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.
$$\begin{array}{l}{H}_0:\mu \ne 98.6\\ H_a:\mu =98.6\end{array}$$
(uE.
$$\begin{array}{l}{H}_0\cdot \mu =98.6\\ H_a\cdot \mu <98.6\end{array}$$
C.
$$\begin{array}{l}{H}_0\mu =98.6\\ H_{\mathrm{a}}:\mu >98.6\end{array}$$
$$\begin{array}{l}{H}_0,\mu =98.6\\ H_a\mu \ne 98.6\end{array}$$

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
$$p\text{-value }=◻$$
(Round to three decimal places as needed.)

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 $H_a:\mu \ne 98.6$.

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 $\boxed{{\displaystyle 0.292}}$.