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