Question

a data set includes 103 body temperatures of healthy adult humans having a mean of 98.9°f and a standard deviation of 0.65°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: Determine the critical value

Since the sample size \(n = 103\) (so \(n>30\)) and the confidence level is \(99\%\) (\(\alpha=1 - 0.99=0.01\), \(\alpha/2=0.005\)). Using the standard normal distribution table, the critical value \(z_{\alpha/2}=z_{0.005} = 2.576\)

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a confidence interval for the population mean when the population standard deviation \(\sigma\) is unknown (we use the sample standard deviation \(s\) as an estimate) is \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\)
Given \(s = 0.65\), \(n = 103\), \(z_{\alpha/2}=2.576\)

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Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
Given \(\bar{x}=98.9\)

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Answer:

\(98.735^{\circ}F<\mu<99.065^{\circ}F\)