In order to compare the means of two populations, independent random samples of 395 observations are selected from each population, with the results found in the table to the right. Complete parts a through e below.
\[
\begin{array}{ll}
\hline \text { Sample 1 } & \text { Sample 2 } \\
\hline \mathrm{x}_{1}=5,318 & \bar{x}_{2}=5,284 \\
s_{1}=142 & s_{2}=190 \\
\hline
\end{array}
\]
Reject $\mathrm{H}_{0}$. There is not sufficient evidence that the population means are different.
Do not reject $\mathrm{H}_{0}$. There is sufficient evidence that the population means are different.
c. Suppose the test in part b was conducted with the alternative hypothesis $H_{a}:\left(\mu_{1}-\mu_{2}\right)> 0$. How would your answer to part b change? Select the correct choice below and in the answer box within your choice.
(Round to three decimal places as needed.)
A. The test statistic would be and the null hypothesis would be rejected in favor of the new alternative hypothesis.
B. The observed significance level, or p-value, would be and the null hypothesis would not be rejected in favor of the new alternative hypothesis.
C. The test statistic would be and the null hypothesis would not be rejected in favor of the new alternative hypothesis.
D. The observed significance level, or p-value, would be and the null hypothesis would be rejected in favor of the new alternative hypothesis.

Step 1 ：Given that the sample means are \(\bar{x}_1 = 5318\) and \(\bar{x}_2 = 5284\), the sample standard deviations are \(s_1 = 142\) and \(s_2 = 190\), and the sample sizes are both 395.

Step 2 ：We are asked to perform a hypothesis test to determine if there is a significant difference between the means of two populations. The null hypothesis (H0) is that the means are equal, and the alternative hypothesis (Ha) is that the means are not equal.

Step 3 ：We can use a two-sample t-test to test this hypothesis. The formula for the test statistic for a two-sample t-test is \[t = \frac{{\bar{x}_1 - \bar{x}_2}}{{\sqrt{{s_1^2/n_1 + s_2^2/n_2}}}}\] where \(\bar{x}_1\) and \(\bar{x}_2\) 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 4 ：We can calculate the test statistic using the given data and then compare it to the critical value for a t-distribution with \(n_1 + n_2 - 2\) degrees of freedom to determine whether to reject the null hypothesis.

Step 5 ：Substituting the given values into the formula, we get \(t_{stat} = 2.8488013249525865\).

Step 6 ：The degrees of freedom is \(df = n_1 + n_2 - 2 = 788\).

Step 7 ：The critical value for a t-distribution with 788 degrees of freedom is \(crit_{val} = 1.962979031749005\).

Step 8 ：Since the test statistic is greater than the critical value, we reject the null hypothesis. There is sufficient evidence that the population means are different.

Step 9 ：Final Answer: \(\boxed{\text{Reject H0. There is sufficient evidence that the population means are different.}}\)