Is σ1 and σ

2

known?

σ1 =

σ2

Example #1 – Confidence IntervalA sample of 20 students were asked to fill in a survey on gender, how many hours spent studying for the test and what was the grade earned on the same test that they had studied for.Hours: Male 14, Female 17, Female 3, Female 6, Male 17, Male 3, Male 8, Male 4, Female 20, Male 15, Female 7, Male 9, Male 0, Male 5, Female 11, Female 15, Male 18, Female 13, Male 8, Male 4Construct a 91% confidence interval estimate of the proportion of all female students.a) What parameter are you estimating here? π b) State/check the necessary assumptions required to construct the confidence interval: np > 5? 20 (8/20) = 8 n(1-p) 20 (1 – 8/20) = 12The conditions are met, therefore, it is normally distributed. c) Before you construct a 91% confidence interval estimate of the proportion of all female students, state the critical value: z = ± 1.6954 InvN, Area: 0.91, = 1, σ μ = 0d) The point estimate for a 91% confidence interval estimate of the proportion of all female students is: x = 9.85Example #3A survey is planned to determine the mean annual family medical expenses of employees of a large company. The management of the company wishes to be 95% confident that the sample mean is correct to within ±50 of the population mean annual family medical expenses. A previous study indicates that the standard deviation is approx. $400. a) How large a sample size is necessary? 245.8 246 * In this question, you are estimating (the mean)μz = 1.9599 e = 50 = 400σ n = z² σ ² e²b) If management wants to be correct with ±25, how many employees need to be selected?

To find Z values, use "INVN" function and input "area" (which is the confidence level (usually expressed in %); std dev=1; and mean=0.

To find the t values, you have two options: (1) t table (which will be provided in the quiz/test/exam) and (2) INVt function and input "upper area", "df" (degree of freedom=n-1).

*t-distribution is a continuous probability distribution that resulted from estimating the mean of a small sample size and the population standard deviation is unknown (more spread out than a normal distribution; the larger the degrees of freedom=more accurate normal distribution)Margin of error: measures the uncertainty in estimating the population parameter. Margin of error= (critical value) x (standard error). Point estimate: Using the sample statistic (e.g. or p) to estimate the corresponding Example #2

The director of patient services of a large health maintenance organization wants to evaluate patient waiting time at a local facility. A random sample of 25 patients is selected from the appoint book. The waiting time is defined as the time from when the patient signs in to when he or she is seen by the doctor. The following data represents the waiting times (in minutes). x is known:

19.5

30.5

45.6

39.8

29.6

41.3

13.8

17.4 10.7

25.4

21.8

28.6

52.0

25.4

39.0

36.6

1.9

45.9

42.5

12.1

26.1

4.9 12.7

31.1

43.1

Construct a 95% confidence interval estimate of the pop. avg. waiting time. 1. State the necessary assumption(s)/condition(s) required to construct a 95% confidence interval estimate for the population average waiting time.- It is not normally distributed (n is not equal or greater than 30) - Apply CLT ....no. because n = 25- Assume x is normally distributed 2. State the critical value needed to construct the 95% confidence interval estimate for the population average waiting time. t critical values = ± 2.064 (Calc: Dist., t, Invt, Area: 0.025, df: 24)

3. State the point estimate and its value used to construct a 95% confidence interval estimate for the population average waiting time. x = 27.892 (Calc: INTR, t, 1-s, List, C-level: 0.95)4. The 95% confidence interval estimate for the population average waiting time is: 22.1668939 < μ < 33.6171261Calc: INTR, t, 1-s, List, C-level: 0.95, left and right values

Example #4What proportion of people hit snags with online transactions? According to a poll, 89% hit snags with online transactions. a) To conduct a follow up study that would provide 95% confidence that the point estimate is correct to within ±0.04 of the population proportion, how large a sample size is required? 235.0339 234 *In this question, you are estimating π (the proportion) π = 0.89 z = 1.9599 e = 0.04

Hypothesis Testing - Mean (Z-Test)A new battery has been developed to power laptop computers. It will sell in a certain price range. It is hoped that the battery can be used for more than 4.00 hours before it needs to be recharged. We will assume that the battery lives are normally distributed with a standard deviation of 0.25 hours. A random sample of 50 batteries is tested. The sample batteries lasted an average of 4.12 hours before they required recharging.

Hypothesis Testing - Mean (T-Test)A new battery has been developed to power laptop computers. It will sell in a certain price range. It is hoped that the battery can be used for more than 4.00 hours before it needs to be recharged. We will assume that the battery lives are normally distributed. A random sample of 50 batteries is tested. The sample batteries lasted an average of 4.12 hours with a standard deviation of 0.25 hours before they required recharging. Let Hypothesis Testing for Z- Test – knownσ

A company that makes bolts that are used on an automotive component uses two machines to make these bolts. It has been determined by past studies that the standard deviation of the bolt diameters made by machine 1 is 0.025 mm. and the standard deviation of the bolt diameters of machine 2 is 0.022 mm. Both machines have a dial to set for the desired diameter. Recently they used both machines to fill a large order. The customer found that many of the bolts from a certain package were too large and made a complaint. It was determined that the package in question was made by machine 2. The manufacturer decided to take samples of the bolts from both machines to test to see whether the mean diameter of the bolts from machine 2 was significantly larger than the mean diameter from machine 1 when the dial was set to the same diameter on each machine. The sample of 100 bolts from machine 1 had a mean diameter of 5.023 mm and a

SPSS ExampleThe table below is a random sample of 20 companies whose stock is traded on the New York Stock Exchange. For each company, the number of shares traded on May 25, 1999, and May 26, 1999, is given. a. At the .05 level of significance, is there evidence that the average number of shares traded on May 25 is higher than the average number of shares traded on May 26? b. Determine the p-value in (a)

Ho: µD < 0Ha: µD > 0 Assume the distribution of the differences between the number of shares traded on the two days is at least approximately normally distributed Decision Rule: Reject Ho if p-value is < than the level of significance (0.05), otherwise fail to reject Ho p-value = 0.4645 Calc: TEST, t, 1-s, µ: >µ0, µ0: 0, x: 6090, sx: 301997.99, n: 20Decision: Fail to reject Ho

Statistical Decision:1. If you reject the Ho, you have statistical proof that the alternative hypothesis is correct.2. If you do not reject the Ho, then you have failed to prove the Ha. The failure to prove the alternative hypothesis, does not mean that you have proven the null hypothesis.*we can never prove that H0 is true

Population proportions: π

Hypothesis Testing – One Population

Population Mean: µ

Is σknown?

Use a distribution-free test or if appropriate assume the population is normally distributed and proceed through the flow chart

NO

YES

Z-Test, with test Statistics

NOYES

T-Test, with test statistic

YES

NO

Z-Test, with test Statistics

Convert to underlying binomial distribution

Is normal?Ie. X is or can be assumed normal or n ≥30? (CLT Theorem)

YES

Dependent/Paired Samples

Hypothesis Testing – Two Population Means

Independent Sample

NO

Paired T-test

Is 1 and 2 normal?

Is σ1 and σ

2

known?

Advanced Statistics

Z-Test, with test Statistics

NOYES

σ1 =

σ2

YES

NO

T-Test, with “Pooled Variance”

T-Test, with “No Pooled Variance”

Risks in Decision Making Using Hypothesis-Testing Methodology

- Since our statistical evidence is based on sample data and the corresponding sample variability, there is a risk that we may make the wrong conclusion.

Type 1 error and Type II error- Type 1 error: Reject Ho when Ho is true.

Prob of committing Type I error = (level ofα significance)

You control the Type I error by deciding the risk level that youα are willing to have in rejecting the null hypothesis when it is true.

- Type II error: Did not reject Ho when Ho is false.Prob of committing Type II error = β

It depends on the difference between the hypothesized and the actual value of the population parameter. If the difference between the hypothesized and actual value of the population parameter is large, then is small.β

- In hypothesis testing we begin with a tentative assumption about a population parameter = NULL hypothesis - Denoted as Ho- The null hypothesis is written in terms of the population mean, not

the sample mean

- After specifying Ho, we have to specify Alternative Hypothesis - Denoted Ha OR H1

Example:- Ho : The average age of the population is equal to 45 years old

- Ha : The average age of the population is NOT equal to 45 years old- The alternative hypothesis is also known as research hypothesisImportant note:In any situation that involves testing the validity of a claim, the null hypothesis is based on the assumption that the claim is true. The alternative hypothesis is formulated so that rejection of Ho will

Independent SamplesThe samples chosen at random are not related to each other. We wish to study the mean incomes of companies X and Y. We select a random sample of 28 employees from the Company X and a sample of 19 employees in Company Y. A person cannot be an employee in both companies.

Example: Ho: µ = 4.5 and Ha: µ ≠ 4.5Type 1 error if you say that µ ≠ 4.5 when µ = 4.5

Dependent SampleDependent samples are characterized by a measurement, then some type of intervention, followed by another measurement. Paired samples are also dependent because the same individual or item is a member of both samples.Examples: 10 participants in a marathon were weighed prior to and after competing in the race. We wish to study the mean

Hypothesis Testing for T-Test (equal variance)A work team has developed a new process to assemble a certain component. They would like to know if this new process has significantly reduced the time to assemble the component. They have taken samples of 50 components produced by the existing process and 40 components produced by the new process. The mean and standard deviation of the assembly times for the existing process were 73.2 minutes and 3.6 minutes, respectively. The mean time was 71.4 minutes with a standard deviation of 3.2 minutes for the components assembled by the new process. Assume that the times for

Hypothesis Testing – Proportion (Z-Test)A random sample of 300 retail outlets indicated that 165 outlets included the GST in their prices while the others did not. Can one conclude at the 5% level of significance that more than retail outlets include the GST in their prices than those that do not? What is the corresponding p value?‐

The Actual Data Value

The Predicted Data Value

Regression Analysis(RA) is a statistical forecasting model that is concerned with describing and evaluating the relationship between a given variable (usually called the dependent variable, denoted as Y) and one or more other variable (usually known as the independent/exploratory variable, denoted as X) .• RA can predict the outcome of a given key business indicator (dependent variable) based on the interactions of other related business drivers (independent /exploratory variables)• The relationship can be described as a function of a linear (straight-line) equation <called linear regression> “simple linear regression”

Dependent Variable (Notation: Y) – The variable you wish to predictIndependent Variable (Notation: X) – Variable used to make the predictionSimple Linear Regression – A single numerical independent variable X is used to predict the numerical dependent variable YMultiple Regression – Use several independent variables to predict a numerical dependent variable Y.*When changes in the variable X leads to predictable change in the variable Y then we say “X can be used to explain Y”Regression analysis allows you to identify the type of relationship that exists between a dependent variable (X) and an independent variable (Y).

The simplest relationship is the straight line or linear relationship

Population

Sample

Regression Coefficient

Y intercept

β0 b0

Slope β1 B1

Regression (least squares) line

Y = β0 + β1x + ε

Y = b0 + b1x + e

Random Error (Residual) ε eForecasting Line/ Prediction Line

Y = b0 + b1x

ASSUMPTIONSGeneral Assumptions of the Simple Linear Regression Model <similar to ANOVA> Referring to the residuals1. Linearity - The mean of the model error terms is 0.2. Independence - The model error terms are independent.3. Normality - The regression model errors are normally distributed.4. Equal variance - The model error terms have a constant variance, σε2 , for all combinations of values of the independent variables. The Coefficient of Determination

Testing whether there is a linear relationshipTest for significance of the correlation between x and y.Ho: ρ = 0 (There is no linear relationship) - Ha: ρ ≠ 0 (There is a linear relationship)Test for significance of the regression slope coefficient.Ho: β = 0 (There is no linear relationship) – Ha: β ≠ 0 (There is a linear relationship)Note: For a simple linear regression model (one independent variable), these two are equivalent methods. Example: Simple Linear RegressionThe marketing manager of a large supermarket chain would like to determine the effect of shelf space on the sales of pet food. A random sample of 12 equal-sized stores is selected with the following results:

Store

Shelf Space, X (feet)

Week Sales, Y ($00)

1 5 1.62 5 2.23 5 1.44 10 1.95 10 2.46 10 2.67 15 2.38 15 2.79 15 2.810 20 2.611 20 2.912 20 3.1

**First determine which are the independent and dependent variablesa. Assuming a linear relationship, use the least-squares method to find the regression coefficients b0 and b1. Where b0 = intercept and b1 = slope(stat, F3, TEST, T, REG, ≠, List # x, List # y, Freq-1) b1 = 0.7400 b0 = 1.450 b. Interpret the meaning of the slope b1 in this problem. Positive Relationship - For every increase of one foot in shelf space, there is an expected increase of 0.074 hundred of dollars ($7.40) in weekly sales.c. Use the regression model developed in (a) to predict the average weekly sales (in hundreds of dollars) of pet food for stores with 8 feet of shelf space for pet food(x=8); y= 1.45 + 0.074 (8)= 2.042*100 because sales is in ($00) = $204.20*predict the average weekly sales (in hundreds of dollars) of pet food for stores with (12)20 feet of shelf space for pet food. What is the residual error? (X=20); y=1.45+0.074(20) = 2.93*100 $293; Residual Error = e= y – y = 3.1 – 2.93 = 0.17d. compute the coefficient of determination r2, and interpret its meaning in this problem. NOTE: r2 is called the coefficient of determination. MEANING: The percentage of variation in the dependent variable explained by its relationship to the independent variables in the regression modelr2 = 0.6839 ~ 68.39% of the variation in y (week sales) is explained by x (shelf space)e. Compute the coefficient of correlation r. NOTE: r is the sample correlation coefficient (also called the Pearson coefficient of correlation). It is used to measure the strength of association between two variables.r= 0.82700 (positive relationship)f. At the 0.005 level of significance, is there evidence of a linear relationship between shelf space and sales? T-Test; Reject Ho, since there is evidence of linear rel. between x&yHo: 1 =0β p-value test statistics used to make statistical decisionsHa: 1≠0β tcal 1. Reject Ho or 2. Don’t Reject Ho

Tcal = 4.6517273 p-value = 0.00090566 < =0.05αAnalyzing a Multiple Regression ModelStep 1: Collect sample data. The values of Y, X1, X2, X3, … ,XkStep 2: Hypothesize the form of the model. This includes choosing which independent variables to include in the model. Y = β0 + β1X1 + β2X2 + β3X3 + … + βkXk + ε (Linear)Step 3: Use the method of least squares to estimate the unknown parameters β0 , β1 , β2…., βkStep 4: Specify the probability distribution of the random error component ε and estimates its variance σ2 *σ2 = variance of the random error εStep 5: Statistically evaluate the utility (or usefulness) of the model.Step 6: Check the assumptions on σ are satisfied and make model modifications, if necessary.Step 7: Finally, if the model is deemed adequate, use the fitted model to estimate the mean value of y or to predict a particular value of y for a given values of the independent variables, and to make other inferences.ASSUMPTIONS About the random - Error ε

Rule of Thumb Concepts

X Y X, YType of

RelationshipIncreas

esIncrease

s Move in the same direction

Direct relationshipDecrea

sesDecreas

esIncreas

esDecreas

es Move in the opposite direction

Inverse relationshipDecrea

sesIncrease

sIncreas

esCannot

Tell No apparent relationship

No relationshipDecrea

sesCannot

Tell

Referring to the probability distribution of the random error component ε and estimates its variance σ2.1. Linearity The mean of the model error terms is 0.(E(‐ ε)=0)2. Independence The model error terms are independent.‐3. Normality The regression model errors are normally distributed.‐4. Equal variance The model error terms have a constant variance, ‐ σ2 , for all combinations of values of the independent variables.Estimator of 2 for multiple regression with kσ independent variables

Caution: A rejection of null hypothesis:Ho: β1 = β 2 =…..= β k =0 in the global F test‐ leads to the conclusion that the model is statistically useful.- However, statistically “useful” does not necessarily mean “best”.- Another model may prove even more useful in terms of providing more reliable estimates and predictions. - The global F test is usually regarded as a test‐ the model must pass to merit further consideration.

Example: Multiple RegressionThe following are data on horsepower x1, time from zero to 60 miles per hour (x2), top speed (x3), miles per gallon (x4), and price (y) in thousands of dollars for 10 sports cars (Road & Track, October 1994).

X1 X2 X3 X4 Y BMW M3 24

06.0

120

24.6

38.4

Corvette 300

5.7

170

16.8

41.4

Dodge Viper 400

4.8

160

14.0

54.8

Ford Mustang

240

6.9

140

18.0

25.8

Honda Prelude

190

7.1

139

24.0

25.6

Misubishi GT 320

5.7

159

16.3

43.7

Toyota Supra

320

5.3

155

18.8

48.2

Nissan 300ZX

300

6.0

155

18.7

0.8

Alfa Romeo 320

7.6

150

17.5

38.1

Mazda RX - 7

255

5.5

158

17.0

35.0

a. Develop an estimated regression equation with horsepower, time from zero to 60 miles per hour, top speed, and miles per gallon as the four independent variables to predict

H ypothesis Testing for Dependent Samples The process improvement team selects 12 cars at random and uses both procedures on each car. There are two procedures: A and B. We record the time (in mins) for each procedure to oil and filter change. The results are shown in the next slide. At the 1% level of significance, can we conclude that there is a difference in the average time for an oil change and filter change? The times are normally distributed.Automobile

Time (mins) for an oil and filter change

D=Time A -Time B

Procedure A Procedure B1 28.2 25.4 2.82 27.1 27.0 0.13 26.4 25.5 0.94 27.3 27.1 0.25 24.8 26.5 -1.76 23.4 27.4 -4.07 26.8 26.2 0.68 27.2 26.8 0.49 25.5 28.9 -3.410 25.8 26.1 -0.311 26.0 24.7 1.312 25.4 26.6 -1.2

Hypothesis Two Sample Testing – Proportion:A human resources director decided to investigate employee perception of the fairness of two performance evaluation methods. To test for the differences between the two methods, 160 employees were randomly assigned to be evaluated by one of the methods: 78 were assigned to method 1, where individuals provide feedback to supervisory queries as part of the evaluation process; 82 were assigned to method 2, where individuals provided self-assessments of their work performances. Following the evaluations, employees were asked whether they considered the performance evaluation fair or unfair. Of the 78 employees in method 1, there were 63 fair ratings. Of the 82 employees in method, there were 49 fair ratings. Using a 0.05 level of significance, is there evidence of a significant difference between the two methods in the proportion of fair ratings?a) What type of parameter is being tested here? πb) Define the variable or parameter associated with this test: π1 = population proportion of fair ratings for method 1 π2 = population proportion of fair ratings for method 2c) State the hypotheses: Null Hypothesis: Ho: π1 = π2Alternative Hypothesis:Ha: π1 ≠ π2d) State the condition(s) are required to be true for this procedure to be legitimate: Apply CLT, x1 and x2 are normally distributede) What is the calculator procedure that you used for doing this test? z, 2-p, Var, ≠f) What is the critical value for this test? z = ±1.9599Calc: DIST, Norm, InvN, Tail: Central, Area: 1-0.05, = 1, µ =σ 0g) What is/are the rejection region(s)? Zcal < -1.9599 and Zcal > 1.9599h) What is the value of the test statistic? Zcal = 2.8991 i) What is the p-value for the test? p-value = 0.0037413 Calc: TEST, z, 2-p, p1: ≠p2, x1: 63, n1: 78, x2: 49, n2: 82

j) What was the statistical decision made and why?Reject Ho because the p-value is < than the level of significance (0.05) k) Using specific references to the appropriate population parameters, state the test’s conclusion: There is sufficient sample evidence to indicate there is a significant difference between the two methods in the proportion of fair ratings

Hypothesis Testing for T-Test (unequal variance)We wish to determine if there is a difference in the breaking distances for two types of tires. Use the 5% level of significance and assume that the breaking distances for each type of tire are normally distributed with the unequal variance. Based on the data for the samples of tires shown, at the 5% level of significance, should we conclude that there is a difference in the mean breaking distance?

Breaking Distance (meters)Tire (A) Tire (B)83 7579 8482 7684 8380 8581 78

83

Hypothesis Two Sample Testing – Variances: A carpet manufacturer is studying differences between two of its major outlet stores. The company is particularly interested in the time it takes customers to receive carpeting that was ordered from the plant. Data concerning a sample of delivery times for most popular type of carpet are summarized as follows:

At the 0.01 level of significance, is there evidence in a difference in the variances of the shipping time between the two outlets?a) What type of parameter is being tested here? σ b) Define the variable or parameter associated with this test:

1 ² = population variance in the shipping times for store A σ2 ² = population variance in the shipping times for store Bσ

c) State the hypotheses:Null Hypothesis: Ho: 1 ² = 2 ²σ σAlternative Hypothesis: Ha: 1 ² ≠ 2 ² σ σ d) State the condition(s) are required to be true for this procedure to be legitimate:Assume that the proportions are normally distributede) What is the calculator procedure that you used for doing this test? F, Var f) What is the critical value for this test? To get Fu go to InvF, Area: 0.01/2, n: df = 40, n: df = 30Fu = F0.01/2, 40, 30 = 2.52 FL = F 1/Fu = 0.416g) What is/are the rejection region(s)? F > 2.25 and F < 0.416 h) What is the value of the test statistic? Fcal = 0.5993756i) What is the p-value for the test? p-value = 0.129946737 Calc: TEST, f, Variable, 1: ≠ 2, sx1: 2.4, n1: 41, sx2: 3.1, n2: 31 σ σj) What was the statistical decision made and why? Fail to reject Ho because the p-value is > than the level of significance (0.01)k) Using specific references to the appropriate population parameters, state the test’s conclusion: There is not enough evidence to conclude that the two proportion variances in the shipping times between the two major outlet stores are not different*Note: assuming the underlying normality in the 2 populations is met, based on results above, it is appropriate

Store A Store BX 34.3 days 43.7

daysS 2.4 days 3.1 daysn 41 31

Confidence Interval Estimate

What Population Parameter are you testing?

Is normal?Ie. X is or can be assumed normal or n ≥30? (CLT Theorem)

Is p normal?Ie. np ≥ 5 and n(1-p) ≥ 5

Is σknown?

Non – Standard procedure required

Limits : Limits :

Limits:

Find Sample Size

Limits :

Limits :

(I) Method

(J) Method

MeanDifference (I – J)

Std. Error

Sig. LowerBound

UpperBound

A BCD

10.50*20.00*13.50*

2.442.442.44

.002

.000

.000

3.6613.166.66

17.3426.8420.34

B ACD

-10.50*9.50*3.00

2.442.442.44

.002

.005

.617

-17.342.66-3.84

-3.6616.349.84

C ABD

-20.00*-9.50*-6.50

2.442.442.44

.000

.005

.066

-26.84-16.34-13.34

-13.16-2.66.34

D ABC

-13.50*-3.006.50

2.442.442.44

.000

.617

.066

-20.34-9.84-.34

-6.663.8413.34

Levene Statistic

df1 df2 Sig.

.528 3 20 .668 (p-value approach)

Chi Square x 2 proportion test Qualitative Data “counting” of attribute*When you have a global test after rejection it means that the means are differentOne-Way ANOVA ( AN alysis O f VA riance) - Compare means of more than two groups- One-way ANOVA, deals with one factor of

interest (e.g. performance, salary, etc)- Analyzing the variation “within groups” and

“between groups”

- c groups represent populations whose values are randomly and independently selected, follow a normal distribution

- Have equal variances?o Use Levene’s test (SPSS output)

p-value

- p-value allows you to make direct conclusions

Chi Square Testing: Example #3A survey is taken in three different locations in Nassau County in New York to determine whether there is a relationship between architectual style of houses and georgraphic location. The results for a sample of 233 houses are as follows:

East Medow

Farmingdale

Levittown

Total

Cape 31 14 52 97Expanded ranch

2 1 12 15

Colonial 6 8 9 23Ranch 16 20 24 60Split-level 19 17 2 38Total 74 60 99 233At the 0.05 level of significance, is there evidence of a relationship betweenarchitectural and geographic location?a) What test procedure was used for doing this test? Chi-Squareb) State the hypothesis using statistical symbols: Null Hypothesis: Ho: style is not related to location

Ho: π1 = π2 = π3 Alternative Hypothesis: Ha: style is related to location Ha: at least one πj is different / Ha: not all πj is equalc) State the condition(s) and assumption(s) are required for this procedure to be legitimate: All the fe > 5 for all cells30.806

24.978

41.214

4.7639

3.8626

6.3733

7.3047

5.9927

9.7725

19.055

15.45 25.493

12.068

9.7854

16.145

d) Draw a graph of the most appropriate distribution clearly showing the value(s) for the critical value(s) and the rejection region(s).How to obtain df? df(5-1), df(3-1) = 4 x 2 = 8

What is the test(s) procedure for this set of hypothesis

a. Z test

b. X2 testc. T testd. F teste. A & BMultiple Comparisons: Tukey-Kramer Procedure

• To determine which means differ, you can use Tukey-Kramer procedure.• SPSS output

ANOVA Example #1A snack foods company that supplies stores in a metropolitan area with “healthy” snack products was interested in improving the shelf life of its tortilla chips product. Six batches (each batch containing one pound) of the product were made under each of four different formulations. The batches were then kept under the same conditions of storage. Product condition was checked each day for freshness. The shelf life in days until the product was deemed to be lacking in freshness was as follows: At the .05 level of significance, completely analyze the data to determine whether there is evidence of a difference in the average shelf life among the formulations. If appropriate, determine which groups differ in average shelf life.

Post Hoc Tests – will be given on test Dependent Variable: Shelf LifeTukey HSD * The mean difference is significant at the 0.05 level

Oneway ANOVA – may not be given on testShelf LifeCalculator Input:List 1: A, B, C, DList 2: 111111, 222222, etcFactor A: List 2Dependent: List 1

Define the parameters being tested:μa = population average shelf life for formulation Aμb = population average shelf life for formulation Bμc = population average shelf life for formulation Cμd = population average shelf life for formulation DState the hypothesis:Null Hypothesis: Ho: μa = μb

= μc = μd

Alternative Hypothesis: Ha: at least one μj is different What standardized test statistic is being used by this test? F = 23.274 (given in the “Oneway

What is the p-value for the test? 0.000This is a upper tail test (always an upper tail test)To find Fu: Dist, F, InvF, Area: 0.05, n: df: 3, d: df: 20Fu = 3.098 What condition(s) are required to be true for this procedure to be legitimate? Randomness and independenceNormality visual box whisker plotHomogeneity of variance Ho: σa² = σb² = σc² = σd²

Ha: not all the σj² is equal

Reject Ho since p-value is < than the level of significanceP-value = 0.000, Level of Significance = 0.05

If appropriate, determine which method differs in average shelf life? Use 0.05 level of significance.Since Ho is rejected, it is appropriate to use the Tukey testTukey procedure Post Hoc testsUsing the Tukey output (at the .05 level of significance) formulation A has a longer shelf life than formulations B, C, and D. Formulation B has a longer shelf life than formulation C. At the 5% level of significance formulation A appears to have a longer shelf life than formulations

MEANXa = 95.33Xb = 84.833Xc = 75.33Xd = 81.833

Test of Homogeneity of VariancesLife (days)