Question

a data set includes 108 body temperatures of healthy adult humans having a mean of 98.2°f and a standard deviation of 0.64°f. construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. what does the sample suggest about the use of 98.6°f as the mean body temperature? click here to view a t distribution table. click here to view page 1 of the standard normal distribution table. click here to view page 2 of the standard normal distribution table. what is the confidence interval estimate of the population mean μ? □°f<μ<□°f (round to three decimal places as needed.)

Explanation:

Step1: Find the critical value

For a 99% confidence interval with \(n = 108\) (so degrees of freedom \(df=n - 1=107\)). Using the t - distribution table (or a calculator), the critical value \(t_{\alpha/2}\) for \(\alpha=1 - 0.99 = 0.01\) and \(df = 107\) is approximately \(t_{0.005}= 2.623\) (using a calculator or more precise t - tables).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\alpha/2}\times\frac{s}{\sqrt{n}}\). Given \(s = 0.64\), \(n = 108\), and \(t_{\alpha/2}=2.623\).

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Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\). Given \(\bar{x}=98.2\)

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Answer:

\(98.038^{\circ}F<\mu<98.362^{\circ}F\)