Question

a data set includes 106 body temperatures of healthy adult humans having a mean of 98.9°f and a standard deviation of 0.62°f. construct a 99% confidence inter of the mean body temperature of all healthy humans. 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 \(n = 106\) (so \(df=n - 1=105\)), using a t - distribution table or calculator, the critical value \(t_{\alpha/2}\) is approximately \(t_{0.005}\approx 2.624\) (since \(\alpha=1 - 0.99 = 0.01\) and \(\alpha/2=0.005\)).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(s = 0.62\), \(n = 106\), and \(t_{\alpha/2}=2.624\).

$$E=2.624\times\frac{0.62}{\sqrt{106}}$$

$$E=2.624\times\frac{0.62}{10.2956}$$

$$E=2.624\times0.0602$$

$$E\approx0.158$$

Step3: Construct the confidence interval

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

$$98.9 - 0.158<\mu<98.9+0.158$$

Answer:

\(98.742^{\circ}F<\mu<99.058^{\circ}F\)