Ch 11 實習 (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.

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)?

A Two - Tail Test Solution H0: m = 17.09 Define the rejection region Is the mean different from 17.09? H0: m = 17.09 Define the rejection region

A Two – Tail Test Solution - continued a/2 = 0.025 a/2 = 0.025 17.09 We want this erroneous rejection of H0 to be a rare event, say 5% chance. 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 From the sample we have: 17.55 a/2 = 0.025 17.09 a/2 = 0.025 -za/2 = -1.96 za/2 = 1.96 Rejection region

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 -za/2 = -1.96 za/2 = 1.96

Example 1 A random sample of 100 observations from a normal population whose standard deviation is 50 produced a mean of 75. Does this statistic provide sufficient evidence at the 5% level of significance to infer that the population mean is not 80?

Solution H0: μ=80 vs. H1: μ ≠80 Rejection region: |z| > z0.025=1.96 Test statistic: z = (75-80)/(50/10)=-1.0 Conclusion: Don’t reject . No sufficient evidence at the 5% level of significance to infer that the population mean is not 80.

Example 2 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

Solution

Testing hypotheses and intervals estimators Interval estimators can be used to test hypotheses. Calculate the 1 - a confidence level interval estimator, then if the hypothesized parameter value falls within the interval, do not reject the null hypothesis if the hypothesized parameter value falls outside the interval, conclude that the null hypothesis can be rejected (m is not equal to the hypothesized value).

Drawbacks Two-tail interval estimators may not provide the right answer to the question posed in one-tail hypothesis tests. The interval estimator does not yield a p-value.

Example 3 Using a confidence interval when conducting a two-tail test for m, we do not reject H0 if the hypothesized value for m: a. is to the left of the lower confidence limit (LCL). b. is to the right of the upper confidence limit (UCL). c. falls between the LCL and UCL. d. falls in the rejection region.

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

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) H0: m = 170 H1: m = 180 a=.05 m= 170 m=180 Specify the alternative value under H1. Do not reject H0

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

Calculation of the Probability of a Type II Error H0: m = 170 H1: m = 180 m=180 m= 170

Example 4 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

Solution Rejection region: z>zα

Example 5 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

Solution

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

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.

Judging the Test Increasing the sample size reduces b 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 Note what happens when n increases: a does not change, but b becomes smaller m=180 m= 170

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.

Example 6 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.

Example 7 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)

Solution H0: μ=3 H1: μ>3 Rejection region:z>zα β = P( < 3.52 given that μ = 3.5) = P(z < 0.06) = 0.5239 Power = 1 - β = 0.4761