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Unformatted text preview: Solution to Homework 9 (STAT515, Fall 2011) 9.10. ( 28 points, 4 points/part ) Given information: n 1 = n 2 = 100, ¯ x 1 = 70, ¯ x 2 = 50, σ 2 1 = 100, σ 2 2 = 64. a. σ X 1 X 2 = s σ 2 1 n 1 + σ 2 2 n 2 = r 100 100 + 64 100 ≈ 1 . 2806 . b. (You may sketch a normal pdf curve that centers at 5. You may also mark some points on the horizontal axis that correspond to one (two, three) standard deviation(s) below/above 5. The normality is a result of Central Limit Theorem since both sample sizes are large.) c. Note that the standard error of X 1 X 2 from part a is very small. With the observed sample mean difference ¯ x 1 ¯ x 2 = 20, the corresponding zscore (to be computed in part e ) is very extreme (higher than ten!). This suggests that the observed data strongly contradict with the null hypothesis. d. The rejection region for a twosided test with α = 0 . 05 based on large samples is (∞ , z . 025 ) ∪ ( z . 025 , + ∞ ) = (∞ , 1 . 96) ∪ (1 . 96 , + ∞ ). e. The test statistic is z = 70 50 5 1 . 2806 ≈ 11 . 71 . Because the test statistic falls in the rejection region (in part d ), we reject H at significance level 0.05. That is, the observed data provide sufficient evidence that the difference between two population means in not 5. f. A 95% confidence interval for μ 1 μ 2 is 20 ± 1 . 96 × 1 . 2806 ≈ (17 . 49 , 22 . 51) . Based on this data set, one can be 95% confident that the first population mean is higher than the second population mean by the amount that lies between 17.49 and 22.51. g. I think that the confidence interval provides more information about μ 1 μ 2 because now we have a plausible range of the true value of μ 1 μ 2 . The test of hypothesis simply suggests that μ 1 μ 2 is not 5, and does not provide further indication of what μ 1 μ 2 is likely to be....
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This note was uploaded on 02/21/2012 for the course STAT 515 taught by Professor Zhao during the Fall '10 term at South Carolina.
 Fall '10
 Zhao

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