ps5sol_F08

ps5sol_F08 - The Hong Kong University of Science and...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 p-value = P( z > 2) = 0.0228. Reject H if α ≥ p-value. 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 < p-value < 0.05. Hence, the lower bound on the p-value 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 − p-value = 2 P( z > 2.1) = 2(0.0179) = 0.0358. Since p-value < 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. p-value = 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....
View Full Document

This note was uploaded on 09/18/2010 for the course BBA ISOM111 taught by Professor Hu during the Fall '08 term at HKUST.

Page1 / 12

ps5sol_F08 - The Hong Kong University of Science and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online