tok (cont.)

Step 2: Determine a test statistic and its value. SS, 1 k

F, = - --- MSR F(k,n-p) SS, l(n - p ) MSE

Step 3: Determine a critical value@) for a. f a , k , n - p


f 0 ' f a , k , n - p

R' The coefficient of determination (denoted as R2) indicates the amount of vs. variability in Y explained by the regressors XI , x2, . . . , xk:

Adjusted R~ SSE 2 S S ~ =I--, 0 5 ~ ~ 1 1 R =- ~ S T ~ S T

As discussed in Section 11-8.2, a large value of R2 does necessarily imply that the regression model is adequate. As a new regressor is added to the existing model, R2 will always increase regardless of whether the additional regressor is statistically significant or not; thus, a regression model having a large value of R2 may unsatisfactorily estimate mean responses or predict new observations.

In contrast, the measure adjusted R2 considers the number of regressors (p) included in the model in the computation:

SSE l(n -PI n-1 adjusted R2 = 1 - =I-- (1-R2), 0<adjustedR2 I 1 SS, l(n - 1) n - P

Adjusted R2 is preferred to R~ when comparison of several regression models having different numbers of regressors is needed. While R2 simp1 increases as an Y additional regressor is included in the existing model, adjusted R may decease if the reduction in SSE by addition of a new regressor to the existing model is smaller than the MSE of the existing model. -

Example 12.2 For the power consumption data in Example 12-1, the following ANOVA table is I obtained: Source of Sum of Degrees of Mean Variation Squares Freedom Square P-value

Regression 4,957.2 4 1,239.3 5.1 0.03

Error 1,699.0 7 242.7

Total 6,656.3 11

1. (Test on the Significance of Regression) Test the significance of regression at a = 0.05.

12-8 Chapter




Step 1 : State Ho and HI. Ho: P 1 = P 2 = P 3 = P 4 = 0 HI: P , # O foratleastonej,j= 1 to4

Step 2: Determine a test statistic and its value. SSR 1 k MSR 1,239.3

f o = - -- - --~5.1

SS, l(n - p ) MSE 242.7

Step 3: Determine a critical value(s) for a. f a , k , n - p = f0.05,4,7 = 4.

Step 4: Make a conclusion. Since fo =5.1> fa,, ,,-, =4.12,rejectHoat a = 0.01. This

indicates that at least one of the four regressors has a significant linear relationship with the monthly power consumption at a = 0.05.

2. ( R ~ and Adjusted R') Calculate the coefficient of determination (R2) and adjusted RZ of the regression model.

h - 4 957 2 R' = S S R - z . 7 4 . 5 %

SS, 6,656.3 The four regressors (xl, x2, x3, and x4) explain 74.5% of the variability in the power consumption Y.

2 n - 1 12-1 Adjusted R = 1 - - (1-R2)=1-- (1 - 0.745) = 0.599

n - P 12-5

For the pull strength data in Exercise 12-1, the following ANOVA table is obtained by multiple regression including the four regressors (x2, x,, x4, and x5):

I Source of Sum of BegrePsof Mean Squares Freedom Square

fo P-value Variation

1 Regression 22.31 4 5.58 7.16 cO.01

Error 10.91 14 0.78

Total 33.22 18 -x_l - *a/*--_*1P_ *, -w.w--Y,% m ~ ~ ~ ~ * e _ - ~ ~ n _ ~ ~ X ~ - ~ ~ m a - ~ e e - ~ m ~ - - u r x r

1. Test the significance of regression at a = 0.05. 2. Calculate the coefficient of determination (R2) and adjusted R2 of the

regression model. 1 1

12-2.2 Tests on Individual Regression Coefficients and Subsets of Coefficients

O Test the significance of an individual regressor. Cl Test the significance of a subset of regressors.


Significance of Like the testing on the significance of a regression coefficient in simple Individual regression (see Section 1 1-5. l), the significance of any individual regressor x, is Regressor evaluated by t test (called partial or marginal test).

Inference Parameter of interest: p Context on

Pi Point estimator of p, : p, - ~ ( p , , 02cJ7) , $ is unknown.

p . p . - p . J J - J J Test statistic of P : T = T -

se(P, > (Note) Estimated standard error of p, : se(p , ) = ,/= , where 8' = MSE

Hypothesis Step 1 : State Ho and H I . Test on Ho: pj = pj,0

Pi HI Pj # P,,o (t-test)




It0 1 > tat 2,n-p

Example 12.3 (Hypothesis Test on Pi) Consider the power consumption data in Example 12-1. The following quantities are obtained for production volume ( ~ 4 ) in the regression analysis:

p4 = 0.014, se(b4) = 0.734, n = 12, andp = k + 1 = 5

Test the significance of x4 at a = 0.05. * Step 1 : State Ho and HI.

Ho: p 4 = 0 H,: P 4 f O


0 4 - p4.0 0.014 - 0 to = - - = 0.02

se(b4 > 0.734


ta/2,n-p - t0.0512,12-5 = '0.025,7 = 2.365

Step 4: Make a conclusion. Since )to] = 0.02 2 = 2.365 , fail to reject HO at a = 0.05.

Significance of The significance of a subset of regressors, xl, x2, . . . , x, (r < k), is tested by the the Subset of extra sum of squares method (partial F-test; general regression significance

Regressors test), which is an extension of the ANOVA method explained in Section 12-2.1. Like the test for significance of regression, this test determines if there is at least one xi, out of r regressors, that has a significant linear relationship with Y, i.e.,

Ho: p 1 = p 2 = ... = p r = 0

HI: P j # O foratleastonej,j= 1 t o r

12-1 0 Chapter 12 Multiple Linear Regression

I

To test the significance of the subset of regressors, let the vector of regression coefficients ( p ) be partitioned into $ , for the regressors xl , x2, . . . , xr and fl , for

'I Exercise 12.3

I

the other regressors x,l, x ,+z, . . . , xk:

Consider the wire bond pull strength data in Exercise 12- 1. The following quantities are obtained for bond width on the die (x5) in the regression analysis:

b5 =-2.16, se(b,) =2.39, n = 1 9 , a n d p = k + 1 = 5

Test the significance ofxs at a = 0.05.

Also let X, and X2 denote the columns of the regressors X related to P, and

those related to P, , respectively. Then, the multiple linear model is

J'=X$+E=Xlfll + X 2 p 2 +E

In Section 12-2.1, the regression sum of squares (SSR) and mean error square (MSE) are

where (j = (x'x)-' X'Y . Note that SS, (p) indicates the contribution of the regressors X to the variability in Y.

Now, to identify the contribution of the regressors X, , fit a reduced model for

X 2 , i.e.,

Y =X2P2 + E

The least squares estimate of P, is $, = (x; X, )-' Xiy and the corresponding regression sum of squares is

2

[k i=l yi 1 SS,(P,>= $;x',y -

n

12-2 Hypothesis Tests in Multiple Linear Regression 12-1 1

Significance of the Subset of

Regressors (cont.)

Test Procedure

(F-test)

Then, the regression sum of squares due to given that P 2 is already in the

model is

ssR($l I P ~ ) = ~ ~ R ( P ) - ~ ~ R ( P ~ ) = ~ ' x ' Y -~;x;Y

In other words, SS, (PI I P 2 ) indicates an increase in regression sum of squares

by adding XI to the reduced model, which is called extra sum of squares due to

. The ratio of SS, ( P I I P , ) lr to SSE/(n - p ) follows an F distribution:

Like the full model, the statistic F becomes as Ho: P I = P2 = = Pr = 0 is true.

The variation quantities of the ANOVA analysis for testing the significance of the subset of regressors P1 are summarized in Table 12-3.

Table 12-3 ANOVA Table for Testing the Significance of a Subset of Regressors P I


Regression ( P ) SSR k MSR MS, I MSE

r MSR(P, I P z ) MSR(P1 I P z )

$1 SS, ( P I l P z ) MSE

P z SSR (Pz ) k - r MSR(PZ) Error SSE n - P MSE

Total n - 1

Step 1 : State Ho and HI. Ho: P 1 = O (PI = P 2 = ... = P r = O ) HI : P 1 $ 0 ( P , + O for at leastonej , j= 1 tor )


Step 3: Determine a critical value(s) for a. f a , r , n - p


f o > f a , r , n - p

I I 12-1 2 Chapter 12 Multiple Linear Regression

Example 12.4 (Test on the Subset of Regressors) Consider the power consumption data in I I i Example 12- 1. Suppose that the four regressors (P) are grouped into two 1

b Exercise 12.4

categories: 3 3

(1) Non-production regressor (PI): temperature (x,) I

(2) Production regressors (Pz): number of days (xz), product purity (xg) and , production volume (x4)

Multiple regression is performed separately with the four regressors (P) and the non-production regressor (PI), yielding the ANOVA results below.

I Source of Sum of Degrees of Mean Variation Squares Freedom Square fo

( Regression (P) 4,957.2 4 1,239.3 5.1

I Error 1,699.0 7 242.7


Regression (PI) 3,758.9 1 3,758.9 13.0

Error 2,897.3 10 289.7

Total 6,656.3 11

Test the significance of the production regressors (P2) at a = 0.05. ~ - r r Step I : State Ho and HI.

Ho: p 2 = 0 (p2 = p 3 = p4 = 0) HI: P 2 f 0 (pj#Oforat least o n e j , j = 2 to4)

Step 2: Determine a test statistic and its value. SSR (P2 I PI) = SS,(P) - SSR (PI) = 4,957.2 - 3,758.9 = 1,198.3

Step 3: Determine a critical value(s) for a. fa,r,n-p = f0.05,3,7 = 4.35

I Step 4: Make a conclusion. Since fo = 1.65 > fa,,,,-, = 4.35, fail to reject HO at a = 0.05. This

I indicates that none of the production regressors have a significant linear relationship with the power consumption at a = 0.05.

Consider the wire bond pull strength data in Exercise 12- 1. Suppose that the six regressors (p) are grouped into two categories:

(1) Process regressors (PI): die height (XI), post height (xz), and loop height ( ~ 3 )

(2) Wire-bond regressors (P2): wire length (x4), bond width on the die (x5), and bond width on the post (x6)

12-3 Confidence Intervals in Multiple Regression 12-1 3

Exercise 12.4 Multiple regression is performed separately with the six regressors (P) and the (cont.1 I wire-bond regressors ((Jz), yielding the ANOVA results below.

I . - -

Source of Sum of Degrees of Mean Variation Squares Freedom Square fo

I Regression (P) 23.63 6 3.94 4.93

I Error 9.59 12 0.80

I Total 33.22 18 fl *ljY"UlfX U~j lXU.wm.mw~.m'~- ,~"s~~ e-sm.e-. ~ , a . ~ m ~ " , w ~ ~ ~ " ~ A 8 w ~ m ~ w ~ / ~ ' m m-mm~.*--*.~vaM.B~d. ~ - ' ~ ~ ~ ~ ~ ~ " w ~ w a v ~ ~ # ~ w m " ~ ~ . ~ . w ~ ~ ~ . ~ ~ ~ ~ ~ ~ w > , ~ ~ . a ~ . ~ . ~ , a w*wal in . ..,l*iP. *I_i-Xve.,i *"* a


I Regression (p2) 10.8 1 3 3.61 2.41

1 Error 22.41 15 1.49

Total 33.22 18 w an e-B.-MC_wM*I# Um%#%wmmm -m -- -m- -m-m --A%.w wau#va-%"w -- i * - *msn - ne aau-mxn-aar w r a a ~ ' , r n ~ W ~ ~ - e rn nna nnnw*x-a rxraxx

Test the significance of the process regressors (PI) at a = 0.05.

12-3 Confidence Intervals in Multiple Regression

12-3.1 Confidence Intervals on Individual Regression Coefficients

" . , m ~ . ~ , " . , ~ < ~ ~ ~ r ~ ~ " ~ ~ ~ " ~ ~ . - ~ . d ~ ~ B % ~ , ~ " ~ a ~ ~ ~ ~ , w - " ~ . " , , ~ , w a ~ , " * ~ ~ ~ ' ~ a ~ ~ ~ ~ ~ . b -.~~a%ms".m",m~~m~m mwm.'* ,"* .","ea,.~~.-*.~m.- *-.- "v"vA ,."*"""."v*'"<'

LI Establish a confidence interval on pi.

Inference Context on

Pi

Confidence Interval on

Pi Example 12.5

p . - p . p . - p . J J - J J Test statistic of pJ : T = ------

se(B I - 4- - t(n - P)

(Note) Estimated standard error of B : se(B j ) = d-, where 8' = MSE

A 100(1 - a)% confidence interval on p, is

BJ - t a i 2 , n - p * l ~ 5 p j 5Bj + t . / 2 , n - j p q

(Confidence Interval on pi) Consider the power consumption data in Example 12-1. The following quantities are obtained for production volume (x4) in the regression analysis:

B4 = 0.014, se(B4) = 0.734, n = 12, andp = k + 1 = 5

Establish a 95% two-sided confidence interval on p4.

n- 1 - a = 0 . 9 5 3 a = 0 . 0 5 95% two-sided CI on p4:

B 4 -ta/2,r-pJZric,,5 ~4 5 B 4 +ta/l,n-pJ-

3 0.014 - t ,,,, ,,,,-, x0.7345D4 50.014 + t,,,12,,2-, ~ 0 . 7 3 4

Example 12.5 0.014 - 2.365 x 0.734 5 p4 5 0.014 + 2.365 x 0.734 (cont.) I 3 -1.725P4 11.75

12-3.2 Confidence Interval on the Mean Response


--B w-s-l_l(jll--Y,-w-*wm--%m-- --- wLs-XIX(- I-mx.XIIw eU-ii*-i*D---I

Establish a confidence interval on the mean response at x = xo. **U*VP( m *-*a - - * ' *a (/- s.x_---n I? -n-B"IXwm---m--Bw---Yw M sw- w a w-m e s - a m -wm- . . * s m - s e w _ x amsaaw w ~ w m s m n x ra ;-a *x- ram rx m-ws,,Bmm *- a a x u r n axcen a t a m * r - * a **a* a

Consider the wire bond pull strength data in Exercise 12-1. The following quantities are obtained for bond width on the die (XS) in the regression analysis:

Pi = -2.16, se(p,) = 2.39, n = 19, andp = k + 1 = 5

Establish a 95% two-sided confidence interval on PS

The mean response at x = xo is

Prjxo = E(Y I xo) = E(XP + E I xo) =

Sampling Suppose that the vector xo represents a particular point of the regressors X I , x ~ , . . . , Distribution of x,:

Then, the point estimator of p ylxo is

Mean Response

Estimator a t x = x , X o =

( PYlx=xo

The sampling distribution of fi y,xo is

- - 1

Xo 1

X02

-XOk -

Inference Parameter of interest: p yl%

Context on Mean Point estimator of p ylxo : fi ylxo = x;B - ~ ( p ylxo , 02xb (x'x)-' xo ) , d is

Response unknown. a t x = x , Pyjx0 - C L Y / ~ ,

(cl,lx0 Test statistic of p ylxo : - t(n - P)

--

(Note) Estimated standard error of fi ylxo : se(P ylxo ) = ,/e2x; (X'X)-' xo ,

where e2 = MSE .

12-4 Prediction of New Observations 12-1 5

Confidence A 100(1 - a)% confidence interval on p y,xo is Interval on

p ~ 1 x 0 fiylxo -to/2,n-p d m I ~y~~~ I 6 Y I x 0 +to/2,n-p

"4 Example 12.6


(Confidence Interval on p ) Consider the power consumption data in

Example 12-1. The following quantities are obtained for the mean power consumption when xl = 65"F, x2 = 25 days, x3 = 91%, and x4 = 94 tons in the regression analysis:

P Y I X , =291.81, se(Pylxo) =7.68, n = 1 2 , a n d p = 5 Establish a 95% two-sided confidence interval on the mean response.

~ - r r 1 - a = 0 . 9 5 3 a = 0 . 0 5 95% two-sided CI on p :

Consider the wire bond pull strength data in Exercise 12-1. The following quantities are obtained for the mean pull strength when xz = 18.6, x3 = 28.6, x4 =

86.5, and x5 = 2.0 in the regression analysis:

Py lx0 = 8.66, se(P ) = 0.57, n = 19, andp = 5

Establish a 95% two-sided confidence interval on the mean response.

12-4 Prediction of New Observations

s e p v a * m * " #* ww ws wm MMm-m- wBw a * * * ** * - * -- # 3 v A

Establish a confidence interval on the mean response at x = xo.

Prediction Suppose that the vector x', = [l, x,, , xo2 , ... , x,, ] denote a particular point of the Interval On regressors XI, ~ 2 , . . . , xk. Then, a point estimate of a future observation yo at

Future x=x' , is Observation

A 100(1 - a)% prediction interval on yo at xo is

j,, -to,,,-, de2 (1 + x i (xx)-' xo) I yo I ja +t,,2,n-p \/e2 (1 + x;(xfx)-l xO)

As discussed in prediction of new observations in simple regression (Section 1 1 - 7), a prediction interval (PI) is always wider than the corresponding confidence interval (CI) on the mean response because the PI depends on both the error from the fitted regression model and the inherent variability in the random variable Y, while the CI depends on only the error from the fitted regression model.

I 12-1 6 Chapter 12 Multiple Linear Regression

v Example 12.7


(Prediction Interval on yo) Consider the power consumption data in Example 12-1. The following quantities are obtained in the regression analysis for a future power consumption observation when x2 = 18.6, x3 = 28.6, x4 = 86.5, and x5 =

2.0: Po = 291.81, 6 2 x b ( ~ ' ~ ) - ' x , = 7.682, 62 = 1 5 . 5 8 ~ , n = 12, andp = 5

Establish a 95% two-sided prediction interval on the power consumption.

rrn 1 - a = 0 . 9 5 3 a = 0 . 0 5 95% two-sided CI on yo:

$0 -hi 2,n-p 6 YO 6 F~ + la, 2,n-p 4-

Consider the wire bond pull strength data in Exercise 12-1. The following quantities are obtained in the regression analysis for a future pull strength observation when XI = 65"F, x2 = 25 days, x3 = 91%, and x4 = 94 tons:

Po = 8.66, e2x; (x'x)-'x, = 0 . 5 7 ~ , e2 = 0 . 8 8 ~ , n = 19, andp = 5 Establish a 95% two-sided prediction interval on the pull strength.

12-5 Model Adequacy Checking

12-5.1 Residual Analysis

O Describe the purpose of residual analysis.

Residual Recall the four assumptions of a multiple linear model (see Section 12-1): (1) Analysis linearity between xj's and Y, (2) randomness of error, (3) constant variance of

error, and (4) normality of error. These four assumptions can be checked by analyzing residuals (e i = yi - j i ) , which is called the assessment of model adequacy.

The first three assumptions (linearitv, randomness, and constant variance) can be checked by examining the following residual plots:

(1) ei vs. ji (2) e, vs. xv, j = l , 2 , . . . , r

(3) e, vs. t (if time sequence is known)

Next, the normalitv of error is checked by one of the following: (1) Frequency histogram of residuals

12-5 Model Adequacy Checking 12-1 7

Residual (2) Normal probability plot of residuals Analysis (3) Standardization of residuals: If the residuals are normally distributed,

(cont.) about 95% of the standardized residuals

are in the interval (-2,2). Residuals whose standardized values are beyond f 3 may be outliers.

An undesirable residual pattern (see Section 1 1-8.1) may be due to non-linearity or insignificance of the relationship between the corresponding regressor and response variable (Y). To correct an undesirable residual pattern, the appropriate method out of the following can be applied:

( I ) Transform xj or y into another form (such as lly, f i , 9, and in y).

(2) Add a high order term(s) of xj to the model. (3) Include a new regressor(s) in the model.

12-5.2 Influential Observations

e* a*

O Identify influential observations by using the measure Cook's distance.

Cook's The measure Cook's distance (denoted by D;) is the squared distance between the Distance least squares estimate fi based on all n observations and fie, based on all the data

(Di) except the ith observation:

As the ith observation is unusually influential in determining the estimates of regression coefficients, Di becomes large-a value of Di > 1 would indicate that the point is influential. If an influential observation is due to error during the experiment, this observation should be excluded from the analysis.

Example 12.8 (Influential Observation) Consider the power consumption data in Example 12- I 1. The Cook's distances of the observations are calculated by using software: Power Power

consumption Cook's consumption Cook's No. (v; unit: kW) Distance (Di) No. 01; unit: kW) Distance (Dr)

316 0.20 I2 261 0.01 "%Xw _ U je;j x-W." X(s*mm-w%.- m**As%Mw*.,--* **aw-a- * * v # - A # e w w ma xm-2 w - x w e w & d " . m Y w e - m - ~ B - ~ m ~ *

Check if any influential observations exist in the data. (b No influential observations whose Di > I are found in the data.

12-1 8 Chapter 12 Multiple Linear Regression

Exercise 12.8 Consider the wire bond pull strength data in Exercise 12-1. The Cook's distances (MR 12-40) I of the observations are calculated by using software as follows: -

Pull strength Cook's Pull strength Cook's 6)) Distance (D)) No. 01) Distance (Di)

12-6 Aspects of Multiple Regression Modeling

12-6.1 Polynomial Regression Models

O Establish a polynomial regression model. O Test the significance of a high order term.

Polynomial When the response is curvilinear, a polynomial (a linear combination of powers Regression in one or several variables) can be fit by using a multiple linear regression model.

Model For example, a second-order polynomial in one variable can be transformed to a linear regression model as follows:

Y=PO + p 1 x + P l l x 2 + &

When fitting a polynomial model, the lowest-order model is preferred for simplicity reasons (see Section 12-6.3). The significance of a higher-order

- , term(s) is tested by t test or the extra sum of squares method.

Example 12.9 (Polynomial Model) Consider the power consumption data in Example 12-1. By I including only temperature (xl), the following second-order regression model is 1 developed:

j = P o + Blxl + P2x: = 162.76 + 3 . 5 0 ~ ~ - 0.02~: The corresponding ANOVA table is as follows:

12-6 Aspects of Multiple Regression Modeling 12-1 9

Example 12.9 I Sum of Degrees of Some of Variation SquaRs Mean (cont.) Freedom Square fo


Regression (xl, x: ) 4,362.8 2 2,181.4 8.6

SSR (PI I Po) 3,758.9 1 3,758.9 13.0

SSR(P2 I Pi,Po) 603.8 1 603.8 2.4

Error 2,293.5 9 254.8

Total 6,656.3 11

Test if the quadratic term of temperature (x:) is significant at a = 0.05.

h Step 1: State Ho and HI. Ho: p 2 = 0 HI: P 2 f 0


Step 3: Determine a critical value(s) for a. f a , r , n -p = f0.05,1,9 = 5.12

Step 4: Make a conclusion. Since fo = 2.4 P f ,,,,,-, = 5.12, fail to reject Ho at a = 0.05. This

indicates that the quadratic term does not significantly contribute to the model.

The following data are collected during an experiment to determine the change in thrust efficiency O/, in percent) as the divergence angle of a rocket nozzle (x) changes:

Y x

Y x

A polynomial model is developed for the thrust efficiency data:

>=so + six + p2x2 =-4.46 + 1 . 3 8 ~ + 1.47x2

ding ANOVA table is as follows:

Regression (XI, x; ) 5,740.6 2 2,870.3 1,045.0

s s~(P , IPo) 5,716.3 1 5,7 16.3 1,167.0

s s ~ ( P 2 IP1,Po) 24.3 1 24.3 8.8 Error 24.7 9 2.7

Total 5,765.3 11 - ~ . ~ ~ , ~ ~ ~ ~ _ _ ( _ I . X ~ . ~ ~ ~ ~ % ~ ~ ~ ~ - m . w a ~ ~ ~ B ~ - - ~ . ~ ~ - ~ d ~ - M - - - - ~ ~ ~ . ' ~ " - r miu*l*w--rmelw-res nl.n*-MPxraaa iarxi.e.arrd arr ra*- i=.n nw alar e n am Irrnr--larx - Test if the quadratic term of the nozzle divergence angle ( x 2 ) is significant at a = 0.05.

24.60 4.0

67.1 1 6.5

24.71 4.0

67.24 6.5

23.90 4.0

67.15 6.75

39.50 5.0

77.87 7.0

39.60 5.0

80.1 1 7.1

57.12 6.0

84.67 7.3

r


12-6.2 Categorical Regressors and Indicator Variables

Explain use of indicator variables in regression analysis. w-~"*~~"a~ - - .~ -~ .~ - -m~ s ~ - H B f * ~ " ~ . ~ ~ ~ a ~ ~ * ~ . w # w m , ~ m . a , ~ m " ~ . ~ m ~ ~ ~ ~ u ""*.ms*s*w sB*m-m.*,9s"~B~a*m.w,eaes~.B Be.~.me~a*~-w#a.~*~.-B*.B-~ m"m---B*B*B w"s~m.-",--w*---Bw~~"~m.*-*~~.~~~b~""~m.w~-a"~>w>~~*~"-

Indicator The regression models presented in previous sections are based on continuous Variables (quantitative) variables. However, categorical (qualitative) variables such as

gender and education level are sometimes considered in regression analysis. To incorporate categorical variables in regression, indicator (dummy) variables are used. A categorical variable with c levels can be modeled by c - 1 indicator variables.

r Example 12.10

(e.g.) Indicator variables (1) gender = (male, female}

0 for male

1 for female

(2) education = {high school, college, and postgraduate} 1 for high school for college for postgraduate

Another method of analyzing data including categorical information is to fit a separate regression model to the data of each category. Compared to this approach, the indicator variable approach has the following advantages:

1 . One regression model: One regression model integrates several regression models for different categories.

2. Increased statistical power: By pooling the data of different categories, the error term will have a smaller mean square error due to a larger number of degrees of freedom (n - p ) , which results in increased power in statistical testinn.

(Indicator Variable) A relationship between liver weight @; unit: g) and body surface area (xl; unit: m2) is under study. The following measurements are obtained from 20 cadavers (10 males and 10 females; gender (x2) = 0 for male and 1 for female):

12-6 Aspects of Multiple Regression Modeling 12-21


By assuming that the interaction between xl and x2 is negligible, a fitted regression model is obtained for the liver weight data:

F = P o + Plxl + P2x2 = -108.7 + 9 7 7 . 0 ~ ~ - 1 . 3 ~ ~

I Predictor Coefficient to P-value Standard error


I Predictor Coefficient to P-value Standard error

Constant -108.7 829.2 -0.13 0.897

BSA (XI) 977.0 502.7 1.94 0.069

b Since the P-value for P2 is greater than 0.05, gender (xl) does not have a significant effect on liver weight (y) at a = 0.05.

A mechanical engineer is investigating the surface finish 0/; unit: pm) of metal parts produced on a lathe and its relationship to the speed (xl; unit: RPM) of the lathe. The following measurements are obtained by using two different types of cutting tools (x2 = 0 for tool type 302 and 1 for tool type 416):

Type of Type of Surface Lathe cutting Surface Lathe cutting finish speed tool finish speed tool

No. O ( ~ 1 ) ( ~ 2 ) NO. (0 (XI) ( ~ 2 )

7 52.26 265 0 17 32.13 224 1 8 50.52 259 0 18 35.47 25 1 1 9 45.58 22 1 0 19 33.49 232 1

1 ".",*%"A'*"" ,.., ,w

The following regression model is fit to the surface finish data by using software:

I 5 = Po + Plxl + P2x2 + Ij3X1X2 = 11.50 + 0 . 1 5 ~ ~ - 6 . 0 9 ~ ~ - 0 . 0 3 ~ ~ ~ ~

Constant 11.50 2.50 4.59 cO.01

Lathe speed (xl) 0.15 0.01 14.43 <0.01

Tool type (x2) -6.09 4.02 -1.51 0.15

X1 X X2 -0.03 0.02 -1.79 0.09

required to adequately model the tool life data.


12-6.3 Selection of Variables and Model Building

R Explain the features of a satisfactory regression model. Compare variable selection techniques with each other: (1) all-possible regression evaluation; (2) forced selectiodelimination; (3) forward selection; (4) backward elimination; and (5) stepwise regression.

O Determine the best set of regressors by the all-possible regression evaluation technique. O Select a subset of regressors for model building by the stepwise regression technique.

Variable In regression analysis, the features of a 'good' model include Selection for 1. Validity: The model includes regressors having empirical (engineering or Good Model practical) significance as well as statistical significance.

2. Satisfactory performance: The model provides estimates of the response (given conditions of the regressors) within an acceptable accuracy.

3. Simplicity: The model includes as few regressors as possible for ease of use.

4. Stability: The model does not change sensitively by use of a different set of observations.

These four criteria often conflict in many applications, thus a compromise is necessary to select appropriate regressors in a model. In this variable selection process, experience, judgment, and discussion among the analysts and practitioners involved in the study are important to establish a satisfactory model.

Variable No 'best' technique exists to select the best subset of regressors out of k Selection candidate regressors. Available selection techniques include (1) all-possible

Techniques regression evaluation; (2) forced selectiodelimination; (3) forward selection; (4) backward elimination; and (5) stepwise regression. Each selection technique will show its own trade-off between computational efficiency and model performance in a specific application.

All-Possible The all-possible regression evaluation technique generates all possible regression Regression equations from the combinations of k regressors and then evaluates the equations Evaluation by using selected criteria (which will be presented in the following paragraphs).

This approach finds the best regression equation through comprehensive search and evaluation, but the number of equations to be examined increases geometrically as the number of candidate regressors increases (e.g., if k = 3, there are Z3 = 8 possible equations; if k = 10, there are 2'' = 1,024 possible equations).

To determine a subset of regressors for the best model, four statistical criteria (R:, adjusted R;, MSE (p), and C,; p = number of regression coefficients in the mode1,p I k + 1) are available to use (see Fi ure 12-1). First, the best set of 8 regressors is found where an increment in R, (coefficient of determination) begins smaller than a value predetermined by the analyst. Note that R: ever increases a s p increases (see Section 12-2.1).


All-Possible Regression Evaluation

(cont.) Increase in R: smaller

:, than the value predetermined

P

Adjusted R: Maximum adjusted R;

Minimum MS, (p)

P * * * * * i s * *,"rim w *ijs -*XI I _6--"# Y l f Y I-- - w - - -XI- -wm-m- -* w- sac wee** u am*- ura= *--

Figure 12-1 Plots of variable selection criteria againstp (number of regression coefficients).

Second, the best set o f p regressors is determined at which adjusted R; is at maximum. Note that adjusted R: may decrease a s p increases unless the reduction in error sum of squares (SSE) is greater than the error mean square (MSE) of the existing model (see Section 12-2.1).

Third, the best set o f p regressors is determined where MSE (p) is at minimum. This MSE criterion is equivalent to the adjusted R; criterion in the aspect that MSE (p) will increase if the reduction in SSE is smaller than MSE (p - 1) due to the loss of one degree of freedom in the error term by the addition of a new regressor.

Lastly, the best set o f p regressors is determined where C,, which is a measure of the total mean square error, is at minimum or close top:

If the p-term regression model has negligible bias, the value of Cp is close top; if the model is significantly biased, the value of C, is significantly greater thanp.

iI '

12-24 Chapter 1

t+' Example 12.11



(All-Possible Regression Evaluation) Consider the power consumption data in I Example 12-1. By using software, the values of R;, adjusted R;, MSE(p), and Cp

are obtained as follows:

1 2 X I 56.5 52.1 17.0 3.9 1 2 X3 0.2 <O. 1 25.8 19.4 1 <0.1 <O. 1 25.8 19.4

2 3 X 2 , X 3 64.6 56.8 16.2 3.7 2 3 x z , X 4 64.5 56.6 16.2 3.7 2 3 X l , X 3 64.1 56.2 16.3 3.8

3 4 X I , X ~ , X ~ 73.2 63.1 14.9 3.4 3 4 X 2 , X 3 , X 4 64.7 5 1.4 17.1 5.7

4 5 X I , X ~ , X ~ , X ~ 74.5 59.9 15.6 5.0 ---* m.-.*"-ma u ~ ~ - ~ ~ ~ " " - # " v a ' ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ " ~ ~ ~ ~ . " ~ - " " . ~ w ~ . ~ w - " - " ""-- ."*me-"-,"sL* wm,.-w-,"*-" .-~-2a.,. " "."..-*"" * ,",~----*"" .*"

Discuss which model is preferable for the power consumption data.

U-F A two-regressor model including xl and x 2 is recommended for it has the maximum value of adjusted R; and minimum values of MSE(p) and Cp.

(All-Possible Regression Evaluation) Consider the wire bond pull strength data in Exercise 12-1. By using software, the values of R:, adjusted R;, MSE@), and C, are obtained as follows:

No. regressors

in the Regressors in the R: Ad'usted ME model p model (%) RP (%) @) CP

4 1 2 X 3 47.7 44.6 1.01 6.7 1 2 X 4 31.1 27.0 1.16 13.6

2 3 X 3 , X 5 57.2 51.8 0.94 4.8

3 3 X 3 , X 4 , X 5 67.1 60.6 0.85 2.7

4 4 X I , x 3 , X 4 , X 6 68.6 59.6 0.86 4.1

5 5 x 1 , x 2 , x 3 , x 4 , x 6 69.0 57.0 0.89 5.9

6 6 X I , X ~ , X ~ , X ~ , X ~ , X ~ 71.1 56.7 0.89 7.0 " ma"", sss*aae ,,.a, ma.m -*-" msm.>m --- wv"",-'*,a.w .wsm.Me, *.a*-" *--. ~---" sm-m w , * ~ ~ * % . m ~ " , s a * " "* '%.,,,*'"d"'.~ *,* *.mm,ms ~,,-*?w~,m *.- -,b.#,,m"

Discuss which model is preferable for the pull strength data.


Forced The forced selection/elimination technique includes/excludes regressors Selection1 designated by the analyst inlfrom the model regardless of their statistical

Elimination significance.

Forward The forward selection technique adds to the existing model the regressor xi out of Selection candidate regressors (not in the model) that has the largest partial F-value V;) iff;

exceeds the entry criterionJn (corresponding probability pi, = 0.10 or 0.15; preselected by the analyst):

s s ~ (Pi I B model )

f i = M s ~ (fl rnodeI,Pi

In other words, regressors are added to the model one at a time as long as their partial F-values are greater thanj,,.

Backward The backward elimination technique includes all candidate regressors in the Elimination model and then eliminates from the model the regressor xj that has the smallest

partial F-value G) if& is less than the removal criterion&,, (corresponding probability pout = 0.15 or 0.20; predefined by the analyst):

In other words, regressors are removed from the model one at a time as long as their partial F-values are than&,,.

Stepwise The stepwise regression technique is the mix of the forward selection and Regression backward elimination techniques: forward selection followed by backward

elimination. This mixed variable selection technique is widely used. The procedure of stepwise regression is as follows.

Step 1: Out of candidate regressors, include xi in the model having the largest F-value iff; >J;, (corresponding pin = 0.10 or 0.15).

Step 2: Among the regressors in the model, remove x, from the model having the smallest F-value if& > h u t (corresponding pout = 0.15 or 0.20; pin

5 pout). Step 3: Repeat steps 1 and 2 until no other regressors are added to or

removed from the model.

Example 12.12 (Stepwise Regression) Consider the power consumption data in Example 12-1. Out of the four regressors (XI, x2, xg, and xq), determine an appropriate set of regressors by the stepwise regression technique to build a regression model. Use software withpi, =pout = 0.15 (equivalent toJ;, = h u t =A, 12-, = 2.5). h Two regressors, xl and x2, are selected for model building.

Exercise 12.12 Consider the wire bond pull strength data in Exercise 12-1. Out of the six (MR 12-57) regressors (XI, x2, x3, x4, xg, and x ~ ) , determine an appropriate set of regressors by

the stepwise regression technique to build a regression model. Use software with Pin =pout = 0.15 (equivalent to& = h u t =A, 19_, = 2.3).


12-6.4 Multicollinearity

Explain the terms multicollinearity and variance injlation factor (VIF). O Describe undesirable effects due to multicollinearity in multiple regression analysis.

Identify regressors having multicollinearity based on VIF. O Describe remedial measures to resolve the problem of multicollinearity.

Multicollinearity When there are strong dependencies among the regressors xj7s, it is said that multicollinearity exists. The following undesirable effects can be observed due to multicollinearity:

1. Unstable estimation of regression coefficients: The estimates of

regression coefficients, b, 's, are sensitive to a small change in data.

2. High standard errors of regression coefficient estimates: The standard

errors of regression coefficient estimates. se(b j ) 's, are unreasonably

large. 3. Contradicting test results: Although the full regression model is

significant by F-test, no individual regressors are found significant by t- test.

A regression model with the problem of multicollinearity will produce unsatisfactory estimations of mean responses or predictions of new observations beyond the region of the x-space surveyed (where extrapolation is needed). However, this regression model may produce satisfactory estimations on the response within the region of the x-space surveyed (where interpolation is performed).

Variance The measure variance inflation factor (VIF) indicates the extent of Inflation multicollinearity:

Factor 1 (VIF)

VIF(P,)=- 2 , j = 1 , 2 ,..., n I -R i

where R; is the coefficient of determination between xj and the other k - 1 regressors. The stronger the multicollinearity, the larger the value of R;. If a VIF exceeds 10 (some researchers suggest 4 or 5), the problem of multicollinearity exists.

Remedial Remedial measures against multicollinearity include Measures 1. Inclusion of additional data: Collect additional data to explore the

possibility to break the existing linear dependencies between the regressors.

2. Removal of regressors with large values of VIF: Exclude regressors having large values of VIF from the model.

12-6 Aspects of Multiple Regression Modeling 12-27 'r Example 12.13

& Exercise 12.13

(Multicollinearity) Consider the power consumption data in Example 12-1. The variance inflation factors of the four regressors (x,, x2, x3, and x4) are obtained by using software as follows:

Number of days in the month (x2) 2.2 Product purity (x3) 1.3

1 .o

Interpret the VIF values.

I - Since all the VIF values are less than 10, the regression model does not have the problem of multicollinearity.

Consider the wire bond pull strength data in Exercise 12- 1. The variance inflation factors of the four regressors (XZ, x3, x4, and xg) are obtained by using software as follows:

Loop height (x3) 1.07 Wire length (x4) 1.41 Bond width on t

$"."* "..- %,* ->*~"'.* s".m **~a,r" .--,-

Interpret the VIF values.



sheet.

Examples

12.1-3,5-8,13

(3) Choose Stat * Regression * Regression. In Response select Y and in

(Multiple Linear Regression)


(4) Click Graphs. For Residuals for Plots, select Regular or Standardized. Under Residual Plots, check Normal plot of residuals, Residuals versus fits, andlor Residuals versus order. In Residuals versus the


OK. -

Examples 12.1-3,543, 13

(cont.)

1 (6) Click Results. Under Control the Display of Results, click the level of

(5) Click Options. Check Fit intercept. Under Display, check Variance inflation factors. In Prediction intervals for new observations, enter the values of the regressors under consideration for the confidence interval on the mean response andlor prediction interval of a future observation. In Confidence level, type the level of confidence. Then click

1 (7) Click Storage. Check Fits, Residuals, and Cook's distance. Then click

z f 0 @QQ 1 ~ 1 1 P!z , $6- - -406 - - -9EZ _ __ _ __ ~ ~ 0 € t 0 ~ ~ 8 5 ~ 8 1 - j 8 S L 8 5 Z 001 ,16 PZ R OPZ 1

,f I1>1003 LlS3tl ' 1SlI4 Px T .--.-* ......

m I i n t 9 D n

I TnnpTsaI P ~ Z T P I B P U W ~ S a6lwl w qirn uoraBnxasqo ue saiouap 1 I

Example 12.4

Example 12.9


(Test on the Subset of Regressors)

The


The r e g r e s s i o n equa t i on is 1 Y = 224 + 0.953 xl

Pred i c to r Coef StDev T P Constant I 224.38 15.88 14.13 0.000

0.9525 0.2645 3.60 0.005

........................................................................................................... I ~ n a l y s ~ s 0. va r i ance

: source DF ss MS F p i i Regress ion 1 3758.9 3758.9 12.97 0.005 i i ~ e s i d u a l E r ro r 10 2897.3 289.7 : Tota l 11 6656.31 , .......................................................................................................... w: .. .$": .~"", I#,j , 8 .,

(Polynomial Model)

( I ) Choose Stat * Regression P Fitted Line Plot. In Response select Y, in Predictor select the regressor under consideration, and under Type of Regression Model click the type of the polynomial to be developed. Then click OK.

Polynomial Regression I I Analys is of Var iance

I SOURCE DF SS MS F P Regress ion 2 4362.75 2181.38 8.56002 8.27E-03

Error 9 2293.50 254.83 I Tota l 11 6656.25

SOURCE DF Seq SS F P .............. ..... ,..... .... I .Lanear L 3.7.56..92 12..2?.3.7 LR3&R3 ...:

Quadra t i c 1 603.83 2.36953 0.158109 I


Example 12.10 (Indicator Variables) I (1) Choose File , New, click Minitab Project, and click OK.

I (2) Enter the liver weight data on the worksheet.

I (3) Choose Stat , Regression , Regression. In Response select Y and in Predictors select regressors under consideration.


The regress ion equa t ion is y = - 109 + 977 x l - 1 x2 ...................................................................................................... Pred ic to r Coef StDev T P i Constant -108.7 8 2 9 . 2 -0.13 0 .897 1 x l 977.0 502.7 1 . 9 4 0 .069 i x2 -1.3 126.7 - 0 . 0 1 0.992 i

, ..................................................................................................... S = 281.2 R-Sq = 18.45 R-Sq(ad3) = 8.85

Source DF SS US F P Regression 2 303477 151738 1.92 0 .177

Resldual Error 17 1344478 79087 Tota l 19 1647955

/ source DF Seq SS I

Example 12.11 (All Possible Regression Evaluation) I (1) Choose Stat , Regression , Best Subsets. In Response select Y and in


Example 12.1 1 (cont.)

(2) Click Options. In Minimum and Maximum enter the minimum and maximum numbers of regressors that will be included in regression models, respectively. In Models of each size to print, specify the number (1 to 5) of best regression models to be printed for each number of

Best Subsets Regression

Response is y I

1 64.5 60.9 1.7 15 .381 X 1 56.5 5 2 . 1 3 . 9 17.022 X 1 0.2 0 . 0 19 .4 25.769 X 1 0 . 0 0.0 19.4 25.799 X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 7 3 . 1 67.2 1.4 14.095 X X 2 64.6 56 .8 3 . 7 16.173 X X 2 64.5 56 .6 3 . 7 16.210 X X 2 64 .1 56.2 3.8 16.289 X X 2 56 .5 46 .8 5 . 9 17 .941 X X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 74.5 64.9 3 . 0 14.573 X X X 3 73.2 63 .1 3.4 14.944 X X X 3 64.7 51 .4 5.7 17.149 X X X

-

12-34 Chapter

Example 12.12


(Stepwise Regression)

(1) Choose Stat > Regression > Stepwise. In Response select Y and in

(3) Intemret the analvsis results.

Stepwise Regression

I R e s p o n s e is y o n 4 p r e d i c t o r s , w i t h N = 1 2

S t e p 1 2 C o n s t a n t -30.1607 0.5287 , . . . . . . . . . . . . . , . . . . .

I x2 15.2 10.3 i T - V a l u e ! 4.26 2.36

respectively.



Exercise 12.1

Exercise 12.2

(Calculation of Residual)

The estimate of the pull strength when x2 = 18.6, x3 = 28.6, x4 = 86.5, and x5 = 2.0 is

$=7.46 - 0.03~18.6+ 0 .52~28 .6 - 0.10~86.5 - 2.16~2.0=8.66

Thus, e = y - j = 9.0 - 8.66 = 0.34 (underestimate)

2. (Estimation of c2)

n = 19 and p (number of regression coefficients) = 5 - 2 SSE 10.91 (J =-- -- = 0.78 = 0 .88~

n - p 19-5

' 1. (Test on the Significance of Regression)

Step 1 : State Ho and HI. Ho: p 2 = p3 = p 4 = p5 = 0 H I : p J # O foratleastonej,j= 1 to4

Step 2: Determine a test statistic and its value. SSR l k MSR - 5.58

f o = - -- - 7.16 SS, l(n - p ) MS, 0.78

Step 3: Determine a critical value@) for a. f a , k , n - p = f0.05,4,14 = 3.1

Step 4: Make a conclusion. Since fo =7.16> f a,,,n-, =3.11,reject Hoat a = 0.05. This

indicates that at least one of the four regressors in the model has a significant linear relationship with the pull strength of a wire bond at a = 0.05.

2. (R2 and Adjusted R2)

2 ss R - R-22.31 - - = 67.2% SS, 33.22

The four regressors (x2, x3, x4, and x5) explain 67.2% of the variability in the pull strength Y.

n - 1 2 19-1 Adjusted R~ = 1 - - (1-R )=I-- (1 - 0.672) = 0.578 n - P 19-5


Exercise 12.3 (Hypothesis Test on Pi) Step 1 : State Ho and HI.

H,,: p5 = 0 H1: p5 + 0


b s - P 5 , 0 -2.16-0 to = - - = -0.90

se<b5 1 2.39


ta /2 ,n-p - 05/2,19-5 = 025,14 = 2.145

Step 4: Make a conclusion. Since Ito\ = 0.90 ;a ta12,,-p = 2.145 , fail to reject Ho at a = 0.05. This

indicates that production volume (x4) does not have a significant linear relationship with the pull strength at a = 0.05.

Exercise 12.4

Exercise 12.5

(Test on the Subset of Regressors) Step 1 : State Ho and HI.

Ho: p1= 0 (p1 = p2 = p3 = 0 ) H I : # O (Pj#O for at least o n e j , j = 1 to 3)

I Step 2: Determine a test statistic and its value. ssR(pl ~ ~ 2 ) = ~ ~ R ( ~ ) - ~ ~ R ( ~ z ) = 2 2 . 3 1 - ~ o . 8 1 = 1 1 . 5 0

Step 3: Determine a critical value@) for a. fa,r,n-p = f0.05,3,12 = 3'49

Step 4: Make a conclusion. Since fo = 4.79 > fa,k,n-p = 3.49, reject Ho at a = 0.05. This

indicates that at least one of the process regressors has a significant linear relationship with the pull strength at a = 0.05.

(Confidence Interval on Pi)

1-a=0.95 a=0.05 95% two-sided CI on P 5 :

P 5 -ta12,n-p d87C, 1 p5 5 f i 5 + ta12,n-p JB'C,

Exercise 12.6

Exercise 12.7

Exercise 12.8

Exercise 12.9


(Confidence Interval on yylxo )

1 - a = 0.95 3 a = 0.05 95% two-sided C I on y :

d m PYlxo - t a / 2 , , - , d ~ 5 ~ y l x , 5 f i y lx0 + t a i 2 , n - p XX)

(Prediction Interval on yo)

1 - a = 0.95 a = 0.05 95% two-sided C I on yo:

YO -tai2,n-p J%GizG 5 YO 5 jo + t a / 2 , n - p d m

(Influential Observation) Since all the Di's are less than one, no influential observations are found in the data.

(Polynomial Model) Step 1 : State HO and HI

Ho: p 2 = 0 H1: p 2 2 0


Step 3: Determine a critical value(s) for a. fa,r,n-fj = f0.05,1,9 = 5'12

Step 4: Make a conclusion. Since fo = 9.0 > f a , , ,,-, = 5.12, reject Ho at a = 0.05. This indicates

that the quadratic term significantly contributes to the model.


Exercise 12.10

Exercise 12.11

Exercise 12.12

Exercise 12.13

I (Indicator Variable) For tool type 302 (x2 = O), the fitted model is

j = B, + B,x, = 11.50 + 0 . 1 5 ~ ~

In contrast, for tool type 416 (x2 = I), the fitted model is I

1 j = B, + B1.1 + B 2 + 133x1 =(Do + 82) + (ljl + B3)4 I

I = (1 1.50 - 6.09) + (0.15 - 0 . 0 3 ) ~ ~ =5.41+ 0 . 1 2 ~ ~

However, since the P-values for Pi and P3 are greater than 0.05, it is 1 concluded that two regression equations are not needed to model the tool life I data.

1 (All-Possible Regression Evaluation) Either a two-regressor model including x3 and x4 or three-regressor model

I

I including xl, x3 and x4 is recommended. The two-regressor model has the 1 minimum value of C, and the three-regressor model has the maximum value I of adjusted R; and minimum value of MS&).

(Stepwise Regression)

1 Two regressors, x3 and x4, are selected for model building.

(Multicollinearity) Since all the VIF values are less than 10, the regression model does not have the problem of multicollinearity.

13-2 Chapter 13 Design and Analysis of Single-Factor Experiments: The Analysis of Variance

13-2 The Completely Randomized Single-Factor Experiment

13-2.1 An Example

0 Identify dependent, independent, and extraneous variables in an experiment. 0 Explain the terms treatment and replicate.

Compare the following pairs of concepts each: controllable vs. uncontrollable variables; fixed- effects vs. random-effects factors; balanced vs. unbalanced designs; complete vs. restricted randomizations.

Explain the purpose of randomization.

Tensile A tensile strength experiment is introduced to explain fundamental terminologies Strength of experimental design. Suppose that a manufacturer of paper is interested in

Experiment improving the tensile strength of the product. The manufacturer wishes to (example) identify if the tensile strength is affected by the hardwood concentration in the

pulp used for the paper. The range of hardwood concentration of interest is between 5 and 20%; and, within the range, four levels (a = 4) of hardwood concentration are examined: 5, 10, 15, and 20%. At each concentration level, three specimens (ni = 3, i = 1 to a) are tested in a laboratory. The measurements by, j = 1 to ni, i = 1 to a) of this experiment can be summarized in a table format like Table 13-1.

Table 13-1 Data Structure of a Single Factor Experiment F'wcrar3hWle I

Replicate 1 2 ... I ... a

j Y l j Y2j . . . Y y . . . Y aj

... ... n1 Yln, Y2n2 ... Yin, ... Y a,, 1

**- **s n_(__" aiamla -* - -* rr--arw paa.-aarr----s----~---wM*w- a.-n-xnrtrmaran-n-M m- i Totals Y I . Y2. ... Yl. ... Ya. Y.. 1 1

- - - - - Averages y,.

, Y2. ... Yi, . . . Ya. Y.. \

:a m-invn*_?_f* X*j. _ w w f . i . _ e a a 8*_ X_ n l * ~ ~ 8 x / ~ l _ e _ n ~ ~ . ~ ~ ~ ~ ~ " ~ ~ ' . ~ % ~ . ~ ~ ~ ~ ~ % ~ ~ ~ ~ w ~ ~ ~ ~ ' ~ ~ < ~ w ~ ~ ~ a ~ ~ - ~ ~ . ~ ~ " ~ r ax*-carulxi.in-nr-m---*->--d~w*u"%.r id%-

I

Types of Three categories of variables exist in an experiment: Variables 1. Response (dependent) variable: The outcome (response) of interest (e.g.,

tensile strength) 2. Independent variable (factor): The variable whose effect on the response

variable is under study (e.g., hardwood concentration) 3. Extraneous variable (nuisance factor): The variable whose effect on the

response is not under consideration but may affect the response variable (e.g., difference between specimens, temperature, etc.)

Treatment The term treatment (factor level) of a factor refers to an individual condition of (Factor Level) the factor. For example, the tensile strength experiment has four factor levels: 5,

10, 15, and 20%.

13-2 The Completely Randomized Single-Factor Experiment 13-3

Fixed-Effects VS.

Random-Effects Factors

Controllable VS.

Uncontrollable Variables

Replicate

Balanced VS.

Unbalanced Designs

Block

Complete VS.

Restricted Randomization

Two types of factors are defined depending on the selection method of factor levels: fixed-effects and random-effects factors. While the levels of a fixed- effects factor are specifically selected by the experimenter, those of a random- effects factor are chosen by random mechanism from a population of factor levels. The conclusion of a fixed-effects factor experiment cannot be extended beyond the factor levels selected, while that of a random-effects factor experiment can be extended to all the factor levels. For example, in the tensile strength experiment hardwood concentration is a fixed-effects factor because the four factor levels are specifically (not randomly) selected by the experimenter; thus, the corresponding conclusion is restricted to the four hardwood concentrations examined.

An extraneous variable can be classified into either a controllable or uncontrollable variable depending on if the condition of the variable can be controlled by the experimenter with an acceptable variation. For example, if the tensile strength experiment is conducted in a laboratory where its temperature is under control, temperature is a controllable variable.

The term replicate refers to a repeated measurement under the same experimental condition. As an example, the tensile strength experiment has six replicates for each treatment.

A design of experiment is classified into either a balanced or unbalanced design depending on the equality of replicates. A balanced design has the same number of replicates between the treatments, while an unbalanced design does not. For example, the tensile strength experiment is a balanced design.

Recall that the paired t-test is used to block out the effect of heterogeneity between specimens (experimental units) (see Section 10-4). This heterogeneity between experimental units becomes a nuisance factor; and each experimental unit is called block.

Randomization means that the order of trials is determined by random mechanism (such as random number table and random number generator) and two types of randomization are defined according to restriction in randomization:

(1) Complete randomization: All trials are randomized (see Table 13-2(b)). (2) Restricted randomization: Trials are randomized within each block (see

Table 13-2(c)).

Randomization balances out the effects of extraneous variables on the response. For example, in the tensile strength experiment, suppose that the tensile strength measurement device has a warm-up effect-an increase of two units in error per measurement. Table 13-2 shows errors in measurement due to the warm-up effect for different types of running trials; the warm-up effect becomes balanced out by randomization (from the range [30,48] in sequential experiment to [34,44] in complete randomization and [34,42] in restricted randomization).


Complete Table 13-2 Errors Due to Warm-up Effects VS.

Restricted Randomization

(cont.)

. , . . 3 12(6)

.e,e.-*.-..-a.*s.-L.~ m~-d..BMwm-a.- -.-,. -.a<. ",.."*~ Totals r 38 .,., , . <.-,..,,,wm.-,,*.e,% ,111,1 ,*, ,,., ,,,.,,.,.- ,.,.,,,,,,,,,,,, ,.,,,,,

13-2.2 The Analysis of Variance

Context 1. Factor: Fixed-effects factor with a treatments 2. Blocking: No blocks defined 3. Repetition: n replicates for each factor level (balanced design)

(Note) N (total number of observations) = n x a 4. Randomization: Complete randomization

Model Y.. = p + ' t i + E . . = p i

1/ 1/

where: y = overall mean a

'ti = i" treatment effect, r = 0 i=l

pi = i'h treatment mean 2 E~ = random error - i.i.d. N (0, o )


Mode' ( N 0 t e s ) l . G - ~ ( p , , o ' ) (cont.)

2. In the fixed-effects model, the treatment effects (zi) are defined as deviations from the overall mean; thus, the sum of zi's is zero.

Least Square The least squares method finds the estimates of p and .r;i's (i = 1 to a) which Estimators minimize the sum of the squares of the errors (SSE)

By solving the partial derivatives of L with respect to p and zi's, the least square estimators of p, ~ ~ ' s , and pi's are determined:

P = F.. A - -

~ ~ = y , - y , . , i = 1 , 2 ,..., a

P, =P+? i =y, , , i = 1 ,2 ,..., a

ANOVA The ANOVA method is used to test the significance of a factor of interest. Like the simple regression model, the total variability in Y (SST, total sum of squares; degrees of freedom = N - 1) is partitioned into two components (see Table 13-3):

(1) Variability explained by the factor (SSTreatments, treatment sum of squares; degrees of freedom = a - 1)

(2) Variability due to random error (SSE, error sum of squares; degrees of freedom = N - a)

Table 13-3 Partitioning of the Total Variability in Y (SST) SST ss~reannents SSE

(Total SS) (Treatment SS) (Emr SS)

M - ~ ~ " w A ~ ~ ~ ~ d - w - ~ - ~ ~ a ~ ~ ~ . ~ . a ~ ~ u ~ " - ~ ~ w e ~ ~ ~ ~ ~ ~ ~ ~ a ~ ~ ~ .

+ N - a

(Note) SS: Sum of Squares; DF: Degrees of Freedom

The ratio of SST,atmentsl(a - 1) to SSEI(N - a) follows an F distribution:

F = "Treatments - - - MS~reatments -F (u -1 ,N-a ) SS, l(N - a) MSE

This statistic F is used to test Ho: TI = 7 2 = ... = T, = 0, because F becomes small as the effect of the factor is insignificant. The quantities MS~reatments (treatment mean square) and MSE (error mean square) are adjusted S S T , ~ ~ ~ ~ , ~ ~ and SSE by their degrees of freedom, respectively. Note that


ANOVA The variation quantities from the ANOVA analysis are summarized in Table 13-4 (cont.) to test the significance of the factor under consideration.

Table 13-4 ANOVA Table for Testing the Significance of a Factor: Completely Randomized Design


Treatments SS~reatrnents a-1 MS~reatments MS~reatments M S ~

Error SSE N - a MSE

Total SST N-1 - " " d. - " - - " " .Wt i "TX I

Hypothesis Step 1 : State No and HI. Test H ~ : TI = TI = ... = T O = 0

(F-test) HI: z; # 0 for at least one i, i = 1, 2,. . . , a


F, = SSTreatments - - - MS~reatrnents - F(u-1, N - a ) SS, l(N - a ) MSE

Step 3: Determine a critical value(s) for a. fa ,a-1,~-a


f 0 ' fa,a-1,~-a 4

I Unbalanced The same ANOVA method of the balanced design is applicable to an unbalanced 1

Design design with slight modifications. If n, denotes the number of observations under the ith treatment, the total number of observations is

Then, the sums of squares for the unbalanced design are

a n, a 2 Yi. 1 SS~reatments

i=] j=l

A balanced design has two advantages compared to an unbalanced design: (1) Robustness: The test procedure of the balanced design is less sensitive to

a small departure from the assumption of equal error variance (d). (2) Statistical power: The power of the test of the balanced design is higher

when the sample sizes of the two designs are the same.

1


Confidence A 100(1 - a)% confidence interval on pi is Interval on -

Pi Yi. - t a / 2 , ~ - a JF - pi 5 Ti, + ta,2,N-a 47 Confidence Interval on

Pi-& r Example 13.1

A 100(1 - a)% confidence interval on yi - pi is

A paper manufacturer is examining if the tensile strength of a paper product is affected by the hardwood concentration of the pulp used for the product. Four hardwood concentrations (5 , 10, 15, and 20%; a = 4 ) are selected by the analyst and five specimens (n = 5 ) are tested at each concentration for tensile strength, resulting in the following: . ,

Hardwood Conmtratian (%; 9 , -- 10 (2) 1s (3) 20 (4)

Totals 5 0 79 84 107 320 A v e r a ~ 10.0 15.8 16.8 21.4 16.0

ee *- W W W * *-- -*#* a* " m - wwwana "' m- - w m v ma** A v-s -aa**ra** e = v, A *- m a *w# * e - # Bw" -- The following summary quantities are also provided for the tensile strength data:

a n a 2 N = 2 0 , ~ ~ y , 2 = 5 , 5 7 6 , a n d ~ y , =27,246

1=1 J=I r = l

(Note) The subscripts 1,2, 3 , and 4 correspond to 5 , 10, 15, and 20% of hardwood concentration, respectively.

1. (Test on the Significance of a Fixed-Effects Factor; Completely Randomized Design) Perform the analysis of variance to test if there is any significant difference in tensile strength due to hardwood concentration at a =

0.05.

h The analysis of variance of the tensile strength data is as follows:



Step 1 : State Ho and HI. Ho: 2 1 = 2 2 = ". =2,=0 H I : z i f 0 for at leastone i, i= 1 t o 4


Step 3: Determine a critical value(s) for a. fa ,*-1 ,~-a = f0.05,4-1,20-4 = f0.05,3,16 = 3.24

Step 4: Make a conclusion. Since f0 = 13.8 > fa,,-, ,,-, = 3.24 , reject Ho at a = 0.05.

2. (Confidence Interval on pi) Establish a 95% two-sided confidence interval on p 2 (mean of tensile strength at 10% of hardwood concentration).

1-a=0.95 2 a=0.05 95% two-sided CI on p 2:

dMS, k h + t a 1 2 . N - ~ d ~ 7 2 . - t a 1 2 , ~ - a - n

I 3. (Confidence Interval on pi - &) Establish a 95% two-sided confidence interval onp1-p2 .

5% two-sided CI on fi 1 - y 2:


The effect of cathode ray tube coating type on the tube conductivity is under study. Five different types of coating (a = 5) are selected by the analyst and four replicates (n = 4) are run for each coating type, resulting in the data as follows:



a n a

N = 2 0 , ~ ~ y y 2 =389,115,and Zy i2 =1,555,487 ;=I ;=I i=l

1. Perform the analysis of variance to test if there is any significant difference in conductivity due to coating type at a = 0.05.

2. Establish a 95% two-sided confidence interval on p 1 (mean of tube conductivity for coating type 1).

3. Establish a 95% two-sided confidence interval on p2 - p3.

13-2.3 Multiple Comparisons Following the ANOVA

Multiple Several methods are available to identify which treatment means are different Comparison from each other:

Methods 1 . Scheffe's orthogonal contrast 2. Fisher's least significant difference (LSD) 3. Duncan's multiple range test 4. Tukey-Kramer's honestly significant difference (HSD) test

In this book, only the LSD method is presented.

Least Fisher's least significant difference (LSD) method is used to compare all possible Significant pairs of treatment means, having the following null hypotheses Difference Ho: pi = CL, for all i # j, i and j = 1,2 , . . . , a

(LSD) Method Each null hypothesis Ho: pi = CL, will be rejected if

ItOI > tai2,~-a

In other words,


LSD As shown above, the pair of pi and ~ 1 , will be concluded significantly different if Method the absolute difference of the corresponding sample means is greater than the

(cont.) LSD. Note that for an unbalanced design the LSD is defined as I / \

Example 13.2

The procedure of the LSD method is as follows: Step 1 : Determine the LSD for a. Step 2: Arrange the means of a treatments in descending order. Step 3: Compare the difference between the largest and smallest means

with the LSD. Continue this comparison with the next smallest mean as long as the mean difference is greater than the LSD.

Step 4: Continue Step 3 for the next largest mean until this iteration reaches the second smallest mean.

Step 5: Summarize the painvise comparison results by arranging the a treatments in order of mean and then underlining treatments whose means are not significantly different.

(Least Significant Difference Method) Consider the tensile strength data in Example 13-1. The following summary quantities are obtained from the analysis of variance:

a = 4 , n = 5 , N = 2 0 , MSE =7.9

j,. =10.0, Y2. =15.8, y3, =16.8, and y,, =21.4

Compare the means of the four treatments by using Fisher's LSD method. Use a = 0.05.

I.+V Step 1 : Determine the LSD for a.

Step 2: Arrange the means of a treatments in descending order. y,, =21.4, j,, =16.8, 7,. =15.8, and j,, =10.0

Step 3: Compare the difference between the largest and smallest means with the LSD. Continue this comparison with the next smallest mean as long as the mean difference is greater than the LSD.

4 ~ s . 1=21.4-10.0=11.4>3.8 4 VS. 2 = 21.4- 15.8 = 5.6 > 3.8 4 VS. 3 = 21.4 - 16.8 = 4.6 > 3.8


3 VS. 1 = 16.8 - 10.0 = 6.8 > 3.8 3 VS. 2 = 16.8- 15.8= 1.0<3.8 2 vs. 1 = 15.8 - 10.0 = 5.8 > 3.8


4 3 2 1

13-2 The Completely Randomized Single-Factor Experiment 13-1 1


Consider the tube conductivity data in Exercise 13-1. The following summary quantities are obtained from the analysis of variance:

a = 5 , n = 4 , N = 2 0 , MSE =16.2

J,. =145.0, 7,. =145.3, 7,. =131.5, J4, =129.3, and 7,. =145.3

Compare the means of the five treatments by using Fisher's LSD method. Use a =

0.05.

13-2.5 Residual Analysis and Model Checking

Describe the purpose of residual analysis in the analysis of variance. Calculate residuals.

Residual In Section 13-2.2, the single-factor ANOVA model has three assumptions on the Analysis error term: (1) normality, (2) constant variance, and (3) randomness. These

assumptions can be checked by analyzing residuals (eij = yij - yij = yij - 7;. ), which is called the assessment of model adequacy.

First, the normality of error is checked by one of the following: (1) Frequency histogram of residuals (2) Normal probability plot of residuals (3) Standardization of residuals: If the residuals are normally distributed,

about 95% of the standardized residuals e.. eij

d . . =L-P , i = 1 , 2 ,..., a and j = 1 , 2 ,..., n " - r n are in the interval (-2,2). Residuals whose standardized values are beyond +3 may be outliers.

Next, the constant variance and randomness of error can be checked by examining the following residual plots:

(1) eij vs. factor level i, i = 1,2, . . . , a

(2) eij vs. 7, (3) eij vs. t (if time sequence is known)

An undesirable residual pattern (see Section 11-8.1) in a residual plot may be corrected by transformation of the response variable ('y) into another form such as

J; or logy. r Example 13.3 (Calculation of Residual) Consider the tensile strength data in Example 13-1. The following summary quantities are obtained from the analysis of variance:

J1, =10.0, J,, =15.8, &, =16.8, and L, =21.4 Calculate the residual of yl2 = 8.

~h' el, = yI2 - j12 = y12 - J1, = 8 - 10 = -2 (overestimate)

13-1 2 Chapter 13 Design and Analysis of Single-Factor Experiments: The Analysis of Variance


Consider the tube conductivity data in Exercise 13-1. The following summary quantities are obtained from the analysis of variance:

y,, =145.0, j2, =145.3, y,, =131.5, 7,. =129.3, and 7,. =145.3

Calculate the residual of y24 = 143.

13-2.6 Determining Sample Size

O Determine the sample size of a single-factor experiment by using an appropriate operating characteristic (OC) curve.

Operating Operating characteristic (OC) curves in Figure 13-6 of MR plot required sample Characteristic sizes n for a fixed-effects factor experiment at different levels of the parameters

(OC) a ,p ,@,vl ,andv2, i .e . , Curve n =f (a , S, a, vl, and ~ 2 )

n2.: where: = 1%

v 2 = a(n - 1) (Notes) 1. If the error variance o2 is unknown, an estimate of o2 (such as MSE,

conventional value, and subjective estimate) can be used. I

2. The analyst can define the ratio r' lo2 (or 0: / 02 ) in sample size 1=1

determination.

3. The sample size required increases as a, P, and XI=, r: (or o: ) decrease and I

o increases. 3 r Example 13.4 (Sample Size Determination; Fixed-Effects Factor) Consider the tensile strength experiment in Example 13- 1. Four hardwood concentrations (a = 4) are selected and their tensile strength means tested at a = 0.05. Suppose that the four normal populations have common variance o2 = 32 and means pl = 12, y2 = 15, p3

= 18, and p4 = 20 (i.e., xa T: / o2 = 4.1 ). Determine the number of replicates I

r=l

within each treatment to reject the hypothesis of equality of means with power of at least 0.90 (= 1 - B).

Use the operating characteristic curve of Figure 13-6(a) in MR for a fixed- effects factor with vl = 3 (= a - 1 = 4 - I). From the OC curve, the power of the test according to n is found as follows:

I n Q, v2 = a(n - 1) fi Power (1 - p)


13-3 The Random Effects Factor 13-1 3

I Thus, at least n = 5 replicates are needed to achieve the designated power of the test.

Exercise 13.4 Suppose that five normal populations (a = 5) have common variance o2 = 1 o2 and (MR 13-20)

means p l = 175, p2 = 190, p3 = 160, p4 = 200, and p5 = 215 (i.e., za T: lo2 = r = l

18.3). How many observations per population must be taken so that the probability of rejecting the hypothesis of equality of means is at least 0.95? Use

a = 0.01.

13-3 The Random Effects Factor

Test the significance of a random-effects factor in a completely randomized design. 0 Estimate the variances of error (02), treatment effect (03, and response variable (V(YV)).

ms* - ws *--* * -+ms --- * * * -- *--- % dMFm * * --- * * - * - a a a * a * ww *# A a> e a

Context 1. Factor: Random-effects factor with a treatments 2. Blocking: No blocks defined 3. Repetition: n for each factor level (balanced design)

(Note) N (total number of observations) = n x a 4. Randomization: Complete randomization

Model

{ i=1,2 , ..., a

Y = p + z i +& . . = p i +& . . 'J rJ ' j = 1 , 2 , ..., n

where: p = overall mean

zi = ith treatment effect - i.i.d. N (0, 0:) th p i = i treatment mean

E, = random error - i.i.d. N (0, 02)


Model (Notes) 1. The random variables zi and E,. are assumed independent of each other. (cont.) 2. Y, - ~ ( y , , o ; + a 2 )

3. In the random-effects model, testing the individual treatment effects (zi) are meaningless because the treatments are selected at random.

ANOVA The analysis of variance procedure of the fixed-effects model is valid for the random-effects model. The computation procedure for an ANOVA table in the fixed-effects model (see Section 13-2.2) is identical to that of the random-effects model.

However, the conclusion of ANOVA for the random-effects model can be extended to the entire population of treatments. This generalization of conclusion is due to random selection of treatments in the experiment.

The variance components (0: and 02) of the random-effects model can be estimated by using MSTreatnebts and MSE:

Therefore, the estimate of the variance of YV is

B(y,)=8: +ir2

Hypothesis Step 1 : State Ho and HI. Test Ho: o, 2 = 0

(F-test) HI : 0 : f O

(Note) If all treatment effects are identical, o: = 0 ; but if there is variability

between treatment effects, a: # 0 .


Fo = "Treatments '(a - - - MSTraatments - F(u-1 ,N-a ) SS, l(N - a) MSE

Step 3: Determine a critical value(s) for a. fa,a-l,(\i-a


f 0 ' fa,a-1.~-a

ft Example 13.5 (Estimation of Variance Components) Consider the tensile strength data in Example 13- 1. Assume that the four levels of hardwood concentration are randomly selected. The following summary quantities are obtained from the analysis of variance:

n = 5 , MSTreatments = 109.7 , and MS, = 7.9

Estimate the (1) variability in tensile strength due to random error (02), (2) variability due to hardwood concentration (0,2), and (3) total variability in tensile strength (V(Yg)). Also interpret these variability estimation results.


bf; Exercise 13.5

These variance estimates indicate that about 72% (= 20.4128.3) of the total variability in tensile strength is attributable to hardwood concentration.

Consider the tube conductivity data in Exercise 13-1. Assume that the five coating types are chosen at random. The following summary quantities are obtained from the analysis of variance:

n = 4 , MS,,e,,,ent, =265.1, and MS, =16.2

Estimate the (1) variability in tube conductivity due to random error (02), (2) variability due to coating type (o:), and (3) total variability in tube conductivity (V(Yij)). Also interpret these variability estimation results.

13-4 Randomized Complete Block Design

0 Distinguish between complete block design and incomplete block design. Explain the effects of blocking on the error term. Test the significance of a fixed-effects factor in a randomized complete block design. Perform multiple comparisons by Fisher's least significant difference (LSD) method.

Complete vs. A complete block design contains all the treatments in each block (see Table 13- Incomplete 5), while an incomplete block design may omit one or more treatments in each

Block Designs block. Recall that blocking is used to balance out the effect of a nuisance factor (heterogeneity between experimental units) on the response variable.

Table 13-5 Data Structure of a Complete Block Design Treatments

Blocks 1 2 . . . i . . . a Totals Means

13-4 Randomized Complete Block Design 13-1 5


Context 1 . Factor: Fixed-effects factor with a treatments 2. Blocking: Complete block design with b fixed-effect blocks 3. Repetition: One observation for each combination of factor levels and blocks

(Note) N (total number observations) = a x b x 1 4. Randomization: Restricted randomization (randomization within each block)

Model i = l , 2 , . . . , a Yv = = + T i + P j +Eii = y i +Pi + E . .

j = 1 , 2 , ..., b

where: y = overall mean

ri = fh treatment effect, ri = 0 i=l

b

Pi = jth block effect, P = 0 ;=I

ki = ith treatment mean

cii = random error - i.i.d. N (0, 02)

(Notes) 1. Y, - N(yi, 0 2 )

2. The treatments and blocks are assumed to be independent of (not interacting with) each other.

3. Since the treatments and blocks are assumed to be fixed-effects factors in the model, the treatment effects (ri) and block effects (Pj) are defined as deviations from the overall mean; thus, the sums of ri's and Pj's are zero.

ANOVA The analysis of variance method is used to test the significance of a factor in a randomized complete block design. The total variability in Y (SST, total sum of squares; degrees of freedom = N - 1) is partitioned into three components:

(1) Variability explained by the treatments (SSTreatments, treatment sum of squares; degrees of freedom = a - 1)

(2) Variability explained by the blocks (SSsloCh, block sum of squares; degrees of freedom =b - 1)

(3) Variability due to random error (SSE, error sum of squares; degrees of freedom = (a - l)(b - I))

In other words,

a b a b 2 2 Y.. where: S S , = z z ( Y , - ~ . ) 2 = ~ , ~ Y , --

i=l ~ T I i=1 ,=I ab


ANOVA (cont.)

The ratio of S S T ~ , , ~ ~ , ~ ~ , ~ ( ~ - 1) to SSEI(a - l)(b - 1) follows an F distribution:

This statistic F is used to test Ho: TI = TZ = ... = = 0, because F becomes as the effect of the factor is insignificant. The quantities MSTreatments (treatment mean square), M S B ~ ~ ~ ~ ~ (block mean square), and MSE (error mean square) are adjusted SSTreatments, S S B ~ ~ ~ ~ ~ , and SSE by their corresponding degrees of freedom, respectively. Note that

MS, = s s ~ = 6 2 (a - l)(b - 1)

The variation quantities from the ANOVA analysis are summarized in Table 13-6 to test the significance of a factor under consideration.

Table 13-6 ANOVA Table for Testing the Significance of a Factor: Randomized Com~lete Block Design ,~ - . - ~ - - ~- -- 0.-


Treatments SSTreatmentS a-1 MS~reatments MSTreatmentS / MSE

Blocks S S ~ ~ o ~ k ~ b - 1 M S ~ l ~ ~ k ~

Error SSE (a - l)(b - 1) MSE

Total N-1

Effectiveness Compared to a completely randomized design (presented in Section 13.-2.2), a of Blocking randomized complete block design reduces the sum of squares and degrees of

freedom for error. In the block design the error sum of squares (see Figure 13-1) and corresponding degrees of freedom are decreased by SSslocks and b - 1 (= (N - a) - (a - l)(b - 1); N = ab), respectively.

Figure 13-1 The effect of blocking on error sum of squares (SSE): (a) completely randomized design; (b) randomized complete block design.


Effectiveness The effectiveness of blocking depends on the significance of block effects. If the of Blocking effect of blocks is significant, a complete block design will result in a smaller

(cont.) error mean square (MSE) than a completely randomized design, which makes the test more sensitive to detect the effect of treatments.

In general, when the significance of block effects is uncertain, apply blocking and check if the block effects exist. There would be a slight loss in the degrees of freedom for error, which may be negligible in the test unless the number of trials is very small.

Hypothesis Step 1: State Ho and HI. Test Ho: T I = T ~ = ... =z,=O

(F-test) HI: T ~ # O forat leastonei , i= 1 , 2 ,..., a


Step 3: Determine a critical value(s) for a. fa,u-l,(u-l)(b-1)


f 0 ' fa,a-l,(u-l)(b-1)

Multiple The multiple comparison methods for the completely randomized design (see Comparison Section 13-2.3) are applicable to the complete block design. As an example, the

Methods value of Fisher's LSD is I

Residual The complete block design model has three assumptions on the error term: (1) Analysis normality, (2) constant variance, and (3) randomness. These assumptions can be

checked by analyzing residuals

e, = Y , -5, = Y , -(Y, + Y , -F

These residuals are analyzed by the model adequacy assessment methods ex~lained in Section 13-2.5.

Example 13.6 The effect of grip angle on maximum grip strength (unit: Ib.) is under study. Three grip angles (a = 3; 20°, 50°, and 80" from the horizontal) are selected by the experimenter. Five participants are recruited in the experiment; and each participant exerts hisfher maximum strength at the three different grip angles presented at random. The grip strength measurements are as follows:



5 62 7 1 65 198 66.0 ? X ~ . X ' X / " ~ - ~ n *--. "iW"r*I.w-"UIU - ~ m - a ~ , m a s - - ".-

Totals 297 338 317 952 Averages

~ ~ . B ~ B . . B ~ " . ~ - m " m m a s a e ~,-mv..*m.m,.mmea-

The following sum a b a b

2 N = 15, yx y y 2 = 62,132, y i = 302,942, and Y,j2 = 185,794 i=1 j=l i=l j=l

1. (Test on the Significance of a Fixed-Effects Factor; Complete Block Design) Perform the analysis of variance to test if there is any significant difference in maximum grip strength due to grip angle at a = 0.05.

D T ~ The analysis of variance of the grip strength data are as follows:


Grip Angle 168.1 2 84.1 20.7

Participant 1,511.1 4 377.8

Error 32.5 8 4.1

Total 1,711.7 14

Step 1: State HO and HI. Ho: z1 = z 2 = ... =I-,=()

HI: zi#O forat leastonei , i= 1 t o 3


f o = "~reatments - '1 - - MS~reatments = 841 = 20.7 S ( a - 1 - 1) MS, 4.1

Step 3: Determine a critical value@) for a. fa,a-l,(a-l)(b-1) = f0.05,3-1,(3-1)(5-1) = f0.05,2,8 = 4.46

Step 4: Make a conclusion. Since fo = 20.7 > fa,a-l,(a-l)(b-l) = 4.46, reject HO at a = 0.05.

I/ 1 1 13-20 Chapter 13 Design and Analysis of Single-Factor Experiments: The Analysis of Variance I I



2. (Least Significant Difference Method) Compare these three treatment means by using Fisher's LSD method at a = 0.05.

~h. Step 1: Determine the LSD for a.

I Step 2: Arrange the means of a treatments in descending order. 7,. = 67.6, j7,, = 63.4, and 7,. = 59.4

Step 3: Compare the difference between the largest and smallest means with the LSD. Continue this comparison with the next smallest mean as q

long as the mean difference is greater than the LSD. 2 VS. 1 = 67.6 - 59.4 = 8.2 > 3.0 2 VS. 3 = 67.6 - 63.4 = 4.2 > 3.0


3 vs. 1 = 63.4 - 59.4 = 4.0 > 3.0

Step 5: Summarize the painvise comparison results. 2 3 1 (significantly different from each other)

I 3. (Calculation of Residual) Calculate the residual of y34 = 79.

e34 =y34 - j 3 4 = y 3 4 -(Y3, +J.4 -j7,,)=79-(63.4+77.0-63.5)=2.1

An experiment is conducted to investigate leaking current in a near-micron SOS MOSFETS device. This experiment investigates how leakage current varies as the channel length changes. Four channel lengths (a = 4) are selected by the experimenter; and for each channel length five different channel widths (b = 5) are used. Suppose that channel width is considered a nuisance factor. The data are as follows:

- i = ~ j=l i=l .j=1

1. Perform the analysis of variance to test if there is any significant difference in leaking current due to the difference in channel length at a = 0.05.

I 2. Compare these four treatment means by using Fisher's LSD method at a =

0.05.

1 3. Calculate the residual = 0.8.



Examples (Fixed-Effects Factor; Completely Randomized Design)

13.1 - 3 I (1) Choose File * New, click Minitab Project, and click OK.

orksheet.

Click Comparisons. Check Fisher's, individual er 'ror rate and type '5'

E ' E I

- . . " .- .-..a 1 .--.-. . ,,.-J i =----[ 23 8 L3

SLL'Z SZO'E- SLL'b- SLS'OI- ST

SUOSTlWdUOS aSTl&.lIWd S , I a q S I 4 ...................................................................................... O'SZ O'OZ O'SI 0.01 8 = naaas parood +---------+---------+---------+------

( - _ _ _ * _ _ - _ _ ) I88'Z OOb'IZ S 0 Z

snsJaa slanp!saa u~ ..xapJo snsJaa slanppaa pue ' s ~ g snsJaa slanp!saa 'slanp!sa~ jo $old l a u r ~ o ~ y3ay3 's)old lanp!saa lapun 'sqda~f) y3q3 ( s ) sa1durex3 I -

a3ue!leA jo s!sdleuv aqL :sluaurpadxg .roped-alZu!s jo s!sd1euv pue uZ!saa E 1 laldeq 3 zz-c c


Example 13.4 I (Sample Size Determination; Fixed-Effects Factor)

(1) Choose Stat % Power and Sample Size % One-way ANOVA. Enter the number of treatments (a) in Number of levels. Click Calculate power for each sample size and enter the range of sample size to be examined in Sample sizes and true means of the treatments in Means. Enter also

(2) In Significance level enter the probability of type I error (a).Then click OK twice.

edefined test condition.

I Power and Sample Size I Sigma = 3 Alpha = 0.05 Nunber of Levels = 4 Corrected Sum of Squares of Means = 36.75 Means = 12, 15, 18, 20 ......................... Sample Size Power i

3 0.6253 ; 4 0.8308 I 5 0.9322 j 6 0.9752 1 4 ......................... w


Example 13.6 (Randomized Complete Block Design) I (1) Choose File % New, click Minitab Project, and click OK.

Click Graphs. Under Residual Plots, check Normal plot of residuals. Residuals versus fits, and Residuals versus order. in Re

- 2

msiduals versus


7


Click Results. In Display means corresponding to the terms, select

I Analysis of Variance (Balanced Designs)

Factor Type Levels Values Angle fixed 3 1 2 3 Partlcip fixed 5 1 2 3 4 5

Analysis of Variance for Strength ....................................................................................... :"""'..."""" :Source DF SS BS : Angle 2 168.13 84.07 20.67 0.001 ~Particip 4 1511.07 377.77 :Error 8 32.53 4.07

13.6.1 :Total 14 1711.73 : .........................................................................................................

I Means Angle N Strength

1 5 59.400 2 5 67.600 3 5 63.400



This analysis of variance is summarized in the following table: Source of Sum of Degrees of Mean Variation Squares Freedom Square fo

Exercise 13.1

I Coating type 1,060.5 4 265.1 16.3

1. (Test on the Significance of a Fixed-Effects Factor; Completely Randomized Design)

The sums of squares for the analysis of variance are as follows:

I Error 243.3 15 16.2

I Total 1,303.8 19

Step 1 : State HO and HI. Ho: z l = z2= ... = z a = O

H I : z i#O fora t leas tone i , i= 1 t o 5



f a , a - 1 , ~ - a - f0.05,5-1,20-5 = f0.05,4,15 = 3'06

Step 4: Make a conclusion. Since f , = 16.3 >fa,a-l,N-a = 3.06, reject Ho at a = 0.05.

1 2. (Confidence Interval on pi)

1-a=0 .95 3 a=0 .05 1 95% two-sided CI on LI I :


Exercise 13.1 ( 3. (Confidence Interval on pi - &)

(cont.) 1 95% two-sided CI on p 2 - p 3:

Exercise 13.2

Exercise 13.3

Exercise 13.4

I E

(Least Significant Difference Method)

Step 1 : Determine the LSD for a.

LSD = ta,2,N-a - - - 2.131 x 2.846 = 6.07

Step 2: Arrange the means of a treatments in descending order. y2. = J 5 , =145.3, J,, =145.0, J3. =131.5, and 7,. = 129.3


2, 5 vs. 4 = 145.3 - 129.3 = 16.0 > 6.07 2, 5 vs. 3 = 145.3 - 131.5 = 13.8 > 6.07 2, 5 vs. 1 = 145.3 - 145.0 = 0.3 < 6.07


1 vs. 4 = 145.0 - 129.3 = 15.7 > 6.07 1 vs. 3 = 145.0 - 131.5 = 13.5 > 6.07 3 vs. 4 = 131.5 - 129.3 = 2.2 < 6.07


2 5 1 3 4

(Calculation of Residual)

e2, = y2, - y2, = y2, - J2 , = 143 - 145.3 = -2.3 (overestimate)

(Sample Size Determination; Fixed-Effects Factor) Use the operating characteristic curve of Figure 13-6(b) in MR for a fixed- effects factor with vl = 4 (= a - 1 = 5 - I). From the OC curve, the power of the test according to n is found as follows:



Thus, at least n = 3 replicates are needed to achieve the designated power of the test.

Exercise 13.5 (Estimation of Variance Components; Random-Effects Factor) I I ( I ) B2 = MS, = 16.2

I These variance estimates indicate that about 79% (= 62.2178.4) of the total variability in tube conductivity is attributable to coating type.

Exercise 13.6 1 . (Test on the Significance of a Fixed-Effects Factor; Complete Block Design)

The analysis of variance of the leaking current data is as follows:

1 " 2 Y.. - 219.4 27.72 SS~reatments = - ~ i . - - T - - = 5.5

i=1 20

Answers to Exercises 13-29 I 1

Exercise 13.6 I Source of Sum of Degrees of Mean (cont.) Variation Squares Freedom Square fo

Channel Length 5.5 3 1.8 4.4

Channel Width 3.6 4 0.9

Error 5.0 12 0.4

Total 14.1 19

Step 1: State Ho and HI. Ho: z , = z 2 = ... = ~ , = 0

H I : z i + 0 for at least one i, i = 1 to 4


Step 3: Determine a critical value(s) for a. fa,a-l,(a-l)(b-1) = f0.05,4-1,(4-1)(5-1) = f0.05,3,12 = 3'49

Step 4: Make a conclusion. Since f o =4.4> fa,a-l ,(a-l)(b-l , =3.49, reject Ho at a=0.05.

2. (Least Significant Difference Method)

Step 1 : Determine the LSD for a .

Step 2: Arrange the means of a treatments in descending order. Y3. =1.9, Y4. =1.9, y2, =0.9, and y,, =0.8


3 ,4 VS. 1 = 1.9 - 0.8 = 1.1 > 0.87 3 , 4 vs. 2 = 1.9 - 0.9 = 1.0 > 0.87


2 VS. 1 = 0.9 - 0.8 = 0.1 < 0.87


3 4 2 1

3. (Calculation of Residual)

ezl = yZ1 - j Z 1 = y21 - ( j 2 , +Y,l -j,.)=0.8-(0.9+0.9-1.4)=0.4

14 Design of Experiment with Several Factors

14- 1 Introduction

14-3 Factorial Experiments

14-4 Two-Factor Factorial Experiments

14-5 General Factorial Experiments

14-7 2k Factorial Designs

14-7.1 22 Design

14-7.2 2k Design for k 2 3 Factors

14-7.3 Single Replicate of the 2k

Design

14-8 Blocking and Confounding in the 2k

Design

14-9 Fractional Replication of the 2k Design

14-9.1 One-Half Fraction of the 2k

Design

14-9.2 Smaller Fractions: The 2k-p

Fractional Factorial



14-1 Introduction

This chapter presents statistical concepts and methods to design an experiment with multiple factors (2 2) and analyze the data obtained from the experiment. Most of the techniques introduced in Chapter 13 for a single-factor experiment are extended to this multiple-factor experiment.

14-3 Factorial Design

CI Explain the term factorial design. Distinguish between main and interaction effects. Explain why information of interactions between factors is more meaningful than that of the main effects when the interactions are significant.

O Explain the potential pitfall of a one-factor-at-a-time design when interactions exist.

14-2 Chapter 14 Design of Experiment with Several Factors

Factorial A factorial design is an experimental design where all combinations of factor Design levels under consideration are tested at each replicate. For example, if factors A

and B have two and three levels, respectively, a factorial design of A and B runs six (= 2 x 3) combinations of the levels at each replicate.

Main The effect of a factor indicates the change in response by a change in the level of vs. the factor. Two types of factor effects are defined:

Interaction 1. Main effect: The effect of a factor alone. Effects 2. Interaction effect: The effect of an interaction between factors. An

interaction exists between factors if the effect of one factor depends on the condition of the other factor(s).

Figure 14-1 shows that the main effect of A is independent of the level of B and vice versa, indicating the absence of an interaction between A and B. In other words, the main effects of A at the low and high levels of B are the same as 20 i

1

(i.e., 30 - 10 = 20 at the low level of B and 40 - 20 = 20 at the high level of B). 1

Similarly, the main effects of B at the low and high levels of A are the same as 10. This insignificant AB interaction can be identified by calculating the difference of the diagonal averages:

Notice that the Bhigh and Blow lines in the figure are parallel to each other when the interaction of A and B is absent.

Response

50

0 A low A high

Factor A

Factor A A~OW Ahigh Torais Averages

Blow 10 3 0 40 20 Factor B

Bhizh 20 40 60 3 0 Totals 3 0 70 100

Averages 15 35 25 ~vm-~~*-wea~ B%v---- eBa v-s -aBa"--a.e a - * &V ? - m w v s v ve * d b BA *< B% #* a a - B * vm - e -

Figure 14-1 Two-factor factorial experiment: no interaction.

On the other hand, Figure 14-2 illustrates that the main effect of A (or B) varies depending on the level of the other factor, indicating the presence of an interaction between A and B. The main effects of A at the low and high levels of B are 20 (= 30 - 10) and -20 (= 20 - 40) each, resulting in an average of zero as

14-3 Factorial Experiments 14-3

Main VS.

Interaction Effects (cont.)

Response

50 -- -- - - I

0 A low A high

Factor A

Blow 10 3 0 40 20 Factor B

Bhigh 40 20 60 3 0 Totals 5 0 5 0 100

Figure 14-2 Two-factor factorial experiment: presence of an interaction.

the main effect of A. Similarly, the main effects of B at the low and high levels of A are 30 (= 40 - 10) and -10 (= 20 - 30) each, resulting in an average of 10 as the main effect of B. This significant AB interaction can be identified by a nonzero value of the diagonal average difference:

AB = Ahigh Bhigh + Alow Blow - Alow Bhigh + Ahigh Blow - 20 + 10 40 + 30 ----=-2o

2 2 2 2 Notice that the Bhigh and Blow lines in the figure are not in parallel in the presence of an interaction between A and B.

As illustrated in Figure 14-2, a significant interaction can mask the significance of main effects. Accordingly, when interaction is present, information of the main effects is not meaningful any longer because the main effect of a factor depends on the condition of the other factor(s).

Necessity of A factorial design is the only experimental design to check interactions between Factorial factors. When a significant interaction exists between factors, experimenting one

Design factor at a time can produce inappropriate results. For example, in Figure 14-2, the maximum response is obtained at A =Alo, and B = Bhigh by experimenting all the treatment combinations of A and B. But if the experiment had been designed first for A at B = Blow and then for B at a time, the maximum response would have been concluded at A = Ahigh and B = Blow as follows:

(1) First the maximum response at B = BI, would be found at A = Ahigh. (2) Then by experimenting B at A = Ahigh, the maximum response would be

achieved at B = Blow. To avoid this kind of incorrect conclusion in the one-factor-at-a-time design, the factorial design should be used unless the absence of interactions is certain.


14-4 Two-Factor Factorial Experiments

Explain the model of a factorial design with two fixed-effects factors. R Test the significance of the main and interaction effects in a two-fixed factor factorial design. R Compare individual treatment means with each other by Fisher's LSD method. 0 Calculate residuals in a two-factor factorial design. R Explain a method to analyze a two-factor factorial design with a single replicate.

~ w a ~ e e w a ~ * ~ e m m a ~ ~ - e w ~ ~ ~ - - - m m ~ ~ , ~ " ~ ~ - Context 1 . Factor: Two fixed-effects factors (A with a treatments and B with b

treatments) 2. Repetition: n replicates for each of ab treatment combinations;

N (total number of observations) = nab (balanced design; see Table 14-1)

3. Randomization: Complete randomization

Table 14-1 Data Structure of a Two-Factor Factorial Design

Model

Ylll Y211

Y122 Y 222 . . . Y a22 Y.2. V.2.

Factor B

Ylbl Y2b l Yab l

b Y l b 2 Y2b2 ... yab2 Y.b, Y.b

Ylbn Y2bn < w - M w * a * # w # ~ ~ ~ ~ ~ - ~ ~ w ~ a , ~ w ~ ~ ~ ~

Yabn --- mw---

Totals Y1. Y2.. ... Ya.. Y . - - -

Means 71 .. Y2. ... Yo.. Y .

where: y = overall mean a

r i = effect of the ith level of factor A, r j = 0 i=l

14-4 Two-Factor Factorial Experiments 14-5

Model (cont.)

b

p I . = effect of the jth level of factor B, zp = 0 j=l

( T P ) ~ = effect of interaction between A and B, a b

~ ( T P ) , = O for j = 1 . 2 ,..., b , ~ ( T P ) ~ =O,for i = 1 , 2 ,... , a i=l j=1

2 E~~ = random error - i.i.d. N (0, o )

(Notes) 1. q,k - ~ ( p , . , a ~ )

2. In the fixed-effects model, the treatment effects zi9s, Pj's, and ( T P ) ~ ' ~ are defined as deviations from the overall mean; thus, each sum of 'ti's, Pj's, and ( T P ) ~ ' ~ is zero.

ANOVA In the two-factor factorial design, the analysis of variance decomposes the total variability in Y (SSn total sum of squares; degrees of freedom = N - 1) into four components:

(1) Variability due to factor A (SSA; degrees of freedom = a - 1) (2) Variability due to factor B (SSB; degrees of freedom = b - 1) (3) Variability due to interaction AB (SSA~; degrees of freedom

= ( a - l)(b- 1)) (4) Variability due to random error (SSE; degrees of freedom

=ab(n- l ) = N - a b )

In other words, ssT = ssA + ssB + ssAB + ssE

a b n a b n 2 2 -Y.- where: SST = C Y , X ( Y , ~ -F..)' = C C C Y ~

AT

a b n a 2 2 Y... ssA = C ~ C ( L ~ . . - L . ) ' = z%-,

i=1 j=1 k=l i=l

a b n b 2 2 Y... ssB = ~ ~ ~ ( Y ~ , -y.,)' = zY,-,

i=1 ,=I k=l j=1 an a b n a b 2 2

yij Y. . ss, - SS, S S A B = c C C ( ~ i j , - 5 ) ' =CC----

i=1 j=1 k=l i=1 j=1 n iV

By dividing each sum of squares by the corresponding degrees of freedom, the mean squares of A, B, AB, and error are determined as follows:


ANOVA The expected values of these mean squares are (cont.) a

E(MSE) = E(&] = o2 (Note) MSE is an unbiased estimator of 02.

The ratios MSA/MSE, MSB/MSE, and MSAB/MSE have F distributions:

F=- MSAB - F((a - l)(b - I), N - ab) MSE

These F values become small as the effects of A, B, and AB are insignificant (i.e., the null hypotheses Ho: zi = 0, i = 1 to a, Ho: = 0, j = 1 to b, and Ho: (z& = 0 are true). The ANOVA results of the fixed factorial design are summarized in Table 14-2 to test the significance of the effects.

Table 14-2 ANOVA Table for a Two-Factor Factorial Design: Fixed-Effects Model .


A ss A a - 1 MSA MSA I MSE

B SSB b - 1 MSB MSB I MSE

AB ss AB (a - l)(b - 1) MSAB MSAB I MSE

Error SSE N-ab MSE

Total SST N-1

Hypothesis Step 1 : State Ho and HI. Test Ho: zi=O fora l l i7s , i=1,2 , ..., a

(F-test) HI: zi f 0 for at least one i 1 for A

Ho: & = 0 fora l l j ' s , j=1,2 ,..., b HI: Pi # 0 for at least one j

) for B

Ho: ( ~ f i ) ~ = 0 for all combinations of i's and j's HI: ( T P ) ~ # 0 for at least one combination of i and j

) for AB

1 I


Hypothesis Step 2: Determine a test statistic and its value. Test SS, /(a - 1)

Fo = - --- (cant.)

M S ~ F(a - 1, N - ab) for A SSE/(N-ab) MS,

SSB l(b - 1) Fo = - --- MSB F(b - 1, N - ab) for B

SS, l(N - ab) MS,

SSAB /(a - l)(b - 1) MSAB Fo = - _-- F((a - l)(b - l), N - ab) for AB SS, l(N - ab) MSE

Step 3: Determine a critical value(s) for a. f a , a - 1 , ~ - a b forA

fa ,b-1 ,~-ab forB

fa , (a- l ) (b- l ) ,~-ab for AB


f0 ' fa ,a-1 ,~-ab for A

f 0 ' fa ,b-1 ,~-ab for

fo > fa,@-l)(b-l),N-ab for AB

It is recommended that the test for interaction be conducted first followed by the tests for main effects. If interaction is insignificant, the tests for main effects are meaningful; otherwise, the tests for main effects are meaningless because the effect of a factor is subject to the condition of the other factor(s).

Multiple For significant fixed factors, the means of their individual treatments are Comparison compared by a multiple comparison method such as Fisher's least significant

difference (LSD) method (see Section 13-2.3). When interaction AB is insignificant, the treatment means of each significant main effect are compared. When AB is significant, the treatment means of one factor (say A) at a particular level of the other factor (say B) are compared. For a fixed two-factor factorial design, the LSDs are determined as follows:

2MSE for A LSD = ta,2,N-ub JT - - t a / 2 , ~ - a b .,/% for B

for AB

Residual Like the single-factor model (shown in Section 13-2.2), the two-factor factorial Analysis model has three assumptions on the error term: (1) normality, (2) constant

variance, and (3) randomness. These assumptions can be checked by analyzing residuals (eyk = yijk - jijk = yqk - Yy, ) as explained in Section 13-2.5.


Factorial Sometimes a two-factor factorial experiment has only one observation (n = 1) for Design each combination of the treatments. In this case, the error degrees of freedom

with a Single become zero (= N - ab = nab - ab for n = 1) and thus the error mean square is Replicate unavailable.

In this unreplicated design, the interaction mean square can be used as the error mean square, assuming the interaction effect is negligible. However, this no- interaction assumption can yield an improper conclusion if the interaction is significant; thus, data and residuals should be carefully examined to check the significance of the interaction.

Drying T i m (A; unit: min.) 20 (1) 25 (2) 30 (3) Totals Means

74 7 3 78

Example 14.1 (MR 14-2)

Paint 1 64 6 1 8 5 62 1 69.0

5 0 44 92 Type (B) 2 86 73 45 70 1 77.9

. ~ ~ a * " d ~ 8 w - . * * w ' a . 8 " " ~ w ~ ~ . B . ~ a " ~ . m *"-

Totals

An engineer suspects that the surface finish (Y) of metal parts is influenced by drying time (A) and type of paint (B). Three drying times (a = 3; 20,25, and 30 min.) and two types of paint (b = 2) are specifically selected by the engineer and three parts (n = 3) are tested for each combination of the drying times and paint types, yielding the following:

Means 72.3 72.8 75. 1 mmmwel--___- '*,a" ,m,wa.a" sasw-b ."m*d"e_" -._,.. ' . " ~ . ~ " ~ . ~ , " ~ " ~ " .." The following summary quantities are also provided for the surface finish data:

1

a b n a b 2 N = 18, zzx yVk2 = 101,598, y i = 582,726, z y, j ,2 = 877,042,

and 7'2 y,,' = 298,066

1 . (Test on the Significance of Main and Interaction Effects; Two-Factor Factorial Design) Test if the effects of A, B, and AB are significant each at a = 0.05 by the analysis of variance.

The analysis of variance of the surface finish data is as follows:


-- -


The ANOVA results are summarized in the following table:

Saurce of Sumof Degrees of Mean V W w Squaw Freedom Square h

Drying Time (A) 27.4 2 13.7 0.07

Paint Type (B) 355.6 1 355.6 1.90

AB 1,878.8 2 939.4 5.03

Error 2,242.7 12 186.9

Total 4,504.4 17

Step 1 : State Ho and HI. Ho: zi = 0 for i = 1 to 3 HI: zi # 0 for at least one i

) for A

Ho: P j = O f o r j = l and2 HI: Pi # 0 for at least one j

) for B

Ho: (TP)~ = 0 for all combinations of i's and j's HI: (TP)~ # 0 for at least one combination of i and j

) for AB


0 - MSB - 355.6 - 1

for B f ----- MS, 186.9

MSAB - 939.4 fo = - - - = 5.03 for AB MS, 186.9

Step 3: Determine a critical value(s) for a. fa,a-1,N-ab = f0.05,3-1,18-6 = f0.05,2,12 = 3'89 for A

fa,b-1,N-ab = f0.05,2-1,18-6 = f0.05,1,12 = 4'75 for -

fa,(a-l)(b-l),N-ab - f0.05,(3-1)(2-1),18-6 = f0.05,2,12 = 3'89 for AB

Step 4: Make a conclusion. Since fo = 5.03 > fa,(a-l)(b-l),N-ab = 3.89, reject Ho at a = 0.05 for

AB. Due to the significant interaction, tests on the main effects of A and B are meaningless.




2. (Multiple Comparison) Compare the surface finish means of the three drying times for paint type 1. The corresponding surface means are

YII , =62.7, y2,. =59.3, and =85.0

rm Step 1 : Determine the LSD for a.

Step 2: Arrange the means of a treatments in descending order. y3,, =85.0, TI,, =62.7, and Y21, =59.3


3 VS. 2 = 85.0 - 59.3 = 25.7 > 24.32 3 VS. 1 = 85.0 - 62.7 = 22.3 < 24.32


1 vs. 2 = 62.7 - 59.3 = 3.40 < 24.32

I Step 5: Summarize the painvise comparison results by arranging the a treatments in order of mean and then underlining treatments whose

I means are not significantly different. 3 1 2

3. (Calculation of Residual) Calculate the residual 0fy313 = 92.

h e313 = y,,, - j313 = Y , , ~ - j731, = 92 - 85 = 7 (underestimate)

An experiment is conducted to determine whether either firing temperature ( A ) or furnace position (B) affects the baked density (Y) of a carbon anode. Three firing temperatures (a = 3) and two furnace positions (b = 2) are specifically selected by the experimenter and three measurements (n = 3) are collected for each combination of the treatments. The measurements are as follows:

Temperature (A; unit: O C ) SOO(1) 825 (2) 850 (3) TOMS Means

5.70 10.63 5.65 1 5.65 10.80 5.10 65.69 7.30

Position (B)

2 5.47 10.26 5.38 62.10 6.90 5.21 10.04 5.32

Totals 33.14 62.04 32.61 127.79

I Means 5.52 10.34 5.44 7.10

1 The following summary quantities are also provided for the baked density data: a b n b

2 ( N =18, ~ ~ ~ Y C =1,003.1, gy,' =6,010.6, ZY,~,' =8,171.6,


- - 14-5 General Factorial Experiments 14-1 1

and z C yb,2 = 3,007.7

1 1. Test if the effects of A, B, and AB are significant each at a = 0.05 by the analysis of variance.

2. Test the differences between the baked density means of the three firing temperatures (A): TI,, = 5.52 , y,,, = 10.34, and y,,, = 5.44

3. Calculate the residual ofy213 = 10.43.

14-5 General Factorial Experiments

Factorial An experiment can include more than two factors. The analysis of variance Design procedure for this general factorial experiment is just an extension of that of a

with two-factor factorial design but requires tedious computation; thus, a computer More Than software program is mostly used for the analysis.

Two Factors

14-7 2k ~actorial Designs

14-7.1 2* Design

O Estimate the main and interaction effects of a 22 factorial desi n. 9 0 Test the significance of the main and interaction effects of a 2 factorial design.

2k Design A 2k factorial design is a factorial design with k factors, each having two levels. The low and high levels of each factor are denoted by - (or '-1 ') and + (or ' 1 '), respectively. A complete replicate of this design requires 2k observations, which is the smallest number of observations for a complete factorial design with k factors.

Due to the minimum number of runs, the 2k design is particularly useful for an experiment with many factors for screening purposes. Two treatments are chosen for each factor by assuming a linearity of the response over the range of the factor levels.

ANOVA for The totals of the treatment combinations in a 22 design (2k design where k = 2 22 Design factors) are denoted by a series of lowercase letters: (I), a , b, and ab (see Figure

14-3). The presence (absence) of a letter indicates that the corresponding factor is at the high (low) level in the treatment combination. For example, the letter a indicates that factor A is at the high level and factor B is at the low level.


ANOVA for 2' Design

(cont.)

high (+I; +)

low (-1; -)

low A high (-1; -1 (+I; +)

Figure 14-3 The 22 factorial design.

The main effects A and B and interaction AB with n replicates are estimated as follows:

The linear combinations in the numerators of the main and interaction equations are called orthogonal contrasts (denoted as C) as they have the following two properties:

(1) Each linear combination has coefficients {ci) whose sum is zero

i

(2) Any two contrasts have coefficients {ci) and {dif whose sum of the

products of the coefficients is zero ( x cidi = 0 ). i

By using the orthogonal contrasts, the sums of squares for A, B, and AB are calculated:

C: [a+ab-b-(1)12 - SS, =7- 4n n C ci

i=l

[ab + (1) - a - b12 4n

Note that the total and error sums of squares are calculated by 2 2 2

Y... ssT = c c R Y : -, i=1 j=l k=l

14-7 2k Factorial Designs 14-1 3

ANOVA for Since each contrast sum of squares has a single degree of freedom, the mean 22 Design squares of A, B, and AB are equal to the corresponding sums of squares each. The

(cont.) mean squares of error is SS, - ss, MsE =---

4(n-1) N - 4

The ratios MSA/MSE, MSBIMS~, and MSAB/MSE have F distributions:

These F values become small as the effects of A, B, and AB are insignificant (i.e., Ho: zi = 0 , Ho: P , = 0 , and Ho: ( T P ) ~ = 0 are true). The variation quantities

from ANOVA are summarized in Table 14-3.

B SSB 1 MSB MSB I MSE

AB SSAB 1 M ~ A B MSAB I MS,

Error SSE N - 4 MSE

Total SST N - 1 *"" " - * _ X m * ~ - - ~ ~ * - . ^ I__ - """* ",- X1 L X I---- X1 =-- - XIWII-"- r X1 awemIIX^XXI V

Hypothesis Step 1 : State HO and HI. Test Ho: zi = 0 for i = 1 and 2

(F-test) HI: zi # 0 for at least one i ) for A

Ho: Pi = 0 for j = 1 and 2 H I : & # 0 for at least one j

) for B

Ho: ( T P ) ~ = 0 for all combinations of i's and j's H I : # 0 for at least one combination of i and j

) for AB

Step 2: Determine a test statistic and its value. ss, 11

F,, = - --- MSA F(1,4(n - I)) for A SS, I4(n - 1 ) MSE

F,, = SS, I1 - --- MSB F(1,4(n - 1 ) ) for B

SS, I 4(n - 1) MS,

F,, = --- ' ' A ~ ' I - MSAB F(1,4(n - 1 ) ) for AB SS, I 4(n - 1 ) MSE

14-14 Chapter 14 Design of Experiment wit11 Several Factors

Hypothesis Step 3: Determine a critical value(s) for a. Test

(cont.) fa , l , 4 (n - l ) for A, B, and AB

r Example 14.2

I Thus, the estimates of the main and interaction effects are

Exercise 14.2


fo > fa , l , 4 (n - l ) for A, B, and AB

(Calculation of Effects and Sums of Squares; 22 Design) Consider the surface finish data in Example 14-1. Assume that only two drying times (20 and 30 min.; a = 2) and two types of paint (b = 2) were tested. The following summary quantities are provided for the surface finish data:

n = 3 , (1)=188, a = 2 5 5 , b=246,and ab=196 Estimate the effects A, B, and AB and calculate the corresponding sums of squares.

n- The orthogonal contrasts of the 22 design for the surface finish data are C, =a+ab-b-(1)=255+196-246-188=17

CB =b+ab-a-(1)=246+196-255-188=-1

CAB =ab+(l)-a-b=196+188-255-246=-117

I Then, the corresponding sums of squares are

Consider the carbon anode baked density data in Exercise 14- 1. Assume that only two firing temperatures (800 and 850 "C; a = 2) and two furnace positions (b = 2) were tested. The following summary quantities are provided:

n = 3 , (1)=17.18, a =16.65, b = 15.96, and ab=15.96 Estimate the effects A, B, and AB and calculate the corresponding sums of squares.


14-7.2 2* Design for k 2 3 Factors

O Identify the orthogonal contrasts of a 2k design. Estimate main and interaction effects in a 2k factorial design.

- m m w a "mmeb*a-m-- s m s - - m e *-- ---- m- aswsd- ---*bms m h w * * --%. -- Orthogonal As shown in the anal sis of variance for a 22 design, orthogonal contrasts are Y Contrasts of useful to analyze a 2 design. The effects of factors and their interactions and

2k Design corresponding sums of squares can be calculated by using orthogonal contrasts: Contrast

Effet = n2k-'

To determine the orthogonal contrasts of a 2k design in a systematic manner, a table of plus and minus signs is used (see Table 14-4 for a 23 design). This table includes columns of treatment combinations, identity ( I ) , main effects, and interactions. In a main effect column, a plus (minus) sign indicates the high (low) level of the corresponding factor for the treatment condition (e.g., in the main effect column A, a plus is assigned to the treatment combination ac because the level of A is high. Once the signs of the main effects are established, the signs of each interaction column are determined by multiplying associated columns row by row (e.g., AB =A x B). Except the identity column, each column includes an equal number of plus and minus signs.

By using the sign table, main and interaction effects and their sums of squares are calculated. For example, in Table 14-4, the main effect A and the corresponding sum of squares when n = 2 are determined as follows:

Table 14-4 Signs for Orthogonal Contrasts in the Z 3 Design Treatment Effi,cts

Combinations f A 3 AB C AC BC ABC

C + - - + + - - + ac + + - - + + - -

bc + - + + - + - -

abc + + + + + + + +

Estimate the main and interaction effects of this 23 design and calculate the corresponding sums of squares.

~ - r r From Table 14-4, the orthogonal contrasts of the 23 design for the grip strength data are

C A =a+ab+ac+abc-(1)-b-c-bc

= 4 0 + 3 3 + 4 2 + 3 4 - 3 5 - 3 1 - 3 6 - 3 0 = 1 7 CB = b+ab+bc+abc-(1)-a-c-ac

=31+33+30+34-35-40-36-42~-25 CAB =(l )+ab+c+abc-a-b-ac-bc

=35+33+36+34-40-31-42-30~-5 C , =c+ac+bc+abc-(1)-a-b-ab

=36+42+30+34-35-40-31-33=3 C,, =(l )+b+ac+abc-a-ab-c-bc

=35+31+42+34-40-33-36-30=3 CB, =(l)+a+bc+abc-b-ab-c-ac

=35+40+30+34-31-33-36-42~-3 CAB, =a+b+c+abc-(1)-ab-ac-bc

=40+31+36+34-35-33-42-30~1



Example 14.3

I The corresponding sums of squares are

(Calculation of Effects and Sums of Squares; 2k Design) Power grip strength is measured on a handle to examine the effects of three handle design factors: diameter (A), angle (B), and weight (C). Two treatments are chosen for each factor and a single replicate (n = 1) is run. The grip strength measurements (unit: kg) are as follows:




An engineer is interested in the effects of cutting speed (A), metal hardness (B), and cutting angle (C) on the life of a cutting tool. Two levels are chosen for each factor and two replicates (n = 2) of this 23 factorial design are run. The tool life data (unit: hours) are as follows:

a 325 43 5 760 380.0 b 354 348 702 351.0 ab 552 472 1024 512.0 c 440 453 893 446.5 ac 406 377 783 391.5 bc 605 500 1105 552.5

Estimate the main and interaction effects of the factors and calculate the corresponding sums of squares.

14-7.3 Single Replicate of the 2* Design

Explain a method to analyze a two-factor factorial design with a single replicate. O Identify significant high-order interactions in a 2k design by a normal probability plot. - - ' - - - I X(" -- ni le-w.#i-n^-w-- t --wv*rraixi-lea a," % w^wB ?A ar ra.. *," -*ax- i.i a a i -" xi rx - -.* llxa

2k Design Like the factorial design with a single replicate (presented in Section 14-4), an with a Single assumption is necessary for an unreplicated 2k design to have the error mean

Replicate square. Interactions higher than a second-order interaction are often assumed negligible and then used as the error mean square. Caution should be exercised to check if the interaction(s) used for the error term is significant.

Significant high-order interactions can be identified by constructing a normal probability plot of main and interaction effects. Negligible effects will be normally distributed with mean zero and variance 02 and thus lie along a straight line in the normal probability plot, whereas significant effects deviate from the straight line (see Figure 14-21 of MR).


lr Example 14.4


(Normal Probability Plot of Effects) Consider the grip strength experiment with a single replicate in Example 14-3. The estimates of the main and interaction effects are as follows:

A = 4.25 AC= 0.75 B = -6.25 BC= -0.75 C = 0.75 ABC= 0.25 AB= -1.25

Construct a normal probability of these effects and check if the third-order interaction ABC is significantly important.

w The normal probability plot of the main and interaction effects below indicates that the second- and third-order interactions (including ABC) are not important. Thus, ABC can be used as the error term to test the significance of the main and other second-order interaction effects.

-5 0 5

Effect

An experiment has run a single replicate of a 24 design and calculated the following factor effects:

A = 80.25 AB = 53.25 ABC = -2.95 B = -65.50 AC = 11.00 ABD = -8.00 C = -9.25 AD = 9.75 ACD = 10.25 D = -20.50 BC = 18.36 BCD = -7.95

BD = 15.10 ABCD = -6.25 CD =-1.25

Construct a normal probability of these effects and check if the third- and fourth- order interactions are significantly important.

14-8 Blocking and Confounding in the 2* Design

CI Construct 2P blocks for a 2k design. 0 Estimate main and interaction effects of a 2k design with 2P blocks.

14-8 Blocking and Confounding in the 2k ~ e s i ~ n 14-19

2k Block Sometimes all treatment combinations cannot be run under homogeneous Design conditions for various reasons. For example, suppose that a 2, design (with 4

and treatment combinations) requires a total of 16 hours (4 hours for each Confounding combination) to run a single replicate; thus, this experiment is conducted in two

days (blocks) and randomization is restricted within each block. This kind of a factorial design is called 2k design with 2* blocks, p < k.

Some effects are confounded with blocks in a 2k block design. As an example, Figure 14-4 displays a 2' block design with interaction AB is confounded with the blocks. In this block design, the contrasts (C) to estimate the effects of A, B, and AB are

C, =ab+a-b-(1) C, =ab+b-a- (1) C,, =ab+( I ) - a -b However, the AB contrast is identical to that of the block effect and thus AB is confounded with the blocks.

Block 1 Block 2

Figure 14-4 A 22 block design.

Construction To construct a 2k design with 2' blocks, first,p effects should be selected for of Blocks confounding. In general high-order interactions are chosen to confound with

blocks; main and low-order interaction effects are less likely selected due to their importance. In addition to thep effects, 2P - p - 1 effects (interactions of thep effects; called generalized interactions) are confounded with blocks. For example, in a 23 design with 2' blocks, if AB and BC are selected for confounding, their generalized interaction AC (= AB x BC = AB'C because B~ =

I) is also confounded. Care should be taken not to confound effects of interest.

Second, based on the selectedp effects, defining contrasts (L) are formulated: L = a ,x , +a,,x, +...+ a,xk

0, if the ith factor is not in the counfounded effect where: a; =

1, if the ith factor is in the counfounded effect

0, if the ith factor is not in the treatment combination

1, if the ith factor is in the treatment combinaiton

For example, in case that AB and BC are selected for confounding in a 23 design with 2, blocks, two defining contrasts are established from AB and BC as follows:

L, = X , + x2 and L, = x, + x,

i 0, if the ith factor is not in the treatment combination where: xi =



Construction Lastly, treatment combinations having the same values of L (modulus 2) will be of Blocks placed in the same block. For example, for the 23 block design with L1 = xl + x2

(cont.) and L2 = x2 + xg, the four blocks of the treatment combinations are determined according to their values of Ll (mod 2) and L2 (mod 2) as follows:

(0 + 0) (mod 2) = 0 (0 + 0) (mod 2) = 0 1 a (1 + 0) (mod 2) = 1 (0 + 0) (mod 2) = 0 2 b (0 + 1) (mod 2) = 1 (1 + 0) (mod 2) = 1 3

ac (1 + 0) (mod 2) = 1 (0 + 1) (mod 2) = 1 3 bc (0+ l ) (mod2)= 1 (1 + l ) (mod2)=0 2

abc + lxmod 2J = 0 Q +A(mod 2) = 0 amm*InraarrarP--~2~-ns rannaxrrx- *- smw - - -- --*- mwr ----- 1 -*-

ANOVA The analysis of variance procedure by using orthogonal contrasts presented in Section 14-7.2 is applicable to the 2k block design except treating the confounded effect(s). The confounded effect(s) is redefined as the block effect and the significance of the block effect can be tested by comparing with the error mean square. Recall that in case of a single replicate experiment, a high-order interaction(s) can be used as the error mean square.

Example 14.5 The effects of backrest angle (A) and location of lumbar support (B) on the disc pressure (unit: N ) at the low back are under study. Two backrest angles (90" and 120") and two lumbar support locations (1 and 5 cm) are selected in the experiment. 1. (Construction of Blocks; 2k Block Design) Suppose that the four treatment

combinations cannot be examined under the same condition. Establish two blocks with AB confounded for the 22 design.

I h The defining contrast is formulated from AB: L = x l + X 2

i 0, if the ith factor is not in the treatment combination where: xi =


I The four treatment combinations are assigned to two blocks according to the

(0 + 0) (mod 2) = 0 1 a (1 + 0) (mod 2) = 1 2 b (0 + 1) (mod 2) = 1 2

ab + lL(mod2 = 0 1 1 -,, -,-, * - ----. a

I 2. (Calculation of Effects and Sums of Squares) The following disc pressure data are collected:

I Calculate the sums of squares for this 2' block design with AB confounded.

14-9 Fractional Replication of the 2k Design 14-21

Example 14.5 - The orthogonal contrasts of the 22 design for the disc pressure data are (cont.) C, =a+ab-b-(1)=320+155-380-620=-525

CB =b+ab-a-(1)=380+155-320-620=-405

CAB =ab+(l)-a-b=155+620-320-380=75

I Thus, the estimates of the effects are

75 lock effect = AB = 5 n2k-1 = - 22-1 = 75


Exercise 14.5 Consider the tool life experiment in Exercise 14.3. Suppose that the eight (MR 14-29) treatment combinations cannot be examined under the same condition.

1. Establish four blocks with AB and AC confounded for the 23 design.

2. Calculate the sums of squares for this 23 design with four blocks. The orthogonal contrasts of the tool life length data have been calculated as follows:

C , = 146, C B = 674, Cc = 574,

CAB = -90, CAc = -954, CBc = -194, and CAB, = -278

14-9 Fractional Replication of the 2k Factorial Design

14-9.1 One Half Fraction of the 2" Design

0 Describe the use of a fractional factorial design. Explain the terms generator, defining relation, principal fraction, alternate fraction, aliasing, and dealiasing.

D Construct a 2k-' design.

0 Estimate the main and interaction effects of a 2k-' design.

Identify the alias structure of a 2k-1 design. am- * -* *- #u.mwm s "+ a m a* # -" s m s w w * me-*-* a >---- - w w w w " * -wwA-wm meawawm A " m w e - * d w w w m a - w *a


2 k - p A 1/2P (p < k) fraction of a 2k design (denoted as 2k-p ) can be run when high- Fractional order interactions are negligible. As the number of factors (= k) increases, the

Factorial number of runs required increases geometrically but a large portion of the runs Design are used to estimate high-order interactions, which are often negligible. For

example, a 25 design (k = five factors) which requires 32 runs uses, out of the 3 1 degrees of freedom, 5, 10, and 16 degrees of freedom to estimate the main effects, two-factor interactions, and three-factor and higher order interactions, respectively. When the three-factor and higher order interactions can be assumed

negligible, a 25-' fractional factorial (which requires 16 runs) can be used to analyze the main effects and low-order interactions of the five factors.

A 2k-1 design is constructed by selecting treatment combinations having the Of same sign (say, plus) on an effect column selected (called generator) in the table

2k - I of signs for the 2k design. For example, a 23-1 design can be formed by selecting

Design four treatment combinations (a, b, c, and abc) that have the plus sign on the ABC column in the sign table of the 23 design (Table 14-5). The fraction with the plus sign is called principal fraction and the other fraction is called alternate fraction.

Since the identity column has only the plus sign, a 2k-1 design has the following relations (called defining relation) between the identity and generator columns: the signs of the generator column are equal to those of the identity column for the principal fraction and the opposite becomes true for the alternate fraction. For

example, a 23-1 design with generator ABC has the defining relation I = ABC for the principal fraction and I = -ABC for the alternate fraction.

C + - - + + - - + ac + + - - + + - -

bc + - + - + - + -

abc + + + + + + + +

Another method to construct a 2k-1 design is available as follows: Step 1 : Prepare the sign table of a factorial design with k - 1 factors (called

basic design). (e.g.) For a 23-1 design in Table 14-6, the basic design starts with two factors ( A and B).

Step 2: Add the kth factor and identify the corresponding interaction by using the defining relation of the fractional design.

(e.g.) I = ABC in Table 14-6 3 C.I = ABC~ = C = A B ( f o r c 2 = ~ )

Step 3: Determine the signs of the interaction for the kth factor column.

Step 4: Identify the treatment combinations of the 2k-1 design row by row.

-


Construction Table 14-6 Construct~on of the 23-' Design with I = ABC of

2k - I

.&MI Design , ', ,A , ""r

(cont.) 1 ( 1 ) - - + C

2 a + - - a 3 b - + - b

abc a.r*,aq-~.e*."'Aa-"

In a 23-1 design with I = ABC (principal fraction in Table 14-5), the linear combinations that are used to estimate the main effects and interactions have the

Effects following relationships: I A -1 -2[a-b-c+abc]=l ,c

I , =+[-a+b-c+abc]=IAc

I , =+[-a-b+c+abc]=I, ,

These relationships indicate that each liner combination estimates both a main effect and an interaction:

l ,=A+BC

I ,=B+AC

I , = C + A B

In contrast, a 23-' design with I = -ABC (alternate fraction in Table 14-5) produces the following estimates of the main effects and interactions:

1; =+[-( l )+ab+ac-bc]=A-BC

It is not possible to differentiate between A and BC, B and AC, C and AB. Thus, assuming the second-order interactions are negligible, the linear combinations lA (or I ; ), 1, (or I ; ), and Ic (or I: ) are used to estimate the main effects.

A normal probability plot can be used to identify significant interactions (see Section 14-7.3). Negligible interactions will lie on a straight line in the normal probability plot, whereas significant ones deviate far from the straight line.

Aliasing In a fractional factorial design, some effects cannot be estimated separately and

these effects are called aliases. For example, in a 23-1 design with I = ABC, A and BC, B and AC, and C and AB are aliased with each other.

An alias of an effect is found by multiplying the defining relation to the effect.

For example, a 23-' design with I = ABC has the following alias structure: A = A.I = A.ABC = A2Bc = BC (for A2 = I) B = B.1 = B.ABC = A B ~ C = AC (for B2 = I) C = C.I = C.ABC = A B ~ = AB (for C2 = I)


Aliasing It is a common practice to form a fractional factorial design which aliases the (cont.) main effects and low-order interactions with high-order interactions (which are

mostly negligible).

Dealiasing A sequence of fractional factorial designs can be performed to identify more

detailed information of interactions. For example, a 23-' design with I = ABC aliases the main effects with the two-factor interactions, assuming that the two- factor interactions are negligible. However, if this assumption is uncertain, the

Projection of 2 k - 1

Design

fif Example 14.6

alternate fraction can be run after running the principal fraction. I

By using the data from the principal and alternate fractions, the main effects and interactions can be estimated separately (called dealiasing) as follows:

1 ~ ( 1 , + I > ) = $ ( A + BC+ A - B C ) = A + ( I A - I > ) = B C

When one or more factors are excluded from a 2k-' design due to their lack of significance, the fractional design will project into a full factorial design to

examine the active factors in detail. For example, a z3-' design will project into a 22 design as one factor is dropped.

A study examines the effects of four factors on maximum acceptable weight (MAW; unit: kg) in lifting: object width (A), lifting level (B), lifting distance (C),

and lifting frequency (D). A 24-1 design with I = ABCD is employed, yielding the following data:

' 1. (Alias Structure; 2'-' Design) Identify the alias structure of this 2'-' design 1 with I = ABCD.

[)" The 24-1 design with I = ABCD has the following alias relationships: A = A.ABCD = A ~ B C D = BCD B = B.ABCD = AB~CD = ACD C = C.ABCD = A B ~ D = ABD D = DSABCD = ABCD~ = ABC AB = AB.ABCD = A ~ B ~ C D = CD AC = AC.ABCD = A2B2D = BD AD = AD.ABCD = A ~ B C D ~ = BC




2. (Estimation of Effects) Estimate the factor effects. Which effects appear large?

m- The main effects and interactions are estimated as follows: I , = A + BCD=$(-24+32-24+16-37+15-17+18)=-5.25

I I D = D + ABC=~(-24+32+24-16+37-15-17+18)=9.75

I,, = A B + c D = ~ ( ~ ~ - 3 2 - 2 4 + 1 6 + 3 7 - 1 5 - 1 7 + 1 8 ) = 1 . 7 5

I,, =AD+BC=+(24+32-24-16-37-15+17+18)=-0.25

(Note) The interactions are estimated by forming columns of AB, AC, and AD in the sign table.

I The effects A, B, D, and AC are relatively large.

A 24-' design with I = ABCD is used to study four factors in a chemical process. The factors are A = temperature, B = pressure, C = concentration, and D = stirring rate, and the response is filtration rate. The design and data are as follows:

6 + + - ac 60 -

7 - + + - bc 80 8 + + + + abed 96

---** --w -a -va-*-* r n W * ~ ~ ~ ~ e - * -ww sw *a- -a wv m- d m e M# s a*- * w a -- ms-B&*mm * e * w#m * swaw

1. Estimate the factor effects. Which factor effects appear large? 2. Project this design into a full factorial including only important factors and

provide a practical interpretation of the results.

14-9.2 Smaller Fractions: The 2k-p Fractional Factorial

0 Explain the conditions of design resolutions 111, IV, and V in a fractional factorial design. Construct a 2k-p fractional factorial with a designated design resolution.

0 Identify the alias structure of a fractional factorial design.

Design The term design resolution indicates the pattern of aliases in a fractional factorial Resolution design. The alias patterns of resolution 111, IV, and V designs are compared with

each other in Table 14-7. A resolution I11 design (such as 23-' design with I =

ABC) does not alias any main effects with other main effects but aliases main


Design effects with two-factor interactions and some two-factor interactions with each

other. Next, a resolution IV design (such as 24-' design with I = ABCD) does not (cant.) alias any main effects with other main effects or two-factor interactions but

aliases two-factor interactions with other two-factor interactions. Lastly, a

resolution V design (such as 25-' design with I = ABCDE) does not alias any main effects and two-factor interactions with other main effects and two-factor interactions. The resolution of a fractional factorial design is indicated with a

subscript (e.g., 2;;' denotes a 23-' design with resolution 111).

As the design resolution becomes higher, more information about interactions is produced. For example, a resolution 111 design provides information about main effects only, while a resolution IV design provides information about two-factor interactions as well as main effects. Resolution 111 and 1V designs are often used in experiments to screen factors.

Table 14-7 Alias Patterns of Resolution 111, IV, and V Designs

Main Effects X (Iv) X (Iv)

Two-Factor 0 (111) 0 (111)

Construction To achieve a designated resolution for a fractional design with k factors, a set ofp of generators should be selected properly. Table 14-29 of MR presents

2k-p recommended generators for various 2k-p designs (k I 15), design resolutions Design (111, IV, and V), and runs (n I 128). For example, the table recommends two

design generators (D = AB and E = AC, which are equivalent to the design

relation I = ABD =ACE) for a 2;;' design with n = 8 runs.

With the recommended generators of a fractional design, a complete defining relation is determined. Incorporating the generalized interactions of the generators (identified by multiplying the generators with each other; 2*-p - 1 generalized interactions are produced by p generators), the complete defining relation is formulated. For example, the two (p = 2) recommended generators,

2 ABD and ACE, of the 2;;' design produces one (=2P-p - 1 = 2 - 2 - 1)

generalized interaction: ABD.ACE = A'BCDE = BCDE

Thus, the complete defining relation of the 2:i2 design becomes I = ABD =ACE = BCDE


Like the method to construct a 2k-' design, a 2k-p design is constructed as follows (see Table 14-8 for an example):

2 k - P Step 1: Prepare the sign table of a factorial design with k -p factors (basic Design design). (cont.) Step 2: Addp columns for recommended design generators.

Step 3: Determine the signs of thep columns with the design generators.

Step 4: Identify the treatment combinations of the 2k-p design row by row.

Table 14-8 Construction of the 2&* Design with I= ABD =ACE = BCDE

4 + + - + - abd

6 + - + - + ace 7 - + + - - bc 8 + + + + + abcde

*Iias The alias structure of a 2k-p design is found by multiplying each effect to the Structure of

?k - P complete defining relation. For example, the 2;i2 design with I = ABD =ACE =

A BCDE has the following alias structure:

Design A = BD = CE = ABCDE BC=ACD=ABE=DE B =AD = ABCE = CDE CD=ABC=ADE=BE C=ABCD=AE=BDE D = AB = ACDE = BCE E=ABDE=AC=BCD

When a number of factors under consideration is large (say, k 2 5), high-order (say, third- or higher-order) interactions are assumed to be negligible, which greatly simplifies the alias structure. For example, the alias structure above of the

2&* design with I = ABD = ACE = BCDE can be simplified as follows by

assuming that interactions higher than three factors are assumed negligible: A=BD=CE BC=ACD=ABE=DE B=AD=CDE CD = ABC = ADE = BE C=AE=BDE D=AB=BCE E=AC=BCD

F = BCD, for the 2fG2 design, which yield the defining relation I = ABCE =

BCDF. The two generators produce one generalized interaction: ABCE.BCDF = AB*C?DEF = ADEF

14s7 1. (Construction of a 2 k - p Design) Construct a 2fG2 fractional factorial design.

~ . r r Table 14-29 of MR recommends two (p = 2) design generators, E = ABC and




Thus, the complete defining relation of the 2;;' design is I = ABCE = BCDF = ADEF

I By using the two design generators, the 2:;' design with I = ABCE = BCDF

1 - - - - - -

+ - - - - (1)

2 + ae 3 - + - + + - bef ,

6 + + - - - + - - - -

acf 7 + + bc

10 + - + + + adef -

1 1 - + - + + - bde

13 - - + + + - cde 14 + - + + - - acd

15 - + + + - + bcdf 16 + + + + + + abcdef

~ ~ , " ~ ~ , ~ - ~ - ~ ~ - ~ ~ ~ - ~ # x ~ ~ ~ ~ ~ ~ ~ ~ - ~ - ~ > ~ ~ ~ ~ ~ ~ ~ ~ w ~ ~ * ~ - - m ~ e - ~ ~ ~ ~ ~ . - s m - ~ * . ~ " ~ " ~

2. (Alias Structure) Identify the aliases of the fractional design, assuming that interactions higher than three factors are negligible.

h Assuming that four-factor or higher-order interactions are negligible, the following aliases are found by multiplying each effect to the complete

defining relation of the 2 : ~ ~ design: A = BCE = DEF AB = CE B =ACE = CDF AC=BE C = ABE = BDF AD = EF D=BCF=AEF AE=BC=DF E = ABC = ADF AF = DE F=BCD=ADE BD = CF

BF= CD

1. Construct a 2:i3 fractional factorial design.

2. Identify the aliases of the fractional design, assuming that only the main effects and two-factor interactions are of interest.

- MINITAB Applications 14-29


Example 14.1 (Two-Factor Factorial Design) I (1) Choose File ) New, click Minitab Project, and click OK.

(2) Enter the surface finish data on the worksheet.

---.. L - - L l - - " . + ---- ---- ' -- 1 r l ~ r y l n g T~me Paint Type / Surface ~ l n l s h l g l

(3) Choose Stat > ANOVA ) Two-way. In Response select Surface Finish and in Row factor and Column factor select Drying Time and Paint Type, respectively. Check Display means and Store residuals. Then click OK.

I (4) Interpret the analysis results.

F I Two-way Analysis of Variance I jAnaly91= of Variance for Surface :$ource DT 55 ns F P i iDrylnp- T 2 27 14 0.07 0.930 i :Palnc Ty 1 356 356 1.90 0.193 i Ilnteractxon 2 1879 939 5.03 0.026 i !Error 12 2243 187 :ToCaI 17 4504


Example 14.2 (2' Factorial Design)


(3) Choose Stat * ANOVA ) Two-way. In Response select Surface Finish and in Row factor and Column factor select Drying Time and Paint

I Two-way Analysis of Variance I

I

J ............................................................................................................. :Analysis of Variance f o r Surface ; Source DF SS MS F P i j Drying T 1 24 24 0.13 0.729 i : Pa in t Ty 1 0 0 0.00 0.984 i i I n t e r ac t ion 1 1141 1141 6.08 0.039 : i Error 8 1501 188 Total 11 2666 .. ............................................................................................................ :


Example 14.4 (Normal Probability Plot of Effects) I (1) Choose File * New, click Minitab Project, and click OK.

- .+--. ..-.-. .j,-, J. Effects E-Estimates

1 A 1 4.25 " --I--- - ------.-ma-----.+ J

6 .2q B 4 -----.-I--- 3 AB 1.25 ! - """ "II" 4 C -- ------__ _l_.----.-._l_ l("-.l__X--._--_-

5 AC ---

Choose Graph * Probability Plot. In Variables select E-Estimc and

99 - ML Erf rnstes

95 - Mean 0321428

90 - StOev 2 92072

80 - 70 - 60 - 50 - 40 - 30 - 20

10 -

5 -

1 -

Data



Exercise 14.1 1 . (Test on the Significance of Main and Interaction Effects; Two-Factor I Factorial Design)

I The analysis of variance of the carbon anode density data is as follows:

I a b n 2 127.79~

1,003.1 ------- = 95.87 i=1 j = 1 k=l 18

Source of Sum of Degrees of Mean Variation Squares Freedom Square h

Firino - ----- D I Temperature (A) 94.53 2 47.27 1,056.1 Furnace Position (B) AB

Error 0.54 12 0.04

Total 95.87 t _ w j n ~ , - ~ e w . M w m ~ - * # * . ~ v m ~ B ~ 8 ~ a -an*"~"~.ea;nwAw*~ .e.nriw.-ien.a.na.

17

Step 1 : State HO and HI. Ho: zi = 0 for i = 1 to 3 H I : Ti f 0 for at least one i

) for A

Ho: Pj = 0 fo r j = 1 and 2 H l : Pj # 0 for at least one j

) for B

No: ( ~ ( 3 ) ~ = 0 for all combinations of i's and j's H I : (TP)~ # 0 for at least one combination of i and j

) for AB



- - - --- 7 - ---


Step 3: Determine a critical value(s) for a. fa,,-1,N-ab = f0.05,3-1,18-6 = f0.05,2,12 = 3'89 for A

fa,b-1,N-ab = f0.05,2-1,18-6 = f0.05,1,12 = 4'75 for -

fa,(,-l)(b-l),N-ab - f0.05,(3-1)(2-1),18-6 = f0.05,2,12 = 3'89 for AB

Step 4: Make a conclusion. Since fo = 0.9 > f,,~,-l,(,-l,,N-ub = 3.89, fail to reject Ho at a = 0.05

for AB, which indicates further tests on the main effects A and B are meaningful. Next, since fo = 1,056.1 > f ,,,- ,,,-,, = 3.89 for A and

fo = 16.0 > fa,b-l,N-,b = 4.75 for B, we can conclude that both A

(firing temperature) and B (furnace position) significantly affect the baked density of a carbon anode.

2. (Multiple Comparison)

Step 1 : Determine the LSD for a.

Step 2: Arrange the means of a treatments in descending order y2,, = 10.34, 2,. = 5.52, and &., = 5.44


2 vs. 3 = 10.34 - 5.44 = 4.90 > 0.25 2 VS. 1 = 10.34 - 5.52 = 4.82 > 0.25


1 vs. 3 = 5.52 - 5.44 = 0.08 < 0.25


2 1 3

3. (Calculation of Residual)

J,,, = 10.62

e,, , = y 2 1 , - j 2 1 , = y 2 1 3 - y 2 , ~ =10.43-10.62=-0.19 (overestimate)


Exercise 14.2

Exercise 14.3

(Calculation of Effects and Sums of Squares; 2' Design)

The orthogonal contrasts of the 22 design for the baked density data are C, =a+ab-b-(1)=16.65+15.96-15.96-17.18=-0.53

C, =b+ab-a-(1)=15.96+15.96-16.65-17.18=-1.91

CAB =ab+(l)-a-b=15.96+17.18-16.65-15.96=0.53


1 The corresponding sums of squares are

I C: - (-0.53)~ 2 SS, =1;- = 0.02 SS, =-- C~ - = 0.30

(-1.91)~

n2 3x2' n2k 3 ~ 2 ~

(Calculation of Effects and Sums of Squares; 2k Design)

The orthogonal contrasts of the 23 design for the tool life length data are CA =a+ab+ac+abc-(1)-b-c-bc

= 760 + 1,024 + 783 + 81 1 - 532 - 702 - 893 - 1,105 = 146

CB =b+ab+bc+abc-(1)-a-c-ac

=702+1,024+1,105+811-532-760-893-783=674 CAB =(l )+ab+c+abc-a-b-ac-bc

=532+1,024+893+811-760-702-783-1,105=-90

C, =c+ac+bc+abc-(1)-a-b-ab

= 893 + 783 + 1,105 + 81 1 - 532 - 760 - 702 - 1,024 = 574

CAC =(l )+b+ac+abc-a-ab-c-bc

=532+702+783+811-760-1,024-893-1,105=-954 CBc =(l)+a+bc+abc-b-ab-c-ac

=532+760+1,105+811-702-1,024-893-783 = -194

CAB, =a+b+c+abc-(1)-ab-ac-bc

=760+702+893+811-532-1,024-783-1,105 = -278





Exercise 14.4 1 (Normal Probability Plot of Effects)

CABC - -278 ABC=---=-34.8 n2k-1 2 x 23-1

The normal probability plot of the effects below indicates that all the third- and fourth-order interactions are not significant so that we can use these insignificant interactions to form the error mean square.

Effect

Exercise 14.5 ( 1. (Construction of Blocks; 2' Block Design)

I The defining contrasts are formulated from AB and AC as follows: L1 = x1 + x2 and L2 = X, + x3

The eight treatment combinations are assigned to four blocks according to

a (1 + 0) (mod 2) = 1 (1 + 0) (mod 2) = 1 2 b (0 + 1) (mod 2) = 1 (0 + 0) (mod 2) = 0 3

ac (1 +0) (mod2)- 1 (1 + 1) (mod2)=O 3 bc (0 + 1) (mod 2) = 1 (0 + 1) (mod 2) = 1 2

-- a*- Lmod22 = 0 (1 + 1) (mod 2) = 0 1 l a i-w XX(IIIX " ali-m *X. i X l N I X W V X -- X



Exercise 14.6

2. (Calculation of Effects and Sums of Squares) The generalized interaction of AB and AC is

A B X A C = A ~ B C = B C

I Thus, the effects AB, AC, and BC are confounded with the blocks. The estimates of the effects are

I 5 74 C = % - = = 7 1 . 8 ABC = - ' A B C - - 278 - 34.8

n2k-' 2 x 23-1 dk-' 2 x 23-1

Block effect = AB + AC + BC = CAB + CAC + CBC n2 k-'

- - -90 - 954 - 194 = 154.8

2 x 23-1

The corresponding sums of squares are 2 2

ss - C", =-- A --

CB - 14" - 1,332.3 SS, = 7 - - 6742 = 28,392.3 nzk 2 ~ 2 ~ n2 2 ~ 2 ~

1 . (Estimation of Effects) The main effects and interactions are estimated as follows:

I , =A+BCD=t(-45+100-45+65-75+60-80+96)=19.0

I , =B+ACD=$(-45-100+45+65-75-60+80+96)=1.5

1, = C + ABD=$(-45-100-45-65+75+60+80+96)=14.0

I , =D+ABC=$(-45+100+45-65+75-60-80+96)=16.5

I,, =AB+CD=$(45-100-45+65+75-60-80+96)=-1.0

I,, =AC+BD=$(45-100+45-65-75+60-80+96)=-18.5

I,, =AD+BC=$(45+100-45-65-75-60+80+96)=19.0

Three factors A, C, and D show large effects on filtration rate, but factor B displays a relatively very small effect.

2. (Projection into a Factorial Design)

The 24-1 design projects into a 23 design with a single replicate including factors A, C, and D. The main effects and interactions of this unreplicated 23

design can be easily determined by using the effect estimates of the design as follows:

- -



Exercise 14.7

A = 19.0 AC =-18.5 C = 14.0 AD = 19.0 D = 16.5 CD = -1.0

ACD = 1.5

The effects A, C, D, AC, and AD have relatively large influence on filtration ' rate.

1 . (Construction o f a 2k-P Design)

Table 14-29 of MR recommends three @ = 3) design generators, D = AB, E =

AC, and F = BC, for the 2;i3 design, which yield the defining relation I =

ABD = ACE = BCF. These three generators produce four (= 2P - p - 1 = 23 - 3 - 1) generalized interactions:

ABD.ACE = A~BCDE = BCDE ABD.BCF = AB~CDF = ACDF ACE.BCF = A B ~ E F = ABEF ABD.ACE BCF = A ~ B ~ C L ? D E F = DEF

Thus, the complete defining relation of the 2fi3 design is I=ABD=ACE=BCF=BCDE=ACDF=ABEF=DEF

By using the three design generators, the 2f i3 design with I = ABD =ACE =

abd cd

6 + - + - + - ace 7 - + + - - + bcf

2. (Alias Structure) Assuming that three-factor or higher-order interactions are negligible, the following aliases are found by multiplying each effect to the complete

defining relation of the 2;i3 design: A=BD=CE B=AD=CF C = A E = B F D=AB=EF E = A C = D F F=BC=DE AF=CD=BE

Nonparametric Statistics

15- 1 Introduction 15-5 Nonparametric Methods in the Analysis

15-2 Sign Test of Variance

15-3 Wilcoxon Signed-Rank Test MINITAB Applications

15-4 Wilcoxon Rank-Sum Test Answers to Exercises

15-1 Introduction

0 Explain nonparametric statistics in comparison with parametric statistics. sm.,m,sm.~m,ae ~~>~*ss . ,a~- .~~ ,mM.B.a uw,w.m&"mM- .m.m,mm ,mm,~.-a*,~'%~*,-~d*e~~.d.~**a,e~#whw# ~ . ~ . m ~ ~ . ~ s ~ m . ~ , ~ ~ ~ ~ ~ ~ > ~ . ~ ~ ~ ~ , . w . ~ ~ " ~ ' ~ ~ ~ a ~ ~ ~ v , ~ w m ' s ~ ~ ~ % * ~ - ~ , ~ a ~ ~ ~ d ~ b ~ ~ . - , B ~ a a ~ ~ ~ " ~ " ~ . m . ~ ~ * w ~ a ~ ~ ~ ~ ~ * ~ . ~ . , ~ ".BMBw"v"s" .a.wm, ms.aea.s,"a.*a

Nonparametric Nonparametric methods do not use information of the distribution of the Statistics underlying population, whereas parametric methods (such as t and F tests) are

based on the sampling distribution of a particular statistic for the parameter of interest. The nonparametric methods make no assumptions except the underlying distribution is continuous, using only signs andlor &-the original observations may need to be transformed. Therefore, the nonparametric statistics is considered a distribution-free method and does not include the procedure of parameter estimation (point and interval estimations).

Parametric and nonparametric statistics have their own advantages and disadvantages in terms of ease of use, time efficiency, applicability, and sensitivity (statistical power), as summarized in Table 15- 1. Nonparametric tests are easier and quicker to perform and their applications are extended to the following contexts where parametric tests are inappropriate:

(1) Underlying distribution unknown: Information of the underlying distribution is unavailable.

(2) Normal approximation inapplicable: Normal approximation is not applicable when the underlying distribution is significantly deviated from a normal distribution and the sample size is small (say, n < 30) (see Section 7-5).

(3) Categorical data: Categorical data include nominal and ordinal data (see Section 9-7).

15-2 Chapter 15 Nonparametric Statistics

Nonparametric However, nonparametric tests are less sensitive (statically powerful) in detecting Statistics a small departure from the hypothesized value because they do not use all the

information of the sample. Therefore, the nonparametric methods usually require a larger sample size to achieve the same statistical power.

Table 15-1 Comparison of Parametric and Nonparametric Statistics Ctitepia Parametric Nonpammtric

Ease of use 0 Time efficiency 0 Applicability 0 Sensitivity 0 - --w -mss w m m - * - m w * -*w*ma wa a a#vaa -# * * - # - es - w %*" -# - * &<--"

(Note) 0 : preferred

15-2 Sign Test

Read the sign test table. Test a hypothesis on the median ( p ) of a continuous distribution by the sign test.

O Calculate the P-value of a sign test statistic. Apply the normal approximation to the sign statistic when n is moderately large.

O Test a hypothesis on the median of differences ( ji, ) for paired samples by the sign test.

Sign Test The sign test uses the plus and minus of the differences between the observations and the hypothesized median P o . The sign test assumes that the observations are drawn from a continuous distribution.

The sign test statistics R- , R+ , and R are determined by the following procedure:

Step 1 : Compute the differences X i - y o , i = 1,2,. . . , n. Step 2: Count the numbers of the minus and plus signs of the differences

R- = number of the minus signs

R+ = number of the plus signs

R = m i n ( ~ - , R + )

Note that ties (xi = p,) are not included in the sign test because the probability

that a value of4 is exactly equal to Po is zero in a continuous distribution:

Sign Test The sign test table (see Appendix Table VII in MR) provides critical values

(r: , ) of a sign test statistic for various sample sizes (n) and levels of

significance (a).

-

15-2 Sign Test 15-3

Sign Test Table

(cont.)

Inference Context

for p

P-value Calculation

Sign Test Procedure

for ii

(e.g.) Reading the sign test table

('1 r0*05,10 =

(2) ro:025,30 = 9

(3) ri005,25 =

Parameter of interest: $ (median)

Test statistic of p : The appropriate sign statistic is selected depending on the type of HI as follows:

(1) R-=numberoftheminussinnsof Xi - p , , i = 1,2, ..., n,forHl: p > p 0

(2) R+ = number of the plus signs of Xi - P o , i = 1,2, . . . , n, for H1: p < Po (3) R=min(R+,R-) forH1: p # p o

The sign statistics R- , Rf , and R follow the binomial distribution B(n,0.5) ; therefore, the P-value (significance probability) of a sign statistic is

Recall that the null hypothesis Ho is rejected if P 5 a (see Section 9-1).

Step 1: State HO and H I . H,,: p=po HI: ji ;t Po for two-sided test

p < Po or p > PO for one-sided test

Step 2: Determine a test statistic and its value. R- = number of the minus signs of Xi - Po for upper-sided test

R + = number of the plus signs of Xi -Po for lower-sided test

R = min(R-, R') for two-sided test

Step 3: Determine a critical value(s) for a. r:12,, for two-sided test; ri,, for one-sided test


r- 5 r:,, for upper-sided test + * r 5 r for lower-sided test

r 5 r for two-sided test


Normal Approximation

Comparison to the t-Test r Example 15.1

Recall that a binomial distribution B(n, p ) approximates to a normal distribution with p = np and o2 = np(1 - p ) if np > 5 and n(1 - p) > 5 (see Section 4-7). Since

R+ ( R - or R) follows B(n, 0.5), if n is moderately large (say, n > lo), the sign statistic has approximately a normal distribution with mean and variance

y = 0.5n and 02 = 0.5~12

Therefore, a hypothesis on the median by R+ can be tested by using the z-test statistic

R+ - 0.5n 2, =

0.5&

If the underlying distribution is normal or nonnormal but symmetric, the t-test is preferred to the sign test for better statistical power.

Test scores of 12 students in a statistics class are as follows:

1. (Test on p ; Sign Test) Test if the median of the test scores is equal to 80 at a = 0.05.

r ~ - Step 1: State Ho and HI. Ho: p = 80

HI: p # 80

Sign + - + - + - - * - + + + #w"wAs--w-m -mr-m~e-~--sw~aa"wmw".am.a.~~~z~~"~~~~~ B.-- ".B* 'au%v.w a .m.ammm ~F,*r*ree~*a.s-asd. *.d-"9e.B. B. b.*a,aw~s"B~um-~m-: mwm"m,.-m-..m.,m,-.m.m"s~,

* Excluded since xi = Po. r =min(r-,r') = min (5,6) = 5


r:,2,n = ri025,11 = 1 (Note) number of ties = 1


Since r = 5 5 r,* ,,,, = 1 , fail to reject Ho at a = 0.05.

2. (P-value Calculation) Find the P-value of the sign statistic r = 5 for the two- sided test.

Since P = 1.0 5 a = 0.05, fail to reject Ho at a = 0.05.

-

15-2 Sign Test 15-5



3. (Normal Approximation) Apply the normal approximation to the sign statistic R and draw a conclusion at a = 0.05.

m- Step 1: State Ho and HI. H,: ji = 80

Step 2: Determine a test statistic and its value. r-0.5n 5 - 0 . 5 ~ 1 1

z , = - - = -0.30 0.5& 0 . 5 ~ - f i

Step 3: Determine a critical value(s) for a. '1x12 = '0.0512 = '0.025 =

Step 4: Make a conclusion. Since (z,( = 0.30 2 z,,,,~ = 1.96, fail to reject HO at a = 0.05.

The impurity level (unit: ppm) is routinely measured in an intermediate chemical product. The following data are observed in an impurity test:

" \ @@V# a$fjj\C58 bs*e@ 8 ~ ~ ~ # # ~ ~ ~ ~ 4 ~ ~ ~ & J m c 8 ~ & 7 ~ & ~ ~ $ ~ ~ & ~ ~ $ ~ ~ # # ~ ~ ~ $ f i w # f l ~ ~ ~ @ ~ ~ ~ ~ ~ @ ~ ~ @ ~ ~ d A s ld s 8 b ~srB# 6kee 4, 4 i #&#* Imurit A') 2.4 2.5 1.7 1.6 1.9 2.6 1.3 1.9 2.0 2.5 2.6 ;( $ g # # * & ~ & $ ~ # @ t ~ @ ~ ~ # & ~ g ~ & ~ J ~ I m ~ u r 2 C _ X ) 2.3 2.0 1.8 1.3 1.7 2.0 1.9 2.3 1.9 2.4 1.6

unrnun- a*m w .xn -a ma- iaxr n a n n r p a - a r =xnixana.- *ma rraarin-w-rxlaam--ww- n - r i il s l m r a r ~ n . a x a i . n ~ r a a e m n . x n v ~ - m *r- a n e n a n m a n *.

1. Test if the median of the impurity level is than 2.5 ppm at a = 0.05.

2. Find the P-value of the sign statistic r + = 2 for the lower-sided test. 3. Apply the normal approximation to the sign statistic R+ and draw a conclusion

at a = 0.05.

- Inference Parameter of interest: p, (median of the paired differences Dj = XV - X2j ) Context

Test statistic of ji, : The appropriate sign statistic is selected depending on the for p, type of HI as follows:

(1) R- = number of the minus signs of D j , j = 1,2, . . . , n for HI : ji, > 6,

(2) R + = number of the plus signs of D j , j = 1,2,. . . , n for HI: p, < 6,

(3) ~ = r n i n ( ~ - , R + ) forH,: ji, +6,

Sign Test Step 1 : State Ho and HI. Procedure Ho: Po =6,

for ji, HI: p, + 6 , for two-sided test

ji, < 6, or ji, > 6, for one-sided test


Sign Test Step 2: Determine a test statistic and its value. Procedure R- = number of the minus signs of D j for upper-sided test

for pD (cont.) Rf = number of the plus signs of D j for lower-sided test

R = min(R-, R + ) for two-sided test

Step 3: Determine a critical value(s) for a. r,*,,,, for two-sided test; r , for one-sided test


r 5 , for upper-sided test + * r I ran for lower-sided test

r 5 r for two-sided test

Example 15.2 (Test on p D ; Sign Test) Scores of 10 students for two exams are below. Test I Ho: p D = 0 vs. HI: p D # 0 at a = 0.05.

I No. 1 2 3 4 5 6 7 8 9 1 0

Exam2(X2$ 90 73 98 80 92 72 78 85 75 92 ---em* -am- *ma a -a m--sw---sw * **am" ""-*- -- * ewma%m * -** ex"v- - - --- < * a w*

Step 1: State Ho and HI. Ho: p , = o HI: p , # O

I Steo 2: Determine a test statistic and its value.

I Sign + - * - m.*"we**-,".B,a " . , " m ~ w " w s , ~ " " " w ~ . a , ~ " m . - - # ~ * # ~ ~ " " # ",a,~w.,"s."""*."ww~*".".".~-~~.- " -.---*- s+-"*a"w ".>, "* w.B.w ..w.w 8..s", ".%"*.".""""*,""" -.a .a.s ,-* ~ ~ . - - . ~ " ~ " ~ . , ~ ~ ~ - ~ , * " , ",*"cw

* Excluded since d j = 6,.

Step 3: Determine a critical value@) for a.

r,*,,,, = r;,,,,, = 1 (Note) number of ties = I

Step 4: Make a conclusion. Since r = 1 I r,*,,,, = 1 , reject Ho at a = 0.05.


The diameter of a ball bearing was measured by 12 inspectors. The measurements, corresponding differences, and their signs are presented in the table on next page. Test if there is a significant difference between the medians of the populations of measurements by using two different calipers. Use a = 0.05.

15-3 Wilcoxon Signed-Rank Test 15-7


15-3 Wilcoxon Signed-Rank Test

-- uswe -wm a * a * *>w " a -a # - A #%* a**-- a m e e d.* *a kw mm Ve-B w * *e a*- mr- rs - *".me * . -- Read the Wilcoxon signed-rank test table.

O Test a hypothesis on the mean ( p ) of a continuous distribution by the Wilcoxon signed-rank test. Apply the normal approximation to the signed-rank statistic when n is moderately large.

O Test a hypothesis on the mean of differences (b) for paired samples by the Wilcoxon signed-rank test.

Wilcoxon While the sign test uses only the plus and minus signs of the differences between Signed-Rank the observations X, and the hypothesized median ii, , the Wilcoxon signed-rank

Test test uses both the of the differences between the observations X, and the hypothesized mean po and the & of the absolute magnitudes of the differences. The signed-rank test assumes that the underlying distribution is svrnmetric and continuous, i.e., the mean equals the median.

The Wilcoxon signed-rank statistics W- , W+ , and Ware determined by the following procedure:

Step 1: Compute the differences Xi - p,, i = 1,2, . .. , n.

Step 2: Rank the absolute differences (xi - po 1 in ascending order. For t k ,

use the average of the ranks that would be assigned if they differed. Step 3: Assign the signs of the differences to the corresponding ranks. Step 4: Calculate the sum of the signed ranks:

W- = sum of the negative ranks

W+= sum of the positive ranks

w = min(W-, W ' )


Wilcoxon Like the sign test table, the Wilcoxon signed-rank test table (see Appendix Table Signed-Rank VIII in MR) provides critical values ( wi,, ) of a signed-rank test statistic for

Test Table various sample sizes (n) and levels of significance (a).

(e.g.) Reading the signed-rank test table

(1) w;.05,10 = 10

(2) w;.025,20 = 52

(3) w;.005.,5 = 15

Inference Parameter of interest: p Context Test statistic of p: The appropriate signed-rank statistic is selected depending

for p on the type of HI as follows:

(I) W-=sumofthene.gativeranksof X i - p o , i = 1,2, ..., n forH1: p > p ,

(2) w + = sum of the positive ranks of xi - po , i = 1,2, . . . , n for HI: p < p0

(3) W=min(W-,W+) forHl: p # p 0

Wilcoxon Step 1 : State HO and HI. Signed-Rank Ho: C L = C L ~

Test HI: p # po for two-sided test Procedure

for p p < po or p > p0 for one-sided test

Step 2: Determine a test statistic and its value. W-= sum of the negative ranks of X , - po for upper-sided test

W+ = sum of the positive ranks of Xi - po for lower-sided test

W = min(W-, w+) for two-sided test

Step 3: Determine a critical value(s) for a. w : , ~ , ~ for two-sided test; w;,, for one-sided test


w- I w;,, for upper-sided test

w+ I w;,, for lower-sided test

w 5 w , , for two-sided test

If the sample size is moderately large (say, n > 20), the signed-rank statistic W + Approximation

( W- or I+') has approximately a normal distribution with mean and variance

- n(n + 1) n(n + 1)(2n + 1) Pw+ -- and ow+ = 4 24

Therefore, a hypothesis on the single mean by W+ can be tested by using the following z-statistic:

15-3 Wilcoxon Signed-Rank Test 15-9

Normal Approximation

(cont.)

r Example 15.3 Test scores of 10 students in a statistics class are below.

No. 1 2 3 4 5 6 7 8 9 1 0 Score(X$ 85 75 96 74 87 71 73 80 75 90

* *- *- ww * ww-* --*,-*w+ w- a #-w** -" -4-* aww' weew+ * w - * # - -> * - - 1. (Test on p; Wilcoxon Signed-Rank Test) Test if the mean of the test scores is

equal to 80 at a = 0.05.

m Step 1: State Ho and HI. Ho: p = 80

H,: p +80

Ster, 2: Determine a test statistic and its value.

xi -80 5 -5 16 -6 7 -9 -7 0 -5 10

(xi-801 5 5 16 6 7 9 7 * 5 10

Rank 2 2 9 4 5 . 5 7 5 . 5 * 2 8

Signed 2 -2 9 -4 5.5 -7 -5.5 * -2 8 rank

* Excluded since xi = p, .

w = min(w-, w') = min(20.5,24.5) = 20.5

Step 3: Determine a critical value@) for a. w:,~ ,~ = ~ i , ~ ~ ~ , ~ = 5 (Note) number of ties = 1

Step 4: Make a conclusion. Since w = 20.5 $ w:,,,, = 5 , fail to reject Ho at a = 0.05.

2. (Normal Approximation) Apply the normal approximation to the signed-rank statistic W and draw a conclusion at a = 0.05.

k Step 1 : State Ho and HI. Ha: p = 80

H,: j,l # 80

Step 2: Determine a test statistic and its value. n(n + 1)

W-- 20.5 - 9x(9+1)

za = je n(n+1)(2n+l) = de 9 x ( 9 + 1 ) x ( 2 x 9 + 1 ) = -0.24

1sai pap!s-lamol.~oj fa jo syum ah!l!sod aqljo urns =+A

lsal pap!s-laddn ~ o j 30 syum a~!@au ayljo urns =-A

.anlaa s$! pua 3!p!$aqs qsaq s aupmalaa :Z days

'SO'( ) = x, 1" Uo!Sn~3U03 a ~ s l p pus +A 3!ls!yels yual-pauZ!s aqlol uo!~sur!xo.~dds pm~.~ou aql klddv ' 2

.SO.O = x, 1s wdd s.2 usql35j s! ]anal Lpt~dur! aqljo usaw aqlj! IsaL '1

-

15-3 Wilcoxon Signed-Rank Test 15-1 1

(Test on p D ; Wilcoxon Signed-Rank Test) The exam scores in Exercise 14.2 is

presented again below. Test Ho: p D = 0 VS. H I : p D # 0 at a = 0.05.

, ;%::." "-8 'J $i ' askiwjra #-" 2 %~CW t >~~&~~&~#@f&~j~#~~$~~~$~~j~;~~g&~$~$$~~#gjjj~~#$@&~~$##~&# n &&ej8

Examl(X1) 85 75 96 74 87 71 73 80 75 90 90 73 98 80 92 72 78 85 75 92

*-navasw. * m * & - # * - -** h Step 1 : State Ho and HI.

Ho: p D = O

H,: pD f O

Example

I Rank 6.5 3 3 9 6.5 1 6.5 6.5 * 3

I Signed -6.5 3 -3 -9 -6.5 -1 -6.5 -6.5 * -3 rank

I ~ ----

,~", "s*sm.#*s"- %a%dw-,d,a,a,-a-.#- *e#.~..--e~.*m-.",*a .e-~ ~ " " " ~ , . " v " , " . ~ ' " . ~ ~ e ~ . ~ . ~ " ~ ~ . ~ w , ~ ~ ~ ~ ~ " ~ - ~ . " ~ ~ ' ~ , ~ ~ , " . e . " ~ ~ ~ ~ , ~ ~ s w ~ " . " ~ " e ~ . ~ ~ " " ~ % v " ,~c-ds a.aa""" m,,,m,*, """a,n"*~%v"*b

* Excluded since d j = 6 , .

Step 3: Determine a critical value(s) for a. w : ~ ~ , ~ = w ; , ~ ~ ~ , ~ = 5 (Note) number of ties = 1



Since w = 3 5 ~ i ~ ~ , ~ = 5, reject Ho at a = 0.05.

Reconsider the ball bearing diameter data in Exercise 15.2. The measurements, corresponding differences, and their signed ranks are presented below. Test Ho: p D = 0 vs. H1: p D # 0 at a = 0.05.

Measurement Signed XI 14 Rgok X2 (XI -X2)


15-4 Wilcoxon Rank-Sum Test

0 Read the Wilcoxon rank-sum test table. Test a hypothesis on the difference in mean of two continuous populations (p1 - p2) by the Wilcoxon rank-sum test.

0 Apply the normal approximation to the rank-sum statistic when nl and n2 are moderately large. m . , ~ . ~ , " ~ ~ ~ B - - . m . ~ ~ ~ ~ ~ ~ ~ A ~ * * - . ~ e ' e w a ~ ~ * ~ : ~ - , * - - ~ ~ " ~ ~ ~ ~ ~ ~ ~ u s * ~ m - ~ ~ ~ . , # ~ ~ - ~ ~ . ~ * ~ ' , ~ B a w ~ - ~ . - ~ ~ ~ ~

Wilcoxon The Wilcoxon rank-sum test (often called Mann-Whitney test) uses the ranks of Rank-Sum the observations from two independent continuous populations XI and X2 to test

Test the difference in mean ( p ~ - p2).

Let XI, , X12,. . . , X,,,, and X2, , X2, , . . . , XZn2 denote two independent random

samples of sizes nl and n2 (a1 5 n2) from the populations XI and X2. Then, the Wilcoxon rank-sum statistics Wl and W2 are determined by the following procedure:

Step 1: Rank the n1 + n2 observations in ascending order. For tk, use the average of the ranks that would be assigned if they differed.

Step 2: Calculate the sum of the ranks for each sample: W, = sum of the ranks in the smaller sample (size = nl) W2 = sum of the ranks in the larger sample (size = n2)

Note that Wl and W2 will be nearly equal for both the samples after adjusting for the difference in sample size.

Wilcoxon The Wilcoxon rank-sum test table (see Appendix Table IX in MR) provides Rank-Sum critical values ( w;,,,, ) of a rank-sum statistic for various sample sires (nl and Test Table

n2 where nl I n2) and levels of significance (a = 0.05 and 0.01) for a two-sided test.

(e.g.) Reading the rank-sum test table for a two-sided test

(1) w;.05i2,10,15 = w;;.025,10,15 = 94

(2) w;.01i2,5,,0 = w~.005,5,10 = 19

Inference Context

for PI - CLZ

Wilcoxon Rank-Sum

Test Procedure for PI - Pz

Parameter of interest: p1 - Test statistic of p1 - p2: The appropriate rank-sum statistic is selected

depending on the type of HI as follows: (1) W,=sumoftheranksof X , j , j = 1 , 2 ,..., nl forH1: p , - p 2 <8,

Step 1: State Ho and HI. Ho: p1 - p2 = 8 0

HI: PI - p2 f 80 for two-sided test p1 - 11.2 > 60 or p1 - p2 < 8o for one-sided test

15-4 Wilcoxon Rank-Sum Test 15-1 3

Wilcoxon Step 2: Determine a test statistic and its value. Rank-Sum W, = sum of the ranks of X , ; for lower-sided test

' J

Test Procedure W2 = sum of the ranks of XZj for upper-sided test

(cont.) Wl or W2 for two-sided test

Step 3: Determine a critical value(s) for a. *

W a I 2,n, ,n2 for two-sided test; w 2 for one-sided test

Step 4: Make a conclusion. Reject Ho if 1

W1 Wa,nl ,n2 for lower-sided test

w2 5 w:,nI,n2 for upper-sided test *

W1 5 Wa I 2,n, ,n2 *

Or W2 5 Wa/2,n,,n2 for two-sided test

Normal When nl and n2 are moderately large (say, nl and n2 > 8), the rank-sum statistic Approximation Wl (or W2 ) has approximately a normal distribution with mean and variance

n, (n1 + n2 + 1) 2 n1nz(n1+n2+1) Pw, = and owl = 2 12

Therefore, a hypothesis on the difference in mean by Wl can be tested by using the z-statistic

Y Example 15.5 The life lengths (unit: hour) of INFINITY (XI) and FOREVER (X2) light bulbs are under study. Suppose that no information is available about the underlying distributions of XI and X2. A random sample selected from each brand results in the following:

F O R E V E R ( X ~ L 795 780 820 770 810 765 800 790 750 730 -am.m a.m-- wwaMMM w- r*a-w--m- ma wawam-m-*- wM--s-s ---Maww#%w-m- - *#"B#a*m wwa-s'.8r

1. (Test on p1 - p2; Wilcoxon Rank-Sum Test) Test if the mean life lengths of the two brands are the same at a = 0.05.

o- Step 1 : State Ho and HI. Ho: p1- p 2 = 0 H1: P1 - p 2 2 0

a test statistic and its value. %9pfpB"

##@dk46 146 &$gC~j4sgg. 18 $ (8~d XI 760 810 775 680 730 725 740 700 720 -

Rank 9 17.5 12 1 5.5 4 7 2 3 X2 795 780 820 770 810 765 800 790 750 730

Rank 15 13 19 11 17.5 10 16 14 8 5.5 "mu ~a*-e"--ams.#m*--- * w - ~ # - ~ ~ " ~ a " ~ - ~ ~ - ~ * * ~ w w w - w m M " ~ - w ~ ~ # ~ " w ~ " ~ - ~ - -.#*- *aw'"**" -

wl=61 and w2= 129

15-1 4 Chapter 15 Nonparametric Statistics

Step 4: Make a conclusion. Since w, = 61 I ~ i / ~ , ~ , ~ ~ = 65, reject Ho at a = 0.05.


2. (Normal Approximation) Apply the normal approximation to the rank-sum statistic Wl and draw a conclusion at a = 0.05.

Step 1: State HO and HI. Ho: p1 -p2 = o H1: p1 - p 2 + 0


~ ; / 2 , n , , n , - - ~ ~ . 0 5 1 2 , 9 , 1 0 = 65

( Step 2: Determine a test statistic and its value.

Step 3: Determine a critical value(s) for a. Z,/2 = Z0,025 = 1.96

I Step 4: Make a conclusion. Since lzol = 2.37 > Z0,025 = 1.96 , reject HO at a = 0.05.

( 1. Test if one heating element is superior to the other at a = 0.05.


I 2. Apply the normal approximation to the rank-sum statistic Wl and draw a conclusion at a = 0.05.

The manufacturer of a hot tub is interested in testing two different types of heating elements. The heating element that produces a larger heat gain (unit: OF) after 15 minutes will be preferable. Ten observations of the heat gain for each heating element are obtained as follows:

15-5 Nonparametric Methods in the Analysis of Variance

O Perform the Kruskal-Wallis test in the single-factor analysis of variance (ANOVA). Explain use of ranks in ANOVA.

Kruskal- The Kruskal-Wallis test is a nonparametric alternative to the parametric F-test of Wallis Test the single-factor analysis of variance (ANOVA). While the parametric ANOVA

model, Yi,. = p + Z~ + E~ , i = 1,2,. . . , a and j = 1,2, . . . , n i , assumes the error

Kruskal- Wallis Test

(cont.)

x2 Approximation

Inference Context

Kruskal- Wallis Test Procedure

15-5 Nonparametric Methods in the Analysis of Variance 15-1 5

terms ei, are normally distributed with mean zero and variance 02, the nonparametric ANOVA method assumes that the eY have the same continuous distribution for all the factor levels i = 1,2, . . . , a.

The Kruskal-Wallis statistic H i s determined by the following procedure:

Step 1 : Rank all N (= za ni ) observations in ascending order. For t k , use r=l

the average of the ranks that would be assigned if they differed. Step 2: Calculate the sum ( R,,) of the ranks RY in the ith treatment, i = I,

2, ... , a. Step 3: Calculate the Kruskal-Wallis statistic

H = l 2 gc - 3(N + 1) for observations without ties

N(N + 1) i=l ni

1 for observations with ties 4

*i N ( N + 112 where S 2 =-!-[T~R~ -

N - 1 i=l j=1 4 I The Kruskal-Wallis static approximates to a X2-distribution with a - 1 degrees of freedom when the sample sizes ni are moderately large (ni 2 6 for a = 3; ni 2 5 for a > 3).

Parameters of interest: 2, ,2,, . . . ,z,

Test statistic of T,, 2,, . . . ,2, (1) Case 1: Observations with ties

(2) Case 2: Observations without ties

Step 1: State HO and HI. Ho: 7, = z 2 =...=z, = o ( p l = p 2 = . - . = P o )

HI: zi + 0 foratleastone i, i= 1 , 2 ,..., a


H = l2 g g - 33(N + 1) for observations without ties

N(N + 1) ni

for observations with ties 4

4 I

1 5-1 6 Chapter 15 Nonparametric Statistics

Kruskal- Wallis Test Procedure

(cont.)

F-test on Ranks

r Example 15.6


Step 4: Make a conclusion. Reject Ho if 2

h 2 %a,,-,

If the ordinary F-test were to be applied to the ranks instead of the original data in the single-factor ANOVA, the F-test statistic would be

H /(a - 1) F, =

(N-1 -H) l (N-a ) Note that Fo is closely related to the Kruskal-Wallis statistic H, which indicates that the Kruskal-Wallis test is approximately equivalent to the parametric ANOVA to the ranks. Accordingly, the F-test on the ranks can be applied when a nonparametric alternative to the analysis of variance is unavailable.

It is recommended that the F-test be performed on both the original data and the ranks when the normality assumption andlor the effect of an outlier(s) are concerned. If both the test results are similar, use of the original data is satisfactory; otherwise, use of the ranks is preferred.

1 (Nonpararnetric ANOVA; Kruskal-Wallis Test) Maximum power grip forces (unit: lb.) of a 45-year-old male at four elbow angles (0°, 45", 90°, and 135") are

I obtained below. Test if elbow angle has a significant effect on maximum power 1 grip strength at a = 0.05.

0" (YI) 65 68 62 66 69 45" (y2) 70 72 69 7 5 68 90" (y3) 67 63 68 64 67

5 8 60 5 9 63 60 ., r3XLri) mmw-B, ,*--,, -*-Bww--B - *- - ms--wmB , -** - m = ,. - *-- ev * mw mMw .- -- h Step 1: State HO and HI.

Step 2: Determine a test statistic and its value. Hbanv Wzk Ri.

135" bj) 58 60 5 9 63 60 Rank (r4J 1 3.5 2 6.5 3.5 16.5

-**-"Mwwm a# " * ~ - ~ 6 w * - . m ~ " - ~ r a B ~ ~ " ~ " ~ " ~ w , w ~ ~ a ~ a v - - - -w~w,-e"6a--*"M,vw--*~-,%"v 6---*---m"eMm*-m--w---



15-5 Nonparametric Methods in the Analysis of Variance 15-1 7

Step 3: Determine a critical value(s) for a. 2 - 2 2

Xa,a-1 - X O . O ~ , ~ - 1 = X 0 . 0 5 ~ = 7.81

Step 4: Make a conclusion. Since h = 14.52 > xi.,-, = 7.81, reject Ho at a = 0.05.

The tensile strengths (unit: 1b.1in.~) of cements prepared by four different mixing techniques (A, B, C, and D) are measured below. Test if the difference in mixing technique affects the tensile strength at a = 0.05.

15-1 8 Chapter 15 Nonparametric Statistics


Example 15.1 (Inference on ; Sign Test) I I (1) Choose File * New, click Minitab Project, and click OK.

Score

75-

96

Choose Stat * Nonparametrics > 1-Sample Sign. In Variables select Score. Check Test median, enter '80' (hypothetical median), and in

I Sign Test for Medim

Slgn test of median = 80.00 versus not = 80.00 ............................................................................................... N Below Equal Above P Median 12 r- = 5 1 v+ = 6 1.0000 j 82.501 3: ................................. rr.. ........................................................ :

" I

Example 15.2 (Inference on E D ; Sign Test)

(1 ) Choose File * New, click Minitab Project, and click OK.

(2) Enter the score data of the two exams on the worksheet.

(3) Choose Calc > Calculator. In Store result in variable, enter C3 as a column to store calculation results. In Expression, click Exam 1, minus

.I ( ~ x a m 1 ( X I ) Exam 2 (x2) XI - x2 -4 1 85 - . -5

. . I


Example 15.2 I (5) Choose Stat > Nonparametrics > 1-Sample Sign. In Variables select x l (cont.) - x2. Check Test median, enter '0' (hv~othetical median of the score

Example 15.3

Sign Test for Median I Sign t e s t of median = 0.00000 v e r s u s not = 0.00000 . ..............................................................................................

N Below Equal Ahove P i Hedian 1x1 - x2 10 r- = 8 1 r + = l 0.0391 1 - 3 . 5 0 0 .i ............................................................................................. : - ."..

(Inference on p.; Wilcoxon Signed-Rank Test) (1) Choose File > New, click Minitab Project, and click OK.

(2) Enter the test score data on the worksheet.

(3) Choose Stat > Nonparametrics > 1-Sample Wilcoxon. In Variables select Score. Check Test median, enter '80' (hypothetical mean), and in



Example 15.4

k --

Wilcoxon Signed Rank Test

Test of medzan = 80.00 versus medlan not = 80.00 .......................................................................................

N for Wilcoxon i Estimated N Test Statlstlc P i Nedian

iscore 10 9 w+= 24.5 0.859 i 80.001 7 1 .......................................................................................

%I

(Inference on p, ; Wilcoxon Signed-Rank Test)

(1) Choose File , New, click Minitab Project, and click OK.

(2) Enter the score data of the two exams on the worksheet.

(3) Choose Calc , Calculator. In Store result in variable, enter C3 as a column to store calculation results. In Expression, click Exam 1, minus sign, and Exam 2 in sequence. Then click OK.

(4) Name the C3 column x l - x2.

(5) Choose Stat , Nonparametrics , 1-Sample Wilcoxon. In Variables select X I - x2. Check Test median, enter '0' (hypothetical mean of the

Wilcoxon Signed Rank Test

I Test of medlan = 0.000000 versus median not = 0.000000

N for Wilcoxon i~stimated N Test Statistic P ; Median

jxl - x2 10 9 wf = 3.0 0.024 i -3.000 : ..................................................................................... ..

15-22 Chapter 15 Nonpararnetric Statistics

Example 15.5 (Inference on yl - CL;?; Wilcoxon Rank-Sum Test) (1) Choose File * New, click Minitab Project, and click OK.

J. lnfin~ty / Forever I 1 760 1 795 '

p -

3 8101 7801 I bulb brands on the worksheet.

(3) Choose Stat * Nonparametrics * Mann-Whitney. In First Sample select Infinity, in Second Sample select Forever, and in Alternative

Mann-Whitney Confidence Interval and Test ....................................................................................... Infinity N = 9 Bedian - 730.00 i Forever N = 10 Median = 785.00 i Point estimate for ETA1-ETA2 is -47.50 I 95.5 Percent C I for ETA1-ETA2 is (-79.99,-10.01)i

............................................................... i Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0200 The test 1s significant at 0.0199 (ad]usted for ties11 $5

43

MINITAB Applications 1 5-23

Example 15.6 (Nonparametric A N 0 VA; Kruskal-Wallis Test) I (1) Choose File * New, click Minitab Project, and click OK.

(2) ksheet.

I--- ---.I

r Angle , Force

1 01 65 - * + --- - - - 6ar ------ I . ; ;I . . 0,- ............. 62 ... ..........

:hoose Stat > Nonparametrics ) Kruskal-Wallis. In R

I Kruskal-Wallis Test I Kruskal-Wallis Test on Force

Angle N Median Ave Rank Z 0 5 66.00 10.9 0.17

45 5 70.00 17.5 3.06 90 5 67.00 10.3 -0.09

135 5 60.00 3.3 -3.14 Overall 20 10.5

.esponse select



Exercise 15.1 1. (Test on p ; Sign Test)

Step 1: State Ho and HI. Ho: ji =2.5

I HI: p < 2.5

Step 2: Determine a test statistic and its value. 11 2 " ' - . 3 'B,'?: '.-:..? ';'. .$ #gS f d

- * Excluded since xi = P o .

Step 3: Determine a critical value@) for a. ri,, = r,',,,,, = 5 (Note) number of ties = 2

I Step 4: Make a conclusion. Since r+ = 2 I r& = 5 , reject Ho at a = 0.05.

I 2. (P-value Calculation)

I Since P = 0.0002 < a = 0.05, reject HO at a = 0.05.

3. (Normal Approximation) Step 1 : State Ho and HI.

Ho: ji =2.5 H I : ji < 2.5


Step 3: Determine a critical value@) for a. Z a = Zo,05 = z0,05 = 1.65

I Step 4: Make a conclusion. Since z , = -3.58 < -z0,,, = -1.65, reject Ho at a = 0.05.


Exercise 15.2

Exercise 15.3

(Test on P D ; Sign Test)

Step 1: State Ho and H I . Ho: P D = 0

H,: jiD # O

Step 2: Determine a test statistic and its value. r=min(r-,r ') =min(2,6)=2


ri12,n = r;025,8 = 0 (Note) number of ties = 4


Since r = 2 g riI2,, = 0 , fail to reject Ho at a = 0.05.

1. (Test on p; Wilcoxon Signed-Rank Test) Step 1: State Ho and HI.

Ho: p = 2.5 HI: p < 2.5


N o , 3 - q - 2 . 3 * ' . #xl-ls) . && . 9i@J@d fa&

12 2.3 -0.2 0.2 5.5 -5.5 13 2.0 -0.5 0.5 8.0 -8.0 14 1 .8 -0.7 0.7 14.0 -14.0 15 1.3 -1.2 1.2 19.5 -19.5

17 2.0 -0.5 0.5 8.0 -8.0 18 1.9 -0.6 0.6 11.5 -1 1.5 19 2.3 -0.2 0.2 5.5 -5.5 20 1.9 -0.6 0.6 11.5 -11.5

22 1.6 -0.9 0.9 17.5 -17.5 A m - - s - - - " # - - d . v - - P ---".?--* Mm -m "+a%%* * e--*-- - - * Excluded since x, = p, .



Exercise 15.4

Exercise 15.5

Step 3: Determine a critical value(s) for a. w:,~ = u ' ( ; , ~ ~ , ~ ~ = 60 (Note) number of ties = 2


Since w* = 5 I w:., = 60, reject Ho at a = 0.05.

2. (Normal Approximation)

Step 1 : State Ho and H I . Ho: p =2.5 HI: p < 2.5


1 Step 3: Determine a critical value(s) for a. Z, = z0,05 = 1.65

Step 4: Make a conclusion. Since zo = -3.73 < -z,,,, = -1.65 , reject Ho at a = 0.05.

(Test on y, ; Wilcoxon Signed-Rank Test)

Step 1 : State Ho and H I .

Ho: y, = o HI : y, # O


w = min(w-, w') = min(14.5,21.5) = 14.5


w:/~,, = ~ : , ~ ~ ~ , 8 = 3 (Note) number of ties = 4


Since w = 14.5 $ w:,,,, = 3 , fail to reject Ho at a = 0.05.

1. (Test on pl - p2; Wilcoxon Rank-Sum Test) Step 1 : State Ho and H I .

Ho: - ~ 2 = 0 I H I : p1 -y2.;tO



Exercise 15.6

Rank 2 4 5.5 9.5 7.5 3 1 11.5 13.5 19.5 X2 31 33 32 35 34 29 38 35 37 30

Step 3: Determine a critical value(s) for a. - wi12,n,,n2 - w ~ , ~ ~ ~ ~ , ~ ~ , ~ ~ = 78 for two-sided test

Step 4: Make a conclusion. Since w, = 77 I w;,, ,,,,,, = 78 , reject Ho at a = 0.05.

2. (Normal Approximation) Step 1 : State Ho and H I .

H,: p, - p2 = o H I : p1 - p 2 + 0


Step 3: Determine a critical value(s) for a. z a I 2 = z ~ , ~ ~ ~ = 1.96

Step 4: Make a conclusion. Since lz, 1 = 2.12 > z,,,,, = 1.96, reject Ho at a = 0.05.

(Nonparametric ANOVA; Kruskal-Wallis Test) Step 1 : State Ho and H I .

H,: 7 , = 7 , = .. . = 7 , = 0

H I : zi # 0 for at least one i, i = 1, 2, 3, 4

ZP'E I =

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