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 \(n = 103\) (large, \(n>30\)), we can use the standard normal distribution (z - distribution) for the critical value. For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\).
From the standard - normal distribution table, \(z_{\alpha/2}=z_{0.005} = 2.576\)

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(s = 0.65\), \(n = 103\)

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Step3: Calculate the confidence interval

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

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Answer:

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