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Unformatted text preview: Comparing two populations or processes – Examples (with solutions) 1. A company produces special running shoes. In their advertisement, they claim that using their shoes will make runners faster. Each of the 85 members of the football team at Eastern Michigan University ran the 40 yard dash with their current shoes and ran the 40 yard dash with the special running shoes. Here is a summary of the results: average time with current shoes 5.20 seconds average time with special shoes 5.29 seconds average difference (current - special) in times-0.09 seconds SD of difference 0.70 seconds Let μ 1 be the mean of the distribution of 40-yard dash times with the current shoes and let μ 2 be the mean of the distribution of 40-yard dash times with the special shoes. Define μ = μ 1- μ 2 . Is there statistical evidence at the 10% significance level that μ > 0? (a) State H and H a Solution: H : μ = 0 H a : μ > (b) Just looking at the observed sample mean of the differences, explain why we can immediately fail to reject H at the 10% level. Solution: Since the observed sample mean of differences (current - special) ¯ x =- . 09 is negative, there is no way we can find statistical evidence that μ , the mean of differences, is positive. The p-value we calculate for this alternative will be at least 50%, and thus cannot be less than 10%. (c) What assumptions must be made to conduct this hypothesis test. Solution: We must assume that the measured differences (current - special) of 40-yard dash times, x 1 ,...,x 85 are a realization of a random sample from N ( μ,σ ). (d) Compute the rejection region for this test. Solution: The rejection region is t > t 1- α,n- 1 = t . 90 , 84 = qt (0 . 90 , 84) = 1 . 291711. That is we reject H if t > 1 . 291711. 1 (e) Calculate the test statistic realization. Solution: t = ¯ x- s/ √ n =- . 09- . 70 / √ 85 =- 1 . 18537 (f) What is the distribution of the test statistic when H is true. Specify the value for its parameter. Solution: The test statistic T has the t-distribution with n- 1 = 85- 1 = 84 degrees of freedom when H is true. (g) Compute the p-value for this test and state the conclusion. Solution: The p-value for this right sided alternative is, p-value = P ( T ≥ t ) = P ( T ≥ - 1 . 18537) = 1- pt (- 1 . 18537 , 84) = 0 . 8803931 where T has the t-distribution with 84 degrees of freedom. Since the p-value is greater than 10% we fail to reject H . Notice that this p-value is large because the test statistic is negative and the alternative hypothesis is right sided.the test statistic is negative and the alternative hypothesis is right sided....
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This note was uploaded on 05/06/2011 for the course STAT 5021 taught by Professor Staff during the Spring '08 term at Minnesota.
- Spring '08