A psychology graduate student wants to test the claim that there is a significant difference between the IQs of spouses. To test this claim, she measures the IQs of 9 married couples using a standard IQ test. The results of the IQ tests are listed in the following table. Using a 0.01 level of significance, test the claim that there is a significant difference between the IQs assuming that the population distribution of the paired differences is approximately normal. Let the spouse 1 group be Population 1 and let the spouse 2 group be Population 2. \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|} \hline \multicolumn{8}{|c|}{ IQs of Married Couples } \\ \hline \begin{tabular}{c} Spouse \\ $\mathbf{1}$ \end{tabular} & 128 & 113 & 118 & 111 & 124 & 122 & 122 & 116 & 123 \\ \hline \begin{tabular}{c} Spouse \\ $\mathbf{2}$ \end{tabular} & 126 & 114 & 115 & 107 & 128 & 120 & 118 & 111 & 122 \\ \hline \end{tabular} Copy Data Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places. Answer Tables Keypad Keyboard Shortcuts

Step 1 ：The problem is asking for the value of the test statistic for a paired t-test. The paired t-test is used to compare the means of two related groups to determine if there is a significant difference between them. In this case, the two groups are the IQs of spouse 1 and spouse 2.

Step 2 ：The test statistic for a paired t-test is calculated as follows: \(t = \frac{\overline{d} - \mu_d}{\frac{s_d}{\sqrt{n}}}\) where: \(\overline{d}\) is the mean of the differences between the two groups, \(\mu_d\) is the hypothesized mean difference (in this case, 0, because we're testing if there's a significant difference), \(s_d\) is the standard deviation of the differences, and \(n\) is the number of pairs.

Step 3 ：We need to calculate the mean and standard deviation of the differences between the two groups, and then plug those values into the formula to get the test statistic.

Step 4 ：The IQs of spouse 1 are [128, 113, 118, 111, 124, 122, 122, 116, 123] and the IQs of spouse 2 are [126, 114, 115, 107, 128, 120, 118, 111, 122]. The differences between the two groups are [2, -1, 3, 4, -4, 2, 4, 5, 1].

Step 5 ：The mean of the differences is approximately 1.778 and the standard deviation of the differences is approximately 2.819.

Step 6 ：Substituting these values into the formula, we get \(t = \frac{1.778 - 0}{\frac{2.819}{\sqrt{9}}}\), which simplifies to \(t = 1.892\).

Step 7 ：Final Answer: The value of the test statistic for the paired t-test is \(\boxed{1.892}\).