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Unformatted text preview: Solutions to Final Review Problems 1. This is a paired data problem. The differenced data (first garage minus second garage) is 2 7 16 32 24 14 2 9 This differenced data has mean 7 . 5 and variance 248 . 5714. Test H : μ D = 0 H a : μ D 6 = 0 where μ D is the true mean of the difference population. Test statistic: t = X μ s/ √ n = 7 . 5 q 248 . 5714 / 8 ≈ 1 . 35 The rejection region, based on α = 0 . 01 and n 1 = 7 df, is the twotailed area under the t (7) curve to the right of 3 . 50 and to the left of 3 . 50. Since our test statistic is not in this rejection region, we fail to reject the null hypothesis in favor of the alternative at 0 . 01 level of significance. That is, there is not significant evidence to suggest that the true mean minor repair costs differ for the two garages. 2. This is a two sample Ztest. The mean difference is: 3975 . 2795 . 0 = 1180 . (Maybe you decided to go the other way and you got 1180. This is okay too.) The variance of the difference is: s 2 1 n 1 + s 2 2 n 2 = 245 . 1 2 40 + 293 . 7 2 40 = 3658 . 342 The Z critical value for a twosided 99% confidence interval is 2 . 575. The answer is then 1180 ± (2 . 575) √ 3658 . 342 (1024 . 253 , 1335 . 747) Since this interval does not contain zero, it provides evidence that the true average durability index differs for the two fabrics. 3. This is another paired ttest. The differenced data (before minus after) is 2 8 3 7 12 This differenced data has mean 5 . 2 and variance 33 . 7. Test H : μ D = 0 H a : μ D > where μ D is the true mean of the differenced population “before minus after”....
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This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.
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