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