4. -12 points My Notes A university with a high water bill is interested in esti

Question

4. -12 points My Notes A university with a high water bill is interested in estimating the mean amount of time that students spend in the shower each day. In a sample of 11 students, the average time was 5.33 minutes and the standard deviation was 1.33 minutes. Using this sample information, construct a 99% confidence interval for the mean amount of time that students spend in the shower each day. Assume normality a) what is the lower limit of the 99% interval? Give your answer to three decimal places b) what is the upper limit of the 99% interval? Give your answer to three decimal places.

Explanation / Answer

mean is 5.33 and s is 1.33. thus SE is s/sqrt(N) which is 1.33/sqrt(11)=0.4

for 99%, the z value is 2.576

thus lower limit is mean-z*SE =5.33-2.576*0.4=4.2996

upper limit is mean+z*SE =5.33+2.576*0.4 =6.3604

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