Question

a data set includes 104 body temperatures of healthy adult humans having a mean of 98.7°f and a standard deviation of 0.66°f. construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. what is the confidence interval estimate of the population mean μ? \\(\square ^\circ\text{f} < \mu < \square ^\circ\text{f}\\) (round to three decimal places as needed.)

Explanation:

Step1: Identify the formula for confidence interval

For a confidence interval of the population mean when the population standard deviation \(\sigma\) is unknown (we use sample standard deviation \(s\) instead) and the sample size \(n\) is large (\(n \geq 30\)), we use the formula: \(\bar{x} \pm z_{\alpha/2} \frac{s}{\sqrt{n}}\)

Here, \(\bar{x} = 98.7^{\circ}\text{F}\), \(s = 0.66^{\circ}\text{F}\), \(n = 104\), and for 99% confidence interval, \(\alpha=1 - 0.99 = 0.01\), so \(\alpha/2=0.005\). The critical value \(z_{0.005}\) (from standard normal table) is approximately \(2.576\).

Step2: Calculate the margin of error (E)

\(E = z_{\alpha/2} \frac{s}{\sqrt{n}}\)

Substitute the values: \(z_{\alpha/2}=2.576\), \(s = 0.66\), \(n = 104\)

First, calculate \(\sqrt{n}=\sqrt{104}\approx10.198\)

Then, \(\frac{s}{\sqrt{n}}=\frac{0.66}{10.198}\approx0.0647\)

Then, \(E = 2.576\times0.0647\approx0.1667\)

Step3: Calculate the confidence interval

Lower limit: \(\bar{x}-E = 98.7 - 0.1667\approx98.533\)

Upper limit: \(\bar{x}+E = 98.7 + 0.1667\approx98.867\)

Answer:

\(98.533^{\circ}\text{F}<\mu< 98.867^{\circ}\text{F}\)