Question

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 0.05 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. SAT Scores Before Prep Course 1410 1500 1020 1270 1050 1410 1500 1040 1470 After Prep Course 1410 1700 1250 1420 1310 1670 1680 1070 1530

Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0Ha: μd=60: μd___60

Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.

Step 3 of 3: Draw a conclusion and interpret the decision.

Answer 1

The null hypothesis states that the mean difference in SAT scores before and after completing the prep course is 60, while the alternative hypothesis states that the mean difference is not equal to 60. The test statistic is calculated to be 0.059, which is not significant at the 0.05 level of significance. Therefore, we fail to reject the null hypothesis and conclude that there is not enough evidence to support the claim that the prep course increases student scores by more than 60 points.

Step 1 of 3: The null hypothesis, H0, states that the mean difference in SAT scores before and after completing the prep course is 60. The alternative hypothesis, Ha, states that the mean difference in SAT scores is not equal to 60.

Step 2 of 3: To compute the test statistic, we need to calculate the mean difference and standard deviation of the differences. The mean difference is the average of the differences between before and after scores, which is 60.667. The standard deviation is 199.75.

Step 3 of 3: Using the t-distribution, we calculate the test statistic as (60.667 - 60) / (199.75 / sqrt(9)) = 0.059. The degrees of freedom for this test is 8. Comparing the test statistic to the critical value from the t-distribution, we find that the test statistic is not significant at the 0.05 level of significance. Therefore, we fail to reject the null hypothesis and conclude that there is not enough evidence to support the claim that the prep course increases student scores by more than 60 points.