Problem
Hypothesis Test for Difference in Population Means ( \( \sigma \) Unknown) You wish to test the following claim \( \left(H_{a}\right) \) at a significance level of \( \alpha=0.10 \) \[ \begin{array}{l} H_{o}: \mu_{1}=\mu_{2} \\ H_{a}: \mu_{1} \neq \mu_{2} \end{array} \] You believe both populations are normally distributed, but you do not know either. We will assume that the population variances are not equal. You obtain a sample of size \( n_{1}=14 \) with a mean of \( M_{1}=68.6 \) and a stand from the first population. You obtain a sample of size \( n_{2}=19 \) with a mean c deviation of \( S D_{2}=16.5 \) from the second population. What is the test statistic for this sample? (Report answer accurate to three di test statistic \( = \) What is the \( p \)-value for this sample? For this calculation, use the conservative degrees of freedom. The degrees of freedom is the minimum of \( n_{1}-1 \) and \( n_{2} \) to four decimal places.) \[ \mathrm{p} \text {-value }= \] Q 0 Type here to search
Solution
Step 1 ：1. Compute the sample mean difference: \( \bar{d} = M_{1} - M_{2} = 68.6 - 81 = -12.4 \)
Step 2 ：2. Calculate standard error: \( SE = \sqrt{\frac{S_{1}^{2}}{n_{1}} + \frac{S_{2}^{2}}{n_{2}}} = \sqrt{\frac{23.1^2}{14} + \frac{16.5^2}{19}} = 7.7467 \)
Step 3 ：3. Compute the test statistic: \( t = \frac{\bar{d} - 0}{SE} = \frac{-12.4}{7.7467} = -1.6019 \)
Step 4 ：4. Calculate degrees of freedom: \( df = \min(n_{1}-1, n_{2}-1) = \min(13, 18) = 13 \)
