1. The Gap is considering putting a factory outlet store onMain Street and they
ID: 2954584 • Letter: 1
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1. The Gap is considering putting a factory outlet store onMain Street and they hire you to conduct some requiredresearch. You implement a survey and ask 25 students how muchthey believe they would spend at the store each year. The studentresponses have a mean of $200 and a standard deviation of$150. a. What is the 95% con?dence interval for (true) mean perstudent expenditure? b. Test the hypothesis at 5% signi?cance level that the truemean expenditure exceeds $240 versus it does not. Find thesmallest signi?cance level at which you can reject the null(i.e. the p value)valuebv value). c. Suppose you somehow knew that the true population mean perstudent expenditure was $262. What (approximately) is theprobability that you would see a sample mean of $200 or lowerif this was the case? 1. The Gap is considering putting a factory outlet store onMain Street and they hire you to conduct some requiredresearch. You implement a survey and ask 25 students how muchthey believe they would spend at the store each year. The studentresponses have a mean of $200 and a standard deviation of$150. a. What is the 95% con?dence interval for (true) mean perstudent expenditure? b. Test the hypothesis at 5% signi?cance level that the truemean expenditure exceeds $240 versus it does not. Find thesmallest signi?cance level at which you can reject the null(i.e. the p value)valuebv value). c. Suppose you somehow knew that the true population mean perstudent expenditure was $262. What (approximately) is theprobability that you would see a sample mean of $200 or lowerif this was the case?Explanation / Answer
Given n=25, xbar=200, =150 (a) =0.05, Z(0.025)=1.96 (check normal table) The 95% CI is xbar ± Z*/n --> 200 ± 1.96*150/sqrt(25) --> (141.2, 258.8) ------------------------------------------------------------------------------------------------------------------------ (b) The test hypothesis is Ho:240 The test statistic is Z=(xbar - )/(/n) =(200-240)/(150/sqrt(25)) =-1.33 The p-value is P(Z>-1.33)= 0.9082 > =0.05, we donot reject Ho. ----------------------------------------------------------------------------------------------------------------------- (c) The probability is P(xbar < 200)= P((xbar-)/(s/n)Related Questions
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