Report

Ch 11 實習 (2) A Two - Tail Test Example 11.2 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 longdistance bills for all AT&T residential customers are $17.09 and $3.87 respectively. Jia-Ying Chen A Two - Tail Test Example 11.2 - continued 3 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 A Two - Tail Test Solution Is the mean different from 17.09? H0: m = 17.09 H1 : m 17.09 – Define the rejection region z z / 2 or z z / 2 4 Jia-Ying Chen A Two – Tail Test Solution - continued /2 = 0.025 x /2 = 0.025 17.09 If H0 is true (m =17.09), x can still fall far above or far below 17.09, in which case we erroneously reject H0 in favor of H1 (m 17.09) 5 x We want this erroneous rejection of H0 to be a rare event, say 5% chance. Jia-Ying Chen A Two – Tail Test Solution - continued z= /2 = 0.025 xm = n 17.55 17.09 = 1.19 3.87 100 17.55 x 17.09 x From the sample we have: /2 = 0.025 /2 = 0.025 /2 = 0.025 x = 17.55 -z/2 = -1.96 6 0 z/2 = 1.96 Rejection region Jia-Ying Chen 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 /2 = 0.025 z= 7 xm n /2 = 0.025 -1.19 0 1.19 = 17.55 17.09 = 1.19 -z/2 = -1.96 z/2 = 1.96 3.87 100 Jia-Ying Chen Example 1 8 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 Solution 9 Jia-Ying Chen Calculation of the Probability of a Type II Error To calculate Type II error we need to… 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 10 express the rejection region directly, in terms of the parameter hypothesized (not standardized). specify the alternative value under H1. 單尾 雙尾 Let us revisit Example 11.1 Jia-Ying Chen 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 x 175.34 with = .05. Let the alternative value be m = 180 (rather than just m>170) H : m = 170 0 H1: m = 180 Do not reject H0 =.05 m= 170 xL = Specify the alternative value under H1. m=180 175.34 11 Jia-Ying Chen Calculation of the Probability of a Type II Error A Type II error occurs when a false H0 is not rejected. H0: m = 170 A false H0… …is not rejected H1: m = 180 x 175.34 m= 170 xL = =.05 m=180 175.34 12 Jia-Ying Chen Calculation of the Probability of a Type II Error = P( x 175.34 given that H 0 is false) = P( x 175.34 given that m = 180) = P( z 175.34 180 65 400 ) = .0764 H0: m = 170 H1: m = 180 m= 170 xL = m=180 175.34 13 Jia-Ying Chen Example 2 14 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 Solution 15 Rejection region: z>zα Jia-Ying Chen Example 3 16 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 Solution 17 Jia-Ying Chen Effects on of changing Decreasing the significance level , increases the value of , and vice versa. 2 < 1 m= 170 18 2 > 1 m=180 Jia-Ying Chen Judging the Test A hypothesis test is effectively defined by the significance level and by the sample size n. If the probability of a Type II error is judged to be too large, we can reduce it by 19 increasing , and/or increasing the sample size. Jia-Ying Chen Judging the Test Increasing the sample size reduces xL m Recall : z = , thus xL = m z n n By increasing the sample size the standard deviation of the sampling distribution of the mean decreases. Thus, x Ldecreases. 20 Jia-Ying Chen Judging the Test Increasing the sample size reduces xL m Recall : z = , thus xL = m z n n Note what happens when n increases: does not change, but becomes smaller 21 m= 170 xxxLxLLxLxLL m=180 Jia-Ying Chen Judging the Test Power of a test 22 The power of a test is defined as 1 - . It represents the probability of rejecting the null hypothesis when it is false. Jia-Ying Chen Example 4 23 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 Example 5 24 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 Solution Rejection region:z>zα x 3 z0.05 = 1.645 = x 3.52 1 10 25 β = P( x < 3.52 given that μ = 3.5) = P(z < 0.06) = 0.5239 Power = 1 - β = 0.4761 Jia-Ying Chen