\( 1 \leftarrow \quad \begin{array}{l}\text { A data set includes } 106 \text { body temperatures of healthy adult humans having a mean of } 98.9^{\circ} \mathrm{F} \text { and a standard deviation of } 0.62^{\circ} \mathrm{F} \text {. Construct a } 99 \% \text { confidence interval estimate of the mea } \\ \text { healthy humans. What does the sample suggest about the use of } 98.6^{\circ} \mathrm{F} \text { as the mean body temperature? } \\ \text { Click here to view a t distribution table. } \\ \text { Click here to view page } 1 \text { of the standard normal distribution table. } \\ \text { Click here to view page } 2 \text { of the standard normal distribution table. } \\ \text { What is the confidence interval estimate of the population mean } \mu \text { ? } \\ \square^{\circ} \mathrm{F}<\mu<\square^{\circ} \mathrm{F} \\ \text { (Round to three decimal places as needed.) }\end{array} \) ?

We are given a sample of \( n = 106 \) healthy adult body temperatures with a sample mean of \[ \bar{x} = 98.9^\circ \mathrm{F} \] and a sample standard deviation of \[ s = 0.62^\circ \mathrm{F}. \] Since the population standard deviation is unknown and the sample size is moderate, we use the \( t \)-distribution to construct a \( 99\% \) confidence interval for the population mean \( \mu \). 1. Define the confidence interval formula using the \( t \)-distribution: \[ \bar{x} \pm t_{\alpha/2,\,n-1} \cdot \frac{s}{\sqrt{n}}. \] 2. For a \( 99\% \) confidence level, the significance level is \[ \alpha = 0.01 \quad \text{and} \quad \frac{\alpha}{2} = 0.005. \] With degrees of freedom \( n-1 = 105 \), from a \( t \)-table we find an approximate critical value \[ t_{0.005,\,105} \approx 2.63. \] 3. Compute the standard error (SE): \[ \text{SE} = \frac{s}{\sqrt{n}} = \frac{0.62}{\sqrt{106}} \approx \frac{0.62}{10.2956} \approx 0.0602. \] 4. Compute the margin of error: \[ \text{Margin} = t_{\alpha/2,\,n-1} \times \text{SE} \approx 2.63 \times 0.0602 \approx 0.158. \] 5. Construct the \( 99\% \) confidence interval: \[ \bar{x} \pm 0.158 \quad \Rightarrow \quad 98.9 - 0.158 < \mu < 98.9 + 0.158, \] which gives \[ 98.742^\circ \mathrm{F} < \mu < 99.058^\circ \mathrm{F}. \] 6. Interpretation regarding \( 98.6^\circ \mathrm{F} \): Notice that the entire \( 99\% \) confidence interval is above \( 98.6^\circ \mathrm{F} \). This suggests that the true mean body temperature for healthy adults is very likely higher than \( 98.6^\circ \mathrm{F} \), calling into question the use of \( 98.6^\circ \mathrm{F} \) as the average body temperature. Thus, the \( 99\% \) confidence interval for the population mean is \[ 98.742^\circ \mathrm{F} < \mu < 99.058^\circ \mathrm{F}, \] and the sample suggests that the typical body temperature of healthy adults is greater than \( 98.6^\circ \mathrm{F} \).