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 \( \alpha=0.05 \). \[ \begin{array}{l} H_{o}: \mu_{1}=\mu_{2} \\ H_{a}: \mu_{1} \neq \mu_{2} \end{array} \] You believe both populations are normally distributed, but you do not know the standard deviations for either. We will assume that the population variances are not equal. You obtain a sample of size \( n_{1}=22 \) with a mean of \( M_{1}=76.8 \) and a standard deviation of \( S D_{1}=5 \). from the first population. You obtain a sample of size \( n_{2}=25 \) with a mean of \( M_{2}=74.4 \) and a standar deviation of \( S D_{2}=18.7 \) from the second population. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic \( = \) What is the \( p \)-value for this sample? For this calculation, use the conservative under-estimate for the degrees of freedom. The degrees of freedom is the minimum of \( n_{1}-1 \) and \( n_{2}-1 \) : (Report answer accurate to four decimal places.) \( p \)-value \( = \) The \( p \)-value is... O Type here to search
Solution
Step 1 ：1. Calculate the T-score: \(t = \frac{(M_{1} - M_{2}) - (\mu_{1} - \mu_{2})}{\sqrt{\frac{S D_{1}^{2}}{n_{1}} + \frac{S D_{2}^{2}}{n_{2}}}\) = \frac{(76.8 - 74.4) - 0}{\sqrt{\frac{5^2}{22} + \frac{18.7^2}{25}}}\ = 0.472\)
Step 2 ：2. Calculate the degrees of freedom (DF): \(DF = \min(n_{1}-1, n_{2}-1) = \min(21, 24) = 21\)
Step 3 ：3. Find the p-value: Using a T-distribution table or calculator with T-score = 0.472 and DF = 21, find the p-value (two-tailed) = 0.6416
