Spring09Midterm1BSols

# Spring09Midterm1BSols - Economics 203 Midterm 1 Solutions...

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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 two-sided alternative, if we obtain a p-value 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 one-sided 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 p-value 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 t-distribution 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 p-value 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 non-golfers. The p-value 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 t-distribution. 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 p-value then is 1-.019=.981....
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## 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.

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Spring09Midterm1BSols - Economics 203 Midterm 1 Solutions...

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