1. 1 SMU EMIS 7364 NTU TO-570-N Inferences About Process Quality Updated: 2/3/04 Statistical Quality Control Dr. Jerrell T. Stracener, SAE Fellow

2. 2 Inferences about Process Quality Sampling & Sampling Distributions Inferences Based on Single Random Sample Inferences Based on Two Random Samples Inferences Based on More than Two Random Samples

3. 3 Sampling & Sampling Distributions

4. 4 Population vs. Sample Population the total of all possible values (measurement, counts, etc.) of a particular characteristic for a specific group of objects. Sample a part of a population selected according to some rule or plan. Why sample?

5. 5 Sampling Characteristics that distinguish one type of sample from another: the manner in which the sample was obtained the purpose for which the sample was obtained

6. 6 Simple Random Sample The sample X 1, X 2,...,X n is a random sample if X 1, X 2,..., X n are independent identically distributed random variables. Remark: Each value in the population has an equal and independent chance of being included in the sample.

7. 7 Generating Random Samples using Monte Carlo Simulation

8. 8 Generating Random Numbers f(y) F(y) y y 1.0 0.8 0.6 0.4 0.2 0 riri yiyi

9. 9 Generating Random Numbers Generating values of a random variable using the probability integral transformation to generate a random value y from a given probability density function f(y): 1. Generate a random value r U from a uniform distribution over (0, 1). 2. Set r U = F(y) 3. Solve the resulting expression for y.

10. 10 Generating Random Numbers with Excel From the Tools menu, look for Data Analysis.

11. 11 Generating Random Numbers with Excel If it is not there, you must install it.

12. 12 Generating Random Numbers with Excel Once you select Data Analysis, the following window will appear. Scroll down to “Random Number Generation” and select it, then press “OK”

13. 13 Generating Random Numbers with Excel Choose which distribution you would like. Use uniform for an exponential or weibull distribution or normal for a normal or lognormal distribution

14. 14 Generating Random Numbers with Excel Uniform Distribution, U(0, 1). Select “Uniform” under the “Distribution” menu. Type in “1” for number of variables and 10 for number of random numbers. Then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

15. 15 Generating Random Numbers with Excel Normal Distribution, N(, ). Select “Normal” under the “Distribution” menu. Type in “1” for number of variables and 10 for number of random numbers. Enter the values for the mean (  ) and standard deviation (  ) then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

16. 16 Generating Random Values from an Exponential Distribution E() with Excel First generate n random variables, r 1, r 2, …, r n, from U(0, 1). Select “Uniform” under the “Distribution” menu. Type in “1” for number of variables and 10 for number of random numbers. Then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

17. 17 Generating Random Values from an Exponential Distribution E() with Excel Select a  that you would like to use, we will use  = 5. Type in the equation x i = - ln(1 - r i ), with filling in  as 5, and r i as cell A1 ( =-5*LN(1-A1) ). Now with that cell selected, place the cursor over the bottom right hand corner of the cell. A cross will appear, drag this cross down to B10. This will transfer that equation to the cells below. Now we have n random values from the exponential distribution with parameter  =5 in cells B1 - B10.

18. 18 Generating Random Values from an Weibull Distribution W(,) with Excel First generate n random variables, r 1, r 2, …, r n, from U(0, 1). Select “Uniform” under the “Distribution” menu. Type in “1” for number of variables and 10 for number of random numbers. Then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

19. 19 Generating Random Values from an Weibull Distribution W(,) with Excel Select a  and  that you would like to use, we will use  = 100,  = 20. Type in the equation x i =  [-ln(1 - r i )] 1/ , with filling in  as 100,  as 20, and r i as cell A1 ( =100*(-LN(1-A1))^(1/20) ). Now transfer that equation to the cells below. Now we have n random variables from the Weibull distribution with parameters  =100 and  =20 in cells B1 - B10.

20. 20 Generating Random Values from an Lognormal Distribution LN(, ) with Excel First generate n random variables, r 1, r 2, …, r n, from N(, ). Select “Normal” under the “Distribution” menu. Type in “1” for number of variables and 10 for number of random numbers. Enter 0 for the mean and 1 for standard deviation then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

21. 21 Generating Random Values from an Lognormal Distribution LN(, ) with Excel Select a  and  that you would like to use, we will use  = 2,  = 1. Type in the equation, with filling in  as 2,  as 1, and r i as cell A1 ( =EXP(2+A1*1) ). Now transfer that equation to the cells below. Now we have an Lognormal distribution in cells B1 - B10.

22. 22 Flow Chart of Monte Carlo Simulation method Input 1: Statistical distribution for each component variable. Input 2: Relationship between component variables and system performance Select a random value from each of these distributions Calculate the value of system performance for a system composed of components with the values obtained in the previous step. Output: Summarize and plot resulting values of system performance. This provides an approximation of the distribution of system performance. Repeat many times

23. 23 Distribution of Sample Mean

24. 24 Sampling Distribution of X with known  If X 1, X 2,...,X n is a random sample of size n from a normal distribution with mean  and known standard deviation , and if, then and

25. 25 Central Limit Theorem If X is the mean of a random sample of size n, X 1, X 2, …, X n, from a population with mean  and finite standard deviation , then if n   the limiting distribution of is the standard normal distribution.

26. 26 Central Limit Theorem Remark: The Central Limit Theorem provides the basis for approximating the distribution of X with a normal distribution with mean  and standard deviation The approximation gets better as n gets larger.

27. 27 Sampling Distribution of X with Unknown  Let X 1, X 2,..., X n be independent random variables that have normal distribution with mean  and unknown standard deviation . Let and Then the random variable has a t-distribution with  = n - 1 degrees of freedom.

28. 28 Distribution of Sample Standard Deviation

29. 29 Sampling Distributions of S 2 If S 2 is the variance of a random sample of size n taken from a normal population having the variance  2, then the statistic has a chi-squared distribution with  = n - 1 degrees of freedom.

30. 30 Inferences Based on a Single Random Sample

31. 31 Estimation - Binomial Distribution Estimation of a Proportion, p X 1, X 2, …, X n is a random sample of size n from B(n, p) Point estimate of p: where f s = # of successes

32. 32 Estimation - Binomial Distribution Approximate (1 - ) ·100% confidence interval for p: whereand where, and is the value of the standard normal random variable Z such that

33. 33 Estimation of the Mean - Normal Distribution X 1, X 2, …, X n is a random sample of size n from N(, ), where both  &  are unknown. Point Estimate of  (1 - )  100% Confidence Interval for the mean where, and

34. 34 Estimation of the Mean - Infinite Population - Type Unknown X 1, X 2, …, X n is a random sample of size n Point Estimate of  An approximate (1 - )  100% Confidence Interval for the mean based on the Central Limit Theorem where and

35. 35 Estimation of Means - Finite Populations X 1, X 2,..., X n is a random sample of size n from a population of size N with unknown parameters  and  Point Estimate of : An approximate (1 - ) · 100% Confidence Interval for  is, where and, where,

36. 36 Estimation of Means - Finite Populations where is the value of T ~ t df for which is the finite population correction factor

37. 37 Estimation of Lognormal Distribution Random sample of size n, X 1, X 2,..., X n from LN (, ) Let Y i = ln X i for i = 1, 2,..., n Treat Y 1, Y 2,..., Y n as a random sample from N(, ) Estimate  and  using the Normal Distribution Methods

38. 38 Estimation of Weibull Distribution Random sample of size n, T 1, T 2, …, T n, from W(, ), where both  &  are unknown. Point estimates is the solution of g() = 0 where

39. 39 Estimation of Standard Deviation - Normal Distribution Point Estimate of  (1 - ) · 100% Confidence Interval for  is, where and

40. 40 Testing Hypotheses There are two possible decision errors associated with testing a statistical hypothesis: A Type I error is made when a true hypothesis is rejected. A Type II error is made when a false hypothesis is accepted. True Situation DecisionH 0 trueH 0 false Accept H 0 correctType II error Reject H 0 Type I errorcorrect (Accept H 1 )

41. 41 Testing Hypotheses The decision risks are measured in terms of probability.  = P(Type I error) = P(reject H 0 |H 0 is true) = Producers risk  = P(Type II error) = P(accept H 0 |H 1 is true) = Consumers risk Remark: 100% ·  is commonly referred to as the significance level of a test. Note: For fixed n,  increases as  decreases, and vice versa, as n increases, both  and  decrease.

42. 42 Power Function Before applying a test procedure, i.e., a decision rule, we need to analyze its discriminating power, i.e., how good the test is. A function called the power function enables us to make this analysis. Power Function = P(rejecting H 0 |true parameter value) OC Function= P(accepting H 0 |true parameter value) = 1 - Power Function where OC is Operating Characteristic.

43. 43 Power Function A plot of the power function vs the test parameter value is called the power curve and 1 - power curve is the OC curve. 1 0 PR()PR() ideal power curve H0H0 H1H1 

44. 44 Power Function The power function of a statistical test of hypothesis is the probability of rejecting H0 as a function of the true value of the parameter being tested, say , i.e., PF() = PR() = P(reject H 0 |) = P(test statistic falls in C A |)

45. 45 Operating Characteristic Function The operating characteristic function of a statistical test of hypothesis is the probability of accepting H0 as a function of the true value of the parameter being tested, say , i.e., OC()= P A () = P(accept H 0 |) = P(test statistic falls in C R |)

46. 46 Tests of Proportions Let X 1, X 2,..., X n be a random sample of size n from B(n, p). Case 1: small sample sizes To test the Null Hypothesis H 0 : p = p 0, a specified value, against the appropriate Alternative Hypothesis 1. H A : p < p 0, or 2. H A : p > p 0, or 3. H A : p  p 0,

47. 47 Tests of Proportions at the 100 · % Level of Significance, calculate the value of the test statistic using X ~ B(n, p = p 0 ). Find the number of successes and compute the appropriate P-Value, depending upon the alternative hypothesis and reject H 0 if P  , where 1. P = P(X  x|p = p 0 ), or 2. P = P(X  x|p = p 0 ), or 3. P = 2P(X  x|p = p 0 ) if x < np 0, or P = 2P(X  x|p = p 0 ) if x > np 0,

48. 48 Tests of Proportions Case 2: large sample sizes with p not extremely close to 0 or 1. To test the Null Hypothesis H 0 : p = p 0, a specified value, against the appropriate Alternative Hypothesis 1. H A : p < p 0, or 2. H A : p > p 0, or 3. H A : p  p 0,

49. 49 Tests of Proportions Calculate the value of the test statistic and reject H 0 if 1., or 2., or 3. or, depending on the alternative hypothesis.

50. 50 Test of Means Let X 1, …, X n, be a random sample of size n, from a normal distribution with mean  and standard deviation , both unknown. To test the Null Hypothesis H 0 :  =  0, a given or specified value against the appropriate Alternative Hypothesis 1. H A :  <  0, or 2. H A :  >  0, or 3. H A :    0,

51. 51 Test of Means at the 100  % level of significance. Calculate the value of the test statistic Reject H 0 if 1. t < -t , n-1, 2. t > t , n-1, 3. t t /2, n-1, depending on the Alternative Hypothesis.

52. 52 Test of Variances Let X 1, …, X n, be a random sample of size n, from a normal distribution with mean  and standard deviation , both unknown. To test the Null Hypothesis H 0 :  2 =  2 0, a specified value against the appropriate Alternative Hypothesis 1. H A :  2 <  2 0, or 2. H A :  2 >  2 0, or 3. H A :  2   2 0,

53. 53 Test of Variances at the 100  % level of significance. Calculate the value of the test statistic Reject H 0 if 1.  2 <  2 1-, n-1, 2.  2 >  2 , n-1, 3.  2  2 /2, n-1, depending on the Alternative Hypothesis.

54. 54 Inferences Based on Two Random Samples

55. 55 Estimation - Binomial Populations Estimation of the difference between two proportions Let X 11, X 12, …,, and X 21, X 22, …,, be random samples from B(n 1, p 1 ) and B(n 2, p 2 ) respectively Point estimation of p 1 - p 2

56. 56 Estimation - Binomial Populations Approximate (1 - ) · 100% confidence interval for where and

57. 57 Estimation of Difference Between Two Means - Normal Distribution Let X 11, X 12, …,, and X 21, X 22, …, be random samples from N( 1,  1 ) and N( 2,  2 ), respectively, where  ,  ,   and   are all unknown Point estimation of  =  1 -  2

58. 58 Estimation of Difference Between Two Means - Normal Distribution An approximate (1 - ) · 100% Confidence Interval for  =  1 -  2 where

59. 59 Estimation of Difference Between Two Means - Normal Distribution where  = degrees of freedom

60. 60 Estimation of Ratio of Two Standard Deviations - Normal Distribution Let X 11, X 12, …,, and X 21, X 22, …, be random samples from n( 1,  1 ) and n( 2,  2 ), respectively Point estimation of where for i = 1, 2

61. 61 Estimation of Ratio of Two Standard Deviations - Normal Distribution (1 - ) · 100% Confidence Interval for where and

62. 62 Estimation of Ratio of Two Standard Deviations - Normal Distribution where is the value of the F-Distribution with anddegrees of freedom for which

63. 63 Test on Two Means Let X 11, X 12, …, X 1n 1 be a random sample of size n 1 from N( 1,  1 ) and X 21, X 22, …, X 2n 2 be a random sample of size n 2 from N( 2,  2 ), where  1,  1,  2 and  2 are all unknown. To test H 0:  1 -  2 = d o, where d o  0, against the appropriate alternative hypothesis

64. 64 Test on Two Means 1. H 1:  1 -  2 < d o, where d o  0, or 2. H 1:  1 -  2 > d o, where d o  0, or 3. H 1:  1 -  2  d o, where d o  0, at the   100% level of significance, calculate the value of the test statistic.

65. 65 Test on Two Means Reject H o if 1. t' < t  or2. t' > t  or3. t' t  depending on the alternative hypothesis.

66. 66 Test on Two Variances Let X 11, X 12, …, X 1n 1 be a random sample of size n 1 from N( 1,  1 ) and X 21, X 22, …, X 2n 2 be a random sample of size n 2 from N( 2,  2 ), where  1,  1,  2 and  2 are all unknown. To test H 0: against the appropriate alternative hypothesis

67. 67 Test on Two Variances 1. H 1: or 2. H 1: or 3. H 1: at the   100% level of significance, calculate the value of the test statistic.

68. 68 Test on Two Variances Reject H o if or depending on the alternative hypothesis.

69. 69 Inferences Based on More than Two Random Samples

70. 70 Normal Distribution - Estimation of  X 1, X 2, …, X n is a random sample of size n from N(, ), where both  &  are unknown. Point Estimate of  (1 - )·100% Confidence Interval for  is, where and

71. 71 Normal Distribution - Estimation of  where is the value of the t-distribution with parameter = n-1 which P(T> ) = /2 and may be obtained from the table t-distribution ( Located in the resource section on the website ).

72. 72 Estimation of Lognormal Distribution Random sample of size n, X 1, X 2,..., X n from LN (, ) Let Y i = ln X i for i = 1, 2,..., n Treat Y 1, Y 2,..., Y n as a random sample from N(, ) Estimate  and  using the Normal Distribution Methods

73. 73 Estimation of Weibull Distribution Random sample of size n, T 1, T 2, …, T n, from W(, ), where both  &  are unknown. Point estimates is the solution of g() = 0 where