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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.

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JeanaShupp

Answer: [tex](4.0845,\ 6.5755)[/tex]

Step-by-step explanation:

As per given , we have

n = 11

[tex]\overline{x}=5.33\\\\ s=1.33[/tex]

Since population standard deviation is missing, so we use t-test.

Critical t-value for 99% confidence :

[tex]t_{\alpha/2,\ n-1}=3.106[/tex]   [using two-tailed t-value table]

Confidence interval :

[tex]\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}\\\\= 5.33\pm (3.1060)\dfrac{1.33}{\sqrt{11}}\\\\\approx5.33\pm1.2455\\\\=(5.33-1.2455,\ 5.33+1.2455)\\\\=(4.0845,\ 6.5755)[/tex]

Hence,  99% confidence interval for the mean amount of time that students spend in the shower each day.= [tex](4.0845,\ 6.5755)[/tex]

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