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Unformatted text preview: Problem Set 2 ECON 231W Spring 2010 Suggested Solutions 1. In the setup of the problem, we are told that Y is distributed as a binomial with parameter p (i.e., for a given p, Y has a binomial distribution). We then know that E ( Y  p ) = np, where n is the number of draws (or size of the sample in this case), which is given here as n = 10 . We are also told that p is distributed uniformly on the interval , 1 4 so that E ( p ) = 0+ 1 4 2 = 1 8 . Now, by the Law of Iterated Expectations, we know that E ( Y ) = E [ E ( Y  p )] . Therefore, E ( Y ) = E [ E ( Y  p )] = E ( np ) = nE ( p ) = n ( 1 8 ) = n 8 . Since n = 10 ,E ( Y ) = 10 8 = 1 . 25 2. (a) New Jersey sample size n 1 = 100, sample average ¯ Y 1 = 58, sample standard deviation s 1 = 8. The standard error of ¯ Y 1 is 8 √ 100 = 0 . 8. The 95% confidence interval for the mean score of all New Jersey third graders is: μ 1 = ¯ Y 1 ± 1 . 96 · s 1 √ n 1 = 58 ± 1 . 96 · . 8 = (56 . 432 , 59 . 568) (b) Iowa sample size n 2 = 200, sample average ¯ Y 2 = 62, sample standard deviation s 2 = 11. The standard error of ¯ Y 1 ¯ Y 2 is q s 2 1 n 1 + s 2 2 n 2 = q 64 100 + 121 200 ≈ 1 . 1158. The 90% confidence interval for the difference in mean score between the two states is: μ 1 μ 2 = (58 62) ± 1 . 64 · 1 . 1158 = ( 5 . 8299 , 2 . 1701) (c) The hypothesis tests for the difference in mean scores is:...
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This note was uploaded on 04/19/2010 for the course ECO 231W taught by Professor Joshuakinsler during the Spring '10 term at Rochester.
 Spring '10
 JoshuaKinsler

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