An SAT prep course claims to increase student scores by more than 60 points, on average. To test this claim, 9 students who have previously taken the SAT are randomly chosen to take the prep course. Their SAT scores before and after completing the prep course are listed in the following table. Test the claim at the 0.10 level of significance assuming that the population distribution of the paired differences is approximately normal. Let scores before completing the prep course be Population 1 and let scores after completing the prep course be Population 2. \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|} \hline \multicolumn{9}{|c|}{ SAT Scores } \\ \hline \begin{tabular}{c} Before \\ Prep \\ Course \end{tabular} & 1060 & 1410 & 1170 & 1130 & 1180 & 1190 & 1280 & 1290 & 1200 \\ \hline \begin{tabular}{c} After \\ Prep \\ Course \end{tabular} & 1170 & 1550 & 1360 & 1410 & 1390 & 1300 & 1520 & 1320 & 1210 \\ \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
Solution
Step 1 ：Calculate the differences between the scores before and after the prep course for each student. The differences are as follows: \(1170 - 1060 = 110\), \(1550 - 1410 = 140\), \(1360 - 1170 = 190\), \(1410 - 1130 = 280\), \(1390 - 1180 = 210\), \(1300 - 1190 = 110\), \(1520 - 1280 = 240\), \(1320 - 1290 = 30\), \(1210 - 1200 = 10\)
Step 2 ：Calculate the mean difference: \(\frac{110 + 140 + 190 + 280 + 210 + 110 + 240 + 30 + 10}{9} = \frac{1420}{9} = 157.78\)
Step 3 ：Calculate the standard deviation: \(\sqrt{\frac{(110-157.78)^2 + (140-157.78)^2 + (190-157.78)^2 + (280-157.78)^2 + (210-157.78)^2 + (110-157.78)^2 + (240-157.78)^2 + (30-157.78)^2 + (10-157.78)^2}{8}} = \sqrt{\frac{64128.56}{8}} = \sqrt{8016.07} = 89.53\)
Step 4 ：Calculate the test statistic: \(\frac{157.78 - 60}{89.53 / \sqrt{9}} = \frac{97.78}{29.84} = 3.276\)
Step 5 ：\(\boxed{\text{The value of the test statistic is 3.276}}\)
