HW4 - 36-226 HOMEWORK 4 Due Friday at 9:30am in class...

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36-226 HOMEWORK 4 Due: Friday 2/12/16 at 9:30am in class Remember to show all work ; just writing down the answer will not receive credit. 1. We’ve seen that ˆ μ = ¯ Y and ˆ p = Y n are unbiased estimates of μ, p respectively with V ar μ ] = σ 2 n , V ar p ] = p (1 - p ) n . For large samples, ˆ μ, ˆ p are approx. normally distributed. An Atlantic City tourism group hires you to help them determine visitor satisfaction. They randomly sample people on vacation and ask them questions about their vacation cost and whether or not they would return for another visit. You’re given summary information on n = 200 subjects: on average, they spent $785 with an estimated variance of s 2 = $272 , 300; 63% would return for another trip. (a) Calculate and interpret 95% confidence intervals for the true average vacation cost amount and the true percentage of potential return visitors. It has been hypothesized that the true average vacation cost is $740. Do you agree? Why/why not? (b) The tourism group contacts you to tell you that there was a typo in the information they sent. It was really n = 2000 subjects (all other numbers were correct). Recalculate your confidence intervals. Does the new interval change your decision about the hypothesized true average vacation cost of $740? Why/why not? (c) You get another apologetic email from the tourism group saying that the correct number of subjects is n = 20 (they’ve now fired their intern). The other information is still correct. Recalculate your confidence intervals. What is your final decision regarding the hypothesized true average cost of $740? (d) With respect to return visitors, the tourism group interested in re-estimating the true percentage within 5% with probability 90%. How many people would they need to sample to get the desired level of precision? (e) You’re then informed that budget constraints indicate that the tourism group can only afford to sample 100 people in this new survey. Your boss is obsessed with finding a 95% interval. What margin of error can you offer using 100 people? Someone else points out that the CNN polls often use +/- 3%. What probability percentage can you get with 100 people and 3% margin of error?
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