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Unformatted text preview: The Hong Kong University of Science and Technology IS O M 111  Business statistics Problem Sheet 5 Solutions 1. H 0 : μ 1 μ 2 = 0 H 1 : μ 1 μ 2 ≠ 0 Test statistic t =4.19, And t α /2, n1 + n2 – 2 = t 0.005, 9 = 3.25. Since t > t 0.005, 9 , reject H at 0.05 level of significance. 2. H 0 : μ = $33.99 H 1 : μ > $33.99 Test statistic . . 2 = z pvalue = P( z > 2) = 0.0228. Reject H if α ≥ pvalue. Hence, the range is 1 ≥ α ≥ 0.0228. 3. (a) H 0 : μ d = 0 H 1 : μ d > 0 Test statistic t = . 95 . 1 And t α , n– 1 = t 0.025, 7 = 2.365. Since t < t 0.025, 7 , we cannot reject H . (b) As t 0.025, 7 = 2.365 & t 0.05, 7 = 1.895 ⇒ 0.025 < pvalue < 0.05. Hence, the lower bound on the pvalue is 0.025 4. (a) H 0 : p = 0.5 H 1 : p > 0.5 Test statistic z =1. And z α = z 0.05 = 1.645. Since z < z 0.05 , cannot reject H at 0.05 level of significance. (b) Let μ 1 be the population average monthly salary among those with one job offer. Let μ 2 be the population average monthly salary among those with three jobs offer. H 0 : μ 1 μ 2 = 0 H 1 : μ 1 μ 2 ≠ 0 Test statistic z = . 1 . 2 − pvalue = 2 P( z > 2.1) = 2(0.0179) = 0.0358. Since pvalue < α = 0.05, reject H at 0.05 level of significance. 5. (a) Let μ 1 be the mean salary per hours of male factory workers. Let μ 2 be the mean salary per hours of female factory workers. H 0 : μ 1 μ 2 = 1.5 H 1 : μ 1 μ 2 > 1.5  1  Test statistic t = 2 1 2 1 2 1 1 1 ) ( ) ( n n s x x + − − − μ μ , where s = ) 2 ( ) 1 ( ) 1 ( 2 1 2 2 2 2 1 1 − + − + − n n S n S n = 2319 . 6 2 12 11 ) 6 . 6 )( 1 12 ( ) 8 . 5 )( 1 11 ( 2 2 = − + − + − Then t = 8836 . 1 12 1 11 1 2319 . 6 5 . 1 ) 25 . 35 65 . 41 ( = + − − . And t α , n1 + n2 – 2 = t 0.05, 21 = 1.721. Since t > t 0.05, 21 , reject H . Therefore, it is reasonable to conclude that male factory workers earn over $1.50 more per hour on average than female factory workers do. (b) Let d be the required difference. Then t = 2 1 1 1 5 . 1 n n s d + − = t 0.01, 21 = 2.518 ⇒ d = 8.0502 (c) A 90% C.I. for μ 1 μ 2 is ( x 1 x 2 ) ± t α /2, n1 + n2 – 2 s 2 1 1 1 n n + , where t 0.05, 21 =1.721 90% C.I. is 6.4 ± (1.721)(2.6013) = (1.9231, 10.8769) Hence, we have 90% chance for this interval to enclose the true difference of the means μ 1 μ 2 . (d) H 0 : μ 1 μ 2 = 0 H 1 : μ 1 μ 2 ≠ 0 Test statistic z = 3878 . 2 40 73400 50 74600 580750 618200 ) ( ) ( 2 2 2 2 2 1 2 1 2 1 2 1 = + − = + − − − n S n S x x μ μ . And z α /2 = z 0.01 = 2.33. Since z > z 0.01 , we reject H at 0.02 level of significance. We have sufficient evidence to conclude that male and female brokers have unequal mean salaries. pvalue = 2P( z > 2.3878) = 2(0.0084) = 0.0168 (e) P(committing Type I error) = 0.02 The true difference when Type I error is made = 0....
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This note was uploaded on 09/18/2010 for the course BBA ISOM111 taught by Professor Hu during the Fall '08 term at HKUST.
 Fall '08
 HU
 Business

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