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You wish to test the following claim $\left(H_{a}\right)$ at a significance level of $\alpha=0.10$.
\[
\begin{array}{l}
H_{o}: \mu_{1}=\mu_{2} \\
H_{a}: \mu_{1}< \mu_{2}
\end{array}
\]

You obtain a sample of size $n_{1}=58$ with a mean of $M_{1}=62.1$ and a standard deviation of $S D_{1}=15.9$ from the first population. You obtain a sample of size $n_{2}=39$ with a mean of $M_{2}=66.4$ and a standard deviation of $S D_{2}=15.9$ from the second population.

What is the critical value for this test? For this calculation, use the conservative under-estimate for the degrees of freedom as mentioned in the textbook. (Report answer accurate to three decimal places.) critical value $=$

What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic $=$

The test statistic iş...
in the critical region

Step 1 ：Given the following values: \(n_1 = 58\), \(M_1 = 62.1\), \(SD_1 = 15.9\), \(n_2 = 39\), \(M_2 = 66.4\), \(SD_2 = 15.9\), and \(\alpha = 0.1\).

Step 2 ：The degrees of freedom is the smaller of \(n_1 - 1\) and \(n_2 - 1\), which is \(38\).

Step 3 ：Using the t-distribution table with a significance level of \(0.10\) and degrees of freedom \(38\), we find the critical value to be approximately \(1.304\).

Step 4 ：The test statistic can be calculated using the formula for the t-statistic in a two-sample t-test, which is \((M_1 - M_2) / \sqrt{(SD_1^2/n_1) + (SD_2^2/n_2)}\). Substituting the given values, we find the test statistic to be approximately \(-1.306\).

Step 5 ：Since the test statistic is less than the critical value, it is in the critical region. Therefore, we reject the null hypothesis and conclude that there is enough evidence to support the claim that \(\mu_{1}<\mu_{2}\).

Step 6 ：Final Answer: The critical value for this test is \(\boxed{1.304}\) and the test statistic is \(\boxed{-1.306}\). The test statistic is in the critical region.