A random sample of 10 independent healthy people showed the body temperatures given below (in degrees Fahrenheit). Test the hypothesis that the population mean is not $98.6^{\circ} \mathrm{F}$, using a significance level of 0.05 . $\begin{array}{llllllllll}98.7 & 98.8 & 99.0 & 96.8 & 98.5 & 98.6 & 97.5 & 99.1 & 98.7 & 97.6\end{array}$ Determine the null and alternative hypotheses. Choose the correct answer below A. \[ \begin{array}{l} H_{0}: \mu>98.6 \\ H_{a}: \mu=98.6 \end{array} \] D. \[ \begin{array}{l} H_{0}: \mu<98.6 \\ H_{a}: \mu=98.6 \end{array} \] B. \[ \begin{array}{l} H_{0}: \mu \neq 98.6 \\ H_{a}: \mu=98.6 \end{array} \] (uE. \[ \begin{array}{l} H_{0} \cdot \mu=98.6 \\ H_{a} \cdot \mu<98.6 \end{array} \] C. \[ \begin{array}{l} H_{0} \mu=98.6 \\ H_{\mathrm{a}}: \mu>98.6 \end{array} \] \[ \begin{array}{l} H_{0}, \mu=98.6 \\ H_{a} \mu \neq 98.6 \end{array} \] Check the conditions to see whether the test statistic will follow a t-distribution. The sample is random and the observations are independent The distribution of the sample is approximately Normal. Find the test statistic. \[ t=-1.12 \] (Round to two decimal places as needed.) Find the $p$-value \[ p \text {-value }=\square \] (Round to three decimal places as needed.)
Solution
Step 1 ：First, we need to determine the null and alternative hypotheses. The null hypothesis \(H_{0}\) is that the population mean is 98.6 degrees Fahrenheit, and the alternative hypothesis \(H_{a}\) is that the population mean is not 98.6 degrees Fahrenheit. So, we have \(H_{0}: \mu=98.6\) and \(H_{a}: \mu \neq 98.6\).
Step 2 ：Next, we check the conditions to see whether the test statistic will follow a t-distribution. The sample is random and the observations are independent. The distribution of the sample is approximately Normal.
Step 3 ：Then, we find the test statistic. The test statistic is \(t=-1.12\).
Step 4 ：Finally, we find the p-value. The p-value is the probability of obtaining a test statistic as extreme as -1.12, or more extreme, under the assumption that the null hypothesis is true. Using the formula for the p-value with a t-distribution, we find that the p-value is approximately 0.292.
Step 5 ：The final answer is that the p-value is approximately \(\boxed{0.292}\).
