Question

a data set includes 103 body temperatures of healthy adult humans having a mean of 98.2°f and a standard deviation of 0.62°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?
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 \(z_{\alpha/2}\)

For a \(99\%\) confidence interval, \(\alpha = 1 - 0.99=0.01\), 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 \(E\)

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)
Given \(n = 109\), \(\sigma=0.62\), and \(z_{\alpha/2}=2.576\)
\(E=2.576\times\frac{0.62}{\sqrt{109}}\)
First, \(\sqrt{109}\approx10.44\)
\(E = 2.576\times\frac{0.62}{10.44}\)
\(E=2.576\times0.0594\approx0.153\)

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is given by \(\bar{x}-E<\mu <\bar{x} + E\)
Given \(\bar{x}=98.2\)
\(\bar{x}-E=98.2 - 0.153=98.047\)
\(\bar{x}+E=98.2+0.153 = 98.353\)

Answer:

\(98.047<\mu<98.353\)