Ch 11 實習 (2)

1. Ch 11 實習 (2)

2. A Two - Tail Test • Example 11.2 • AT&T has been challenged by competitors who argued that their rates resulted in lower bills. • A statistics practitioner determines that the mean and standard deviation of monthly long-distance bills for all AT&T residential customers are $17.09 and $3.87 respectively. Jia-Ying Chen

3. A Two - Tail Test • Example 11.2 - continued • A random sample of 100 customers is selected and customers’ bills recalculated using a leading competitor’s rates (see Xm11-02). • Assuming the standard deviation is the same (3.87), can we infer that there is a difference between AT&T’s bills and the competitor’s bills (on the average)? Jia-Ying Chen

4. A Two - Tail Test • Solution • Is the mean different from 17.09? H0: m = 17.09 • Define the rejection region Jia-Ying Chen

5. a/2 = 0.025 a/2 = 0.025 If H0 is true (m =17.09), can still fall far above or far below 17.09, in which case we erroneously reject H0 in favor of H1 A Two – Tail Test Solution - continued 17.09 We want this erroneous rejection of H0 to be a rare event, say 5% chance. Jia-Ying Chen

6. a/2 = 0.025 a/2 = 0.025 17.55 a/2 = 0.025 a/2 = 0.025 From the sample we have: 0 za/2= 1.96 -za/2= -1.96 Rejection region A Two – Tail Test Solution - continued 17.09 Jia-Ying Chen

7. 0 za/2= 1.96 -za/2= -1.96 A Two – Tail Test There is insufficient evidence to infer that there is a difference between the bills of AT&T and the competitor. Also, by the p value approach: The p-value = P(Z< -1.19)+P(Z >1.19) = 2(.1173) = .2346 > .05 a/2 = 0.025 a/2 = 0.025 -1.19 1.19 Jia-Ying Chen

8. Example 1 • A machine that produces ball bearings is set so that the average diameter is 0.5 inch. A sample of 10 ball bearings was measured with the results shown here. Assuming that the standard deviation is 0.05 inch, can we conclude that at the 5% significance level that the mean diameter is not 0.5 inch? • 0.48 0.50 0.49 0.52 0.53 0.48 0.49 0.47 0.46 0.51 Jia-Ying Chen

9. Solution Jia-Ying Chen

10. Calculation of the Probability of a Type II Error • To calculate Type II error we need to… • express the rejection region directly, in terms of the parameter hypothesized (not standardized). • specify the alternative value under H1. • 型二誤差的定義是，H1正確卻無法拒絕H0 • 在什麼規則下你無法拒絕H0 • 單尾 • 雙尾 • Let us revisit Example 11.1 Jia-Ying Chen

11. a=.05 H0: m = 170 H1: m = 180 m= 170 m=180 Calculation of the Probability of a Type II Error Express the rejection region directly, not in standardized terms • Let us revisit Example 11.1 • The rejection region was with a = .05. • Let the alternative value be m = 180 (rather than just m>170) Specify the alternative value under H1. Do not reject H0 Jia-Ying Chen

12. a=.05 H0: m = 170 H1: m = 180 m= 170 m=180 Calculation of the Probability of a Type II Error • A Type II error occurs when a false H0 is not rejected. A false H0… …is not rejected Jia-Ying Chen

13. H0: m = 170 H1: m = 180 m=180 Calculation of the Probability of a Type II Error m= 170 Jia-Ying Chen

14. Example 2 • A statistics practitioner wants to test the following hypotheses with σ=20 and n=100: • H0: μ=100 • H1: μ>100 • Using α=0.1 find the probability of a Type II error when μ=102 Jia-Ying Chen

15. Solution • Rejection region: z>zα Jia-Ying Chen

16. Example 3 • Calculate the probability of a Type II error for the following test of hypothesis, given that μ=203. • H0: μ=200 • H1: μ≠200 • α=0.05, σ=10, n=100 Jia-Ying Chen

17. Solution Jia-Ying Chen

18. a2 > b2 < Effects on b of changing a • Decreasing the significance level a, increases the value of b, and vice versa. a1 b1 m= 170 m=180 Jia-Ying Chen

19. Judging the Test • A hypothesis test is effectively defined by the significance level a and by the sample size n. • If the probability of a Type II error b is judged to be too large, we can reduce it by • increasing a, and/or • increasing the sample size. Jia-Ying Chen

20. By increasing the sample size the standard deviation of the sampling distribution of the mean decreases. Thus, decreases. Judging the Test • Increasing the sample size reduces b Jia-Ying Chen

21. m= 170 m=180 Judging the Test • Increasing the sample size reduces b Note what happens when n increases: a does not change, but b becomes smaller Jia-Ying Chen

22. Judging the Test • Power of a test • The power of a test is defined as 1 - b. • It represents the probability of rejecting the null hypothesis when it is false. Jia-Ying Chen

23. Example 4 • For a given sample size n, if the level of significance α is decreased, the power of the test will: • a.increase. • b.decrease. • c.remain the same. • d.Not enough information to tell. Jia-Ying Chen

24. Example 5 • During the last energy crisis, a government official claimed that the average car owner refills the tank when there is more than 3 gallons left. To check the claim, 10 cars were surveyed as they entered a gas station. The amount of gas remaining before refill was measured and recorded as follows (in gallons): 3, 5, 3, 2, 3, 3, 2, 6, 4, and 1. Assume that the amount of gas remaining in tanks is normally distributed with a standard deviation of 1 gallon. Compute the probability of a Type II error and the power of the test if the true average amount of gas remaining in tanks is 3.5 gallons. (α=0.05) Jia-Ying Chen

25. Solution • Rejection region:z>zα • β = P( < 3.52 given that μ = 3.5) = P(z < 0.06) = 0.5239 • Power = 1 - β = 0.4761 Jia-Ying Chen