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.)
Answer
The final answer is that the p-value is approximately \(\boxed{0.292}\).
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 \(\boxed{0.292}\).
