Statistics Homework Solutions 160

Statistics Homework Solutions 160 - 200 would be greater...

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160 CHAPTER 6 Therefore the level is 0.4364. (e) Let n be the required number of tires. The null distribution is normal with μ 50 000 and σ X 5000 n . The alternate distribution is normal with μ 49 500 and σ X 5000 n . Let x 0 denote the boundary of the rejection region. Since the level is 5%, the z -score of x 0 is z 1 645 under the null distribution. Therefore x 0 50 000 1 645 5000 n . Since the power is 0.80, the z -score of x 0 is z 0 84 under the alternate distribution. Therefore x 0 49 500 0 84 5000 n . It follows that 50 000 1 645 5000 n 49 500 0 84 5000 n . Solving for n yields n 618. 7. (ii). Since 7 is farther from the null mean of 10 than 8 is, the power against the alternative μ 7 will be greater than the power against the alternative μ 8. 9. (a) Two-tailed. The alternate hypothesis is of the form p p 0 . (b) p 0 5 (c) p 0 4 (d) Less than 0.7. The power for a sample size of 150 is 0.691332, and the power for a smaller sample size of 100 would be less than this. (e) Greater than 0.6. The power for a sample size of 150 is 0.691332, and the power for a larger sample size of
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Unformatted text preview: 200 would be greater than this. (f) Greater than 0.65. The power against the alternative p 4 is 0.691332, and the alternative p 3 is farther from the null than p 4. So the power against the alternative p 3 is greater than 0.691332. (g) Its impossible to tell from the output. The power against the alternative p 45 will be less than the power against p 4, which is 0.691332. But we cannot tell without calculating whether the power will be less than 0.65. 11. (a) Two-tailed. The alternate hypothesis is of the form 1 2 . (b) Less than 0.9. The sample size of 60 is the smallest that will produce power greater than or equal to the target power of 0.9....
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