Assignment 12 solutions - Fall 2009 Hansen Assignment 12...

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Unformatted text preview: Fall 2009 Hansen Assignment 12 Due 12/1/2009 9:10 AM 1. Problem 8.5 in Ross. : = 200 : < 200 =5 =8 We can find = 199.11 5 − = 8 199.11 − 200 = −0.502 5 = 5% level of significance: = . = −1.645 Since > −1.645, we fail to reject at the 5% level 10% level of significance: = . = −1.28 Since > −1.28, we fail to reject at the 10% level either 2. Problem 8.20 in Ross. : = 30 : < 30 unknown = 10 We can find = 26.4 and = = 3.5024 − = 10 26.4 − 30 = −3.25 3.5024 Could test at 5% level of significance: − , = − . , = −1.833 However, students might have also tested at other levels of significance: 1%, 5%, or 10% are all reasonable values. − . , = −2.821 − . , = −1.383 Since < −1.833, −2.821, − 1.383, we reject at the 5%, 1% and 10% levels. Based on the sample, we cannot believe the advertisement. Fall 2009 Hansen We are assuming that average gas mileage rating is normally distributed. 3. Problem 8.24 in Ross. : = 30 : < 30 unknown = 16 We can find = 28.25 and = 3.7907 = − = 16 28.25 − 30 = −1.8466 3.7907 5% level of significance: − , = − . , = −1.7531 Since < −1.7531, we reject at the 5% level. Based on the sample, the claim is unjustified. 4. Graduate students. = 14, = 21 = 31.64, = 24.38 = 11.44, = 12.488 = = 146.07 = a) , : Since : b) , Since c) , : Since = = vs. . , : − 1 + 1 = vs. . , : 1 1 146.07 ∗ 14 + 21 at the 5% level. ≠ , 10% significance = 1.6924 > 1.6924 we reject = 31.64 − 24.38 ≠ , 5% significance = 2.0345 < 2.0345 we cannot reject = = at the 10% level. :> , 5% significance = . , = 1.6924 > 1.6924 we reject at the 5% level. vs. = 1.741 ...
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This note was uploaded on 09/19/2011 for the course CE 93 taught by Professor Hansen during the Fall '10 term at University of California, Berkeley.

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