Step 1 :Given that we have two samples from two different populations, we want to test the claim that the means of these two populations are not equal. The null hypothesis \(H_{0}\) is that the means are equal, \(\mu_{1} = \mu_{2}\), and the alternative hypothesis \(H_{a}\) is that the means are not equal, \(\mu_{1} \neq \mu_{2}\).
Step 2 :The sample from the first population has a size of \(n_{1} = 109\), a mean of \(M_{1} = 65.1\), and a standard deviation of \(SD_{1} = 14.1\). The sample from the second population has a size of \(n_{2} = 72\), a mean of \(M_{2} = 67.1\), and a standard deviation of \(SD_{2} = 12.9\).
Step 3 :We are conducting this test at a significance level of \(\alpha = 0.02\). Since this is a two-tailed test, we will divide the alpha level by 2.
Step 4 :To find the critical value for this test, we need to use the t-distribution table. The degrees of freedom is the smaller of \(n_{1} - 1\) and \(n_{2} - 1\), which is \(df = 71\).
Step 5 :Looking up the t-distribution table with \(df = 71\) and \(\alpha/2 = 0.01\) (for a two-tailed test), we find that the critical value is approximately 2.380.
Step 6 :\(\boxed{2.380}\) is the critical value for this test.