1. Horng-Chyi HorngStatistics II 91 Inference on the Variance of a Normal Population (I) H 0 :  2 =  0  H 1 :  2   0 , where  0  is a specified constant. Sampling from a normal distribution with unknown mean  and unknown variance  2, the quantity has a Chi-square distribution with n-1 degrees of freedom. That is,

2. Horng-Chyi HorngStatistics II 92 Inference on the Variance of a Normal Population (II) Let X 1, X 2, …, X n be a random sample for a normal distribution with unknown mean  and unknown variance  2. To test the hypothesis H 0 :  2 =  0  H 1 :  2   0 , where  0  is a specified constant. We use the statistic If H 0 is true, then the statistic has a chi-square distribution with n-1 d.f..

3. Horng-Chyi HorngStatistics II 93

4. Horng-Chyi HorngStatistics II 94 The Reasoning For H 0 to be true, the value of  0 2 can not be too large or too small. What values of  0 2 should we reject H 0 ? (based on  value) What values of  0 2 should we conclude that there is not enough evidence to reject H 0 ?

5. Horng-Chyi HorngStatistics II 95

6. Horng-Chyi HorngStatistics II 96 Example 8-11 An automatic filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in a sample variance of fill volume of s 2 = 0.0153 (fluid ounces) 2. If the variance of fill volume exceeds 0.01 (fluid ounces) 2, an unacceptable proportion of bottles will be underfilled and overfilled. Is there evidence in the sample data to suggest that the manufacturer has a problem with underfilled and overfilled bottles? Use  = 0.05, and assume that fill volume has a normal distribution.

7. Horng-Chyi HorngStatistics II 97

8. Horng-Chyi HorngStatistics II 98 Hypothesis Testing on Variance - Normal Population

9. Horng-Chyi HorngStatistics II 99 Finding P-Values Steps: 1. Find the degrees of freedom (k = n-1)in the the  2 -table. 2. Compare  0 2 to the values in that row and find the closest one. 3. Look the  value associated with the one you pick. The p-value of your test is equal to this  value. In example 8-11,  0 2 = 29.07, k = n-1 = 19, 0.05 < P-Value < 0.10 because the  2 value associated with (k = 19,  = 0.10) is 27.20 while the  2 value associated with (k = 19,  = 0.05) is 30.14

10. Horng-Chyi HorngStatistics II 100 P-Values of Hypothesis Testing on Variance

11. Horng-Chyi HorngStatistics II 101 The Operating Characteristic Curves - Chi-square test Use to performing sample size or type II error calculations. The parameter is defined as: for various sample sizes n, where  denotes the true value of the standard deviation. Chart VI I,j,k,l are used in chi-square test. (pp. A16-A17)

12. Horng-Chyi HorngStatistics II 102 Example 8-12

13. Horng-Chyi HorngStatistics II 103

14. Horng-Chyi HorngStatistics II 104 Construction of the C.I. on the Variance In general, the distribution of is chi-square with n-1 d.f. Use the properties of t with n-1 d.f.,

15. Horng-Chyi HorngStatistics II 105 Formula for C.I. on the Variance

16. Horng-Chyi HorngStatistics II 106 Formula for C.I. on the Variance - One-sided

17. Horng-Chyi HorngStatistics II 107 Example 8-13 Reconsider the bottle filling machine problem in Example 8- 11. Find a 95% upper-C.I. on the variance? N = 20, s 2 = 0.0153 Therefore,  2  (20-1)0.0153/10.117 = 0.0287 The 95% upper-C.I. on the variance is 0.0287. In addition, the 95% upper-C.I. on the standard deviation is  0.0287 = 0.17.

18. Horng-Chyi HorngStatistics II 108 Inference on a Population Proportion(I) H 0 : p = p 0 H 1 : p  p 0, where p 0 is a specified constant. P = X/n, in which X is a binomial variable, i.e., the number of success in n trials. E(X) = np, V(X) = npq = np(1-p) If H 0 is true, then using the normal approximation to the binomial, the quantity follows the standard normal distribution(Z).

19. Horng-Chyi HorngStatistics II 109 Inference on a Population Proportion (II) Let x be the number of observations in a random sample of size n that belongs to the class associated with p. To test the hypothesis H 0 : p = p 0 H 1 : p  p 0, where p 0 is a specified constant. We use the statistic

20. Horng-Chyi HorngStatistics II 110 The Reasoning For H 0 to be true, the value of Z 0 can not be too large or too small. What values of Z 0 should we reject H 0 ? (based on  value) What values of Z 0 should we conclude that there is not enough evidence to reject H 0 ?

21. Horng-Chyi HorngStatistics II 111 Example 8-14 A semiconductor manufacturing produces controllers used in automobile engine applications. The customer erquires that the process fallout or fraction defective at a critical manufacturing step not exceed 0.05 and that the manufacturing demonstrate process capacity at this level of quality using  = 0.05. The semiconductor manufacturer takes a random sample of 200 devices and finds that four of them are defective. Can the manufacturer demonstrate process capacity for the customer?

22. Horng-Chyi HorngStatistics II 112 The parameter of interest is the process fraction defective. H 0 : p = 0.05 H 1 : p < 0.05 This formulation of the problem will allow the manufacturer to make a strong claim about process capacity if the null hypothesis H 0 : p = 0.05 is rejected.  = 0.05, x = 4, n = 200, and p 0 = 0.05. To reject H 0 : p = 0.05, the test statistic Z 0 must be less than -z 0.05 = -1.645 Conclusion: Since Z 0 = -1.95 < -z 0.05 = -1.645, we reject H 0 and conclude that the process fraction defective p is less than 0.05. The P-value for this value of the test statistic Z 0 is P = 0.0256, which is less than  = 0.05. We conclude that the process is capable.

23. Horng-Chyi HorngStatistics II 113 Hypothesis Testing on a Population Proportion

24. Horng-Chyi HorngStatistics II 114 P-Values of Hypothesis Testing on a Population Proportion

25. Horng-Chyi HorngStatistics II 115 How to calculate Type II Error? (I) (H 0 : p = p 0 Vs. H 1 : p  p 0 ) Under the circumstance of type II error, H 0 is false. Supposed that the true value of the population proportion is p. The distribution of Z 0 is:

26. Horng-Chyi HorngStatistics II 116 How to calculate Type II Error? (II) - refer to section &4.3 (&8.1) Type II error occurred when (fail to reject H 0 while H 0 is false) Therefore,

27. Horng-Chyi HorngStatistics II 117 Formula for Type II Error Two-sided alternative H 1 : p  p 0 One-sided alternative H 1 : p < p 0 One-sided alternative H 1 : p > p 0

28. Horng-Chyi HorngStatistics II 118 The Sample Size (I) Given values of  and p, find the required sample size n to achieve a particular level of .

29. Horng-Chyi HorngStatistics II 119 The Sample Size (II) Two-sided Hypothesis Testing One-sided Hypothesis Testing

30. Horng-Chyi HorngStatistics II 120 Example 8-15 (1) Consider the semiconductor manufacturer from Example 8-14. Suppose that his process fallout is really p = 0.03. What is the  -error for his test of process capacity, which uses n = 200 and  = 0.05? (2) Suppose that the semiconductor manuafcturer was willing to accept a  -error as large as 0.10 if the true value of the process fraction defective was p = 0.03. If the manufacturer continue to use  = 0.05, what sample size would be required?

31. Horng-Chyi HorngStatistics II 121 (1) Since H 1 : p < 0.05, therefore (2) The sample size required for this one-sided alternative is

32. Horng-Chyi HorngStatistics II 122 Formula for C.I. on the Population Proportion

33. Horng-Chyi HorngStatistics II 123 Formula for One-Sided C.I. on the Population Proportion

34. Horng-Chyi HorngStatistics II 124 Example 8-16 In a random sample of 85 automobile engine crankshaft bearings, 10 have a surface finish that is rougher than the specifications allow. Construct a 95% C.I. on the population proportion p? Sol:

35. Horng-Chyi HorngStatistics II 125 Choice of Sample Size for C.I. on a Population Proportion where E is the half-width of the C.I. Since the max value for p(1-p) is 0.25 for 0  p  1, we can use the following formula instead.

36. Horng-Chyi HorngStatistics II 126