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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 =AB X Geometric  Y = AX B Logarithmic  Y = A o  + A 1 (logX) + A 2 (logX) 2 Linear Y = A o  + A 1 X Linear Regression Is the Most Common
THE PURPOSE OF  REGRESSION ANALYSIS Correlation r =1 = perfect correlation r = 0 = no correlation Determination Of  Unknown Parameters r =    (X i  -X) (Y i  - Y)    (X i  -X) 2  (Y i  - Y) 2  1  =    Y i (X i  -X) n i =1    (X i  -X) 2 n i =1 Y =   0  +   1 X ^ ^ ^  0  = Y -  ^  1 X ^
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 Allow You 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 kilos Officer # 2:  57 kilos Officer # 3:  72 kilos Officer # 4:  71 kilos Officer # 5:  100 kilos Officer # 6:  63 kilos Average Weight of Our Sample Is 71.8333 Kilos How Good Is This Statistic? A TYPICAL SITUATION
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],HOW GOOD IS OUR SAMPLE? 59.6 84.1 87.5 52.6 47.3 96.3 90% 95% 99% Average Weight of All Officers In Kilos
HOW DID WE GET OUR  CONFIDENCE INTERVAL? ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
OR OR Single Sided Double-Sided    Known    Unknown CONFIDENCE INTERVAL  EQUATIONS     X  -  U  n      X  +  U  n  X  -  U  n         X  +  U  n      X  - S t    n      X  + S t  n  X  - S t  n         X  + S t  n 
   : 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 Single Sided : Trying to Determine If the Population  Average (  ) Is Less Than or Greater Than the Sample Average  ( X ) Double Sided :  Trying To Determine The Upper& Lower  Boundaries of the Population Average (  )
SOLUTION TO THE  OFFICER WEIGHT PROBLEM X  - S t  n         X  + S t  n  71.833  -  2.015 (14.878) 6         71.833  +  2.015 (14.878) 6 
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)  s 2 ( n  - 1)  s 2        [ ] 1/2 [ ] 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  2  /2;  v  2  (1-  /2;  v) ( n  - 1)  s 2 ( n  - 1)  s 2        [ ] 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
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 DECISIONS 1.  A chemical batch process has yielded average of 802 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 A The Null Hypothesis Is That Process A = Process B H o  : A = B
Table 19.1(page 322  Implementing Six Sigma ) Most Likely:   1 2       2 2  And     Is Unknown 1. Calculate  t 0  = 2.  Calculate    = COMPARISON METHODOLOGY  X 1  - X 2    S 1 2 n 1 n 2 S 2 2 +  S 1 2 n 1 ( ) S 2 2 n 2 ( ) + [ ] 2 (S 1 2 / n 1 ) 2 (S 2 2 / n 2 ) 2 n 1  + 1 n 2  + 1 +
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  t 0  Is Greater  Than Than  t 

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Statistical Tools for Decision Making

  • 2. STATISTICS AND MAKING CORRECT DECISONS
  • 3. 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
  • 5. REGRESSION ANALYSIS No Correlation (R = 0) Strong Positive Correlation ( R = .995) Positive Linear correlation (r=0.85) Negative Linear Correlation (r=-0.85)
  • 6. TYPES OF REGRESSION ANALYSIS (AMONG MANY) Exponential Y =AB X Geometric Y = AX B Logarithmic Y = A o + A 1 (logX) + A 2 (logX) 2 Linear Y = A o + A 1 X Linear Regression Is the Most Common
  • 7. THE PURPOSE OF REGRESSION ANALYSIS Correlation r =1 = perfect correlation r = 0 = no correlation Determination Of Unknown Parameters r =  (X i -X) (Y i - Y)  (X i -X) 2 (Y i - Y) 2  1 =  Y i (X i -X) n i =1  (X i -X) 2 n i =1 Y =  0 +  1 X ^ ^ ^  0 = Y - ^  1 X ^
  • 9. 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 Allow You To Better Evaluate Critical Statistics WHY DO WE NEED CONFIDENCE INTERVALS
  • 10. 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 kilos Officer # 2: 57 kilos Officer # 3: 72 kilos Officer # 4: 71 kilos Officer # 5: 100 kilos Officer # 6: 63 kilos Average Weight of Our Sample Is 71.8333 Kilos How Good Is This Statistic? A TYPICAL SITUATION
  • 11.
  • 12.
  • 13. OR OR Single Sided Double-Sided  Known  Unknown CONFIDENCE INTERVAL EQUATIONS   X -  U  n    X +  U  n  X -  U  n     X +  U  n    X - S t  n    X + S t  n  X - S t  n     X + S t  n 
  • 14. : 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 Single Sided : Trying to Determine If the Population Average (  ) Is Less Than or Greater Than the Sample Average ( X ) Double Sided : Trying To Determine The Upper& Lower Boundaries of the Population Average (  )
  • 15. SOLUTION TO THE OFFICER WEIGHT PROBLEM X - S t  n     X + S t  n  71.833 - 2.015 (14.878) 6     71.833 + 2.015 (14.878) 6 
  • 16. 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) s 2 ( n - 1) s 2    [ ] 1/2 [ ] 1/2
  • 17. 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  2  /2; v  2 (1-  /2; v) ( n - 1) s 2 ( n - 1) s 2    [ ] 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
  • 19. 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
  • 20. 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
  • 21. TYPICAL DECISIONS 1. A chemical batch process has yielded average of 802 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”
  • 22. 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)
  • 23. 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 A The Null Hypothesis Is That Process A = Process B H o : A = B
  • 24. Table 19.1(page 322 Implementing Six Sigma ) Most Likely:  1 2   2 2 And  Is Unknown 1. Calculate t 0 = 2. Calculate  = COMPARISON METHODOLOGY  X 1 - X 2  S 1 2 n 1 n 2 S 2 2 +  S 1 2 n 1 ( ) S 2 2 n 2 ( ) + [ ] 2 (S 1 2 / n 1 ) 2 (S 2 2 / n 2 ) 2 n 1 + 1 n 2 + 1 +
  • 25. 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 t 0 Is Greater Than Than t 