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 12 students, the average time was 5.04 minutes and the standard deviation was 0.96 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.
Accepted Solution
A:
Answer with explanation:As per given , we have[tex]\overline{x}=5.04[/tex][tex]s=0.96[/tex]n=12df = 12-1=11Since population standard deviation is unknown , so we use t-test.Critical t-value for 99% confidence (1% significance): [tex]t_{\alpha/2, df}=t_{0.005,12}=3.1058[/tex]Confidence interval for population mean :_[tex]\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}\\\\=5.04\pm (3.1058)\dfrac{0.96}{\sqrt{12}}\\\\\approx5.04\pm0.861\\\\= (5.04-0.861,\ 5.04+0.861)\\\\=(4.179,\ 5.901)[/tex]Hence, the 99% confidence interval for the mean amount of time that students spend in the shower each day. Assume normality.= [tex](4.179,\ 5.901)[/tex]a) Lower limit of the 99% interval = 4.179b) Upper limit of the 99% interval = 5.901