Question

a data set includes 109 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? 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 \(n = 109\) (large, \(n>30\)), we can use the standard normal distribution. 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 \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\)
Given \(s = 0.62\), \(n = 109\), and \(z_{\alpha/2}=2.576\)
\(E=2.576\times\frac{0.62}{\sqrt{109}}\)
First, \(\sqrt{109}\approx10.440\)
\(E = 2.576\times\frac{0.62}{10.440}\)
\(E=2.576\times0.0594\)
\(E\approx0.153\)

Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
Given \(\bar{x}=98.2\)
\(98.2-0.153<\mu<98.2 + 0.153\)

Answer:

\(98.047^{\circ}F<\mu<98.353^{\circ}F\)