Question

An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first two times they took the verbal SAT. Before taking the SAT for the second time, each student took a course to try to improve his or her verbal SAT scores. Do these results support the claim that the SAT prep course improves the students' verbal SAT scores?

Let d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course)d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.05for the test. Assume that the verbal SAT scores are normally distributed for the population of students both before and after taking the SAT prep course.

Student	1	2	3	4	5	6	7
Score on first SAT	450	470	540	550	570	450	370
Score on second SAT	490	520	590	600	610	470	410

1 of 5: State the null and alternative hypotheses for the test

Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5: Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5: Make the decision for the hypothesis test. Reject or Fail to Reject

Answer 1

First we find the differences in the SAT scores and tabulate it.Then the null and alternate hypothesis is stated. The standard deviation and test statistic are calculated using the formula mentioned. The critical value is obtained from STATKEY (image attached for reference). Then the decision rule for rejecting null hypothesis is stated and as the condition is satisfied we reject null hypothesis and make required conclusion.