Cairo 02 Stat Inference

STATISTICAL INFERENCE

STATISTICS AND MAKING CORRECT DECISONS

STATISTICAL TOOLS USED TO ASSIST DECISION MAKING

Regression Analysis

Determining Confidence Interval

Comparison Tests

Analysis Of Variance

Design Of Experiments

Linear & Non-linear Programming

Queuing Theory

Regression Analysis

REGRESSION ANALYSIS

No Correlation (R = 0)Strong Positive Correlation

( R = .995)

Positive Linear correlation(r=0.85)

Negative Linear Correlation(r=-0.85)

TYPES OF REGRESSION ANALYSIS

(AMONG MANY)

Exponential Y =ABX

Geometric Y = AXB

Logarithmic Y = Ao + A1(logX) + A2(logX)2

Linear Y = Ao + A1X

Linear Regression Is the Most Common

THE PURPOSE OF REGRESSION ANALYSIS

Correlation

r = (Xi -X) (Yi - Y)

(Xi -X)2 (Yi - Y)2

r =1 = perfect correlationr = 0 = no correlation

Determination Of Unknown Parameters

1 =

Yi(Xi -X)n

i=1

(Xi -X)2n

i=1

Y = 0 + 1X^^

^

0 = Y - ^ 1X

^

Confidence Interval

Statistics Usually Do Not Represent Absolute Truth

Very Often They Are A Good Guess

How Good Of A Guess Is Explained By The Confidence Interval

Understanding Confidence Intervals Will AllowYou To Better Evaluate Critical Statistics

WHY DO WE NEED CONFIDENCE INTERVALS

Problem: Commanding General Needs To Know Average Weight of Officers On The Base

1,000 Officers At The Base

Six Officers Selected And Weighed

Officer # 1: 68 kilosOfficer # 2: 57 kilosOfficer # 3: 72 kilosOfficer # 4: 71 kilosOfficer # 5: 100 kilosOfficer # 6: 63 kilos

Average Weight of Our Sample Is 71.8333 Kilos

How Good Is This Statistic?

A TYPICAL SITUATION

• We Can Be 90% Confident That The Average Weight Is Between 59.6 Kilos And 84.1 Kilos

• We Can Be 95% Confident That The Average Weight Is Between 56.2 Kilos And 87.5 Kilos

• We Can Be 99% Confident That The Average Weight Is Between 47.3 Kilos And 96.3 Kilos

59.6 84.1 87.552.647.3 96.3

90%

95%

99%

Average Weight of All Officers In Kilos

HOW GOOD IS OUR SAMPLE?

HOW DID WE GET OUR CONFIDENCE INTERVAL?

Use Table 17.1 (Page 297 of Implementing Six Sigma )

Depends On• Is Known Or Not Known• Sample Variation• Sample Size• Level Of Desired Confidence

X -U

n

X +U

n

X -U

n X +

U

n

X -St

n

X +St

n X -St

n X +

Stn

OR

OR

Single Sided Double-Sided

Known

Unknown

CONFIDENCE INTERVAL EQUATIONS

Single Sided: Trying to Determine If the Population Average () Is Less Than or Greater Than the SampleAverage ( X )Double Sided: Trying To Determine The Upper& Lower Boundaries of the Population Average ()

: Population Average

: Population Standard Deviation

: 1 - The Desired Confidence Level

S : The Sample Standard Deviation

v : Degrees Of Freedom Or n - 1

t : Data Derived From t Distribution

U: Data Derived From Normal Distribution

n: The Number of Units in The Sample

EQUATION HELP

X -St

n X +

Stn

71.833 - 2.015 (14.878)

6 71.833 + 2.015 (14.878)

6

SOLUTION TO THE OFFICER WEIGHT PROBLEM

STANDARD DEVIATION CONFIDENCE INTERVAL

Almost The Same But Different

See Page 300 Of Implementing Six Sigma

We Will Use The Chi Square Distribution ( 2 )

2/2; v 2

(1-/2; v)

(n - 1) s2 (n - 1) s2

[ ]1/2

[ ]1/2

2/2; v 2

(1-/2; v)

(n - 1) s2 (n - 1) s2

[ ]1/2

[ ]1/2

11.07 1.15

(5) (14.878)2 (5) (14.878)2

[ ]1/2

[ ]1/2

(99.9796)1/2

(962.4125)1/2

9.999 31.0227

EXAMPLE USING SIX OFFICER WEIGHTS

We Are 90% Confident That Standard Deviation Of All Officer Weights Is Between 10 Kilos & 31.0 Kilos

Comparison Tests

Is Process B Better Than Process A?

Is Supplier B Better Than Supplier A?

These Questions Are Always Being Asked

COMPARISON TESTS

Comparison Tests Can Give The Right Answers

STEPS INVOLVED IN COMPARISON TESTING

Define Precisely The Problem Objective

Formulate A Null Hypothesis

Evaluation By A One Or Two Tail Test

Choose A Critical Value Of A Test Statistic

Calculate A Test Statistic

Make Inference About The Population

Communicate The Findings

TYPICAL DECISIONS1. A chemical batch process has yielded average of802 tons of product for a long period. Production records for last five batches show following results: 803, 786, 806, 791, and 794. Can we predict with 95% confidence that the process is now at a lower average?

2. The average vial height from an injection molding process has been 5.00 inches with a standard deviation of .12”. A vendor claims to have a new material that will reduce the height variation. An experiment, conducted using the new material, yielded the following results: 5.10, 4.90, 4.92, 4.87, 5.09, 4.89, 4.95, 4.88. The average height of the eight vials is 4.95” and the standard deviation is .093”

Average vial height from an injection molding process has been 5.00 inches with a standard deviation of .12”. Vendor claims to have a new material that will reduce height variation. An experiment, conducted using new material yielded the following results: 5.10, 4.90, 4.92, 4.87, 5.09, 4.89, 4.95, 4.88. Average height of the eight vials is 4.95” and standard deviation is .093”

Is the new material producing shorter vials with the existing molding machine set-up (with 95% confidence)?

Is height variation actually less with the new material (with 95% confidence)

NULL HYPOTHESIS

The Hypothesis To Be Tested

A Null Hypothesis Can Only Be Rejected. It Cannot Be Accepted Because of a Lack of Evidence to Reject It

Example:If A Claim Is That Process B Is Better Than Process AThe Null Hypothesis Is That Process A = Process BHo : A = B

Table 19.1(page 322 Implementing Six Sigma)

Most Likely: 12 2

2 And Is Unknown

1. Calculate t0 =X1 - X2

S12

n1n2

S22+

2. Calculate =

S12

n1( ) S2

2

n2( )+[ ] 2

(S12/ n1)

2 (S22/ n2)

2

n1 + 1 n2 + 1+

COMPARISON METHODOLOGY

COMPARISON METHODOLOGY(Continued)

3. Look Up Value Of t Using Table E Or Table D (Implementing Six Sigma Pages 697 Or 698)

4. Reject The Null Hypothesis If t0 Is Greater Than Than t