Assume that both populations are normally distributed.
(a) Test whether $\mu_{1} \neq \mu_{2}$ at the $\alpha=0.05$ level of significance for the given sample data.
(b) Construct a $95 \%$ confidence interval about $\mu_{1}-\mu_{2}$.
\begin{tabular}{ccc}
& Population 1 & Population 2 \\
\hline $\mathrm{n}$ & 12 & 12 \\
\hline$\overline{\mathrm{x}}$ & 14.6 & 12.1 \\
\hline $\mathrm{s}$ & 3.8 & 3.2
\end{tabular}
(a) Test whether $\mu_{1} \neq \mu_{2}$ at the $\alpha=0.05$ level of significance for the given sample data.

Determine the null and alternative hypothesis for this test.
A.
\[
\begin{array}{l}
H_{0}: \mu_{1} \neq \mu_{2} \\
H_{1}: \mu_{1}=\mu_{2}
\end{array}
\]
B.
\[
\begin{array}{l}
\mathrm{H}_{0}: \mu_{1}=\mu_{2} \\
\mathrm{H}_{1}: \mu_{1}> \mu_{2}
\end{array}
\]
C.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1} \neq \mu_{2}
\end{array}
\]
D.
\[
\begin{array}{l}
\mathrm{H}_{0}: \mu_{1} \neq \mu_{2} \\
\mathrm{H}_{1}: \mu_{1}> \mu_{2}
\end{array}
\]

Detemine the P-value for this hypothesis test.
$P=\square$ (Round to three decimal places as needed.)

Step 1 ：Set up the null and alternative hypotheses. The null hypothesis (H0) is that the means are equal, and the alternative hypothesis (H1) is that the means are not equal. This corresponds to option C in the question: \[\begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array}\]

Step 2 ：Calculate the test statistic for a two-sample t-test using the formula: \[t = \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\] where \(\overline{x}_1\) and \(\overline{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes.

Step 3 ：Substitute the given values into the formula to get the test statistic: \[t = \frac{14.6 - 12.1}{\sqrt{\frac{3.8^2}{12} + \frac{3.2^2}{12}}}\]

Step 4 ：Calculate the P-value using the t-distribution with \(n_1 + n_2 - 2\) degrees of freedom. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.

Step 5 ：Since the P-value is greater than the significance level of 0.05, we do not reject the null hypothesis. This means that we do not have enough evidence to conclude that the means of the two populations are different.

Step 6 ：The correct null and alternative hypotheses are: \[\begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array}\]

Step 7 ：The P-value for this hypothesis test is approximately \(\boxed{0.095}\).