This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Economics 203 Midterm 1 Solutions — Spring 2009 — Form B 1. C. For 95% confidence, we use z . 025 = 1 . 960. Using the fact that we have a prior estimate of the response rate (0.28) we calculate n = 1 . 960 p . 28(0 . 72) . 05 ! 2 = . 880038 . 05 2 = 17 . 60076 2 = 309 . 787 We round up to 310. 2. B. A 98% confidence interval would not include 10, while a 99% confidence interval would include 10. In general if we reject a null hypothesis in favor of its twosided alternative, if we obtain a pvalue of γ , then the 100(1 γ )% confidence interval will have the hypothe sized value as one of its endpoints. A smaller confidence interval will then not include the hypothesized value, while a larger one will include it. 3. D. For 90% confidence, we use z . 05 = 1 . 645. Using the fact that we have a prior estimate of the standard deviation ($12000) we calculate n = 1 . 645($12000 $3000 2 = 19740 3000 2 = 6 . 58 2 = 43 . 2964 We round up to 44. 4. C. Your point estimate of 2.96 lies above the hypothesized value. You have a onesided alternative, with the rejection region on the opposite side of the hypothesized value to your point estimate. The chance of getting a sample mean this “extreme” given that the null hypothesis is true is greater than 50%; so the pvalue is greater than 0.50. It cannot be greater than 1. 5. E. Calculate t = 189 . 75 180 25 / √ 36 = 2 . 340. 6. D. The tdistribution is appropriate here because we have a sample standard deviation. The test statistic calculated in the previous question was 2.34, which lies between t . 01 , 35 = 2 . 438 and t . 025 , 35 = 2 . 030, so the pvalue is between 0.01 and 0.025. 7. E. You would like to prove H 1 : μ 1 μ 2 > 0, however, your sample data suggests nothing of the fact and that the sample mean for golfers is in fact less than the sample mean for nongolfers. The pvalue then has to be greater than .5. To find the appropriate value, take the test statistic of 2.091 and go to the far right end of the tdistribution. Excel assumes you are performing appropriate hypothesis tests and is in fact going from 2.091 to negative infinity and finds that area as .019. The correct pvalue then is 1.019=.981....
View
Full
Document
This note was uploaded on 07/19/2010 for the course ECON ECON203 taught by Professor Petry during the Spring '10 term at University of Illinois at Urbana–Champaign.
 Spring '10
 Petry
 Economics

Click to edit the document details