\( 1 \leftarrow \quad \begin{array}{l}\text { A data set includes } 106 \text { body temperatures of healthy adult humans having a mean of } 98.9^{\circ} \mathrm{F} \text { and a standard deviation of } 0.62^{\circ} \mathrm{F} \text {. Construct a } 99 \% \text { confidence interval estimate of the mea } \\ \text { healthy humans. What does the sample suggest about the use of } 98.6^{\circ} \mathrm{F} \text { as the mean body temperature? } \\ \text { Click here to view a t distribution table. } \\ \text { Click here to view page } 1 \text { of the standard normal distribution table. } \\ \text { Click here to view page } 2 \text { of the standard normal distribution table. } \\ \text { What is the confidence interval estimate of the population mean } \mu \text { ? } \\ \square^{\circ} \mathrm{F}<\mu<\square^{\circ} \mathrm{F} \\ \text { (Round to three decimal places as needed.) }\end{array} \) ?
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To construct a \(99\%\) confidence interval for the mean body temperature, you'll use the formula: \[ \text{CI} = \bar{x} \pm t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right) \] where \(\bar{x} = 98.9^{\circ}F\), \(s = 0.62^{\circ}F\), and \(n = 106\). Checking the t-table for \(n - 1 = 105\) degrees of freedom at \(99\%\) confidence gives a critical value of approximately \(2.626\). So, the margin of error (ME) is: \[ \text{ME} = 2.626 \left(\frac{0.62}{\sqrt{106}}\right) \approx 0.162 \] Thus, the confidence interval is: \[ 98.9 - 0.162 < \mu < 98.9 + 0.162 \Rightarrow 98.738 < \mu < 99.062 \] So the confidence interval estimate of the population mean \(\mu\) is \(98.738^{\circ}F < \mu < 99.062^{\circ}F\). Now, regarding the reference point of \(98.6^{\circ}F\): since \(98.6^{\circ}F\) falls within this interval, it suggests that using \(98.6^{\circ}F\) as the mean body temperature may still be applicable, but the data indicates that the average temperature of healthy adults could be slightly higher.