-12 points My Notes A university with a high water bill is interested in estimat
ID: 3065126 • Letter: #
Question
-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. I
Explanation / Answer
Solution:
Let be X: amount of time that students spend in the shower each day
n = 11
x = 5.33
s = 1.33
Since the sample is small, we use a t interval. Using the StatCrunch t calculator, we find t0.005,10 = 3.1693.
So the lower limit is 5.33-3.1693 * 1.33/sqrt(11) = 4.0591
Upper limit is 5.33+3.1693 * 1.33/sqrt(11) = 6.6009
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