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). Use a significance level of α=0.1 for 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   Score on first SAT   Score on second SAT
1 480 530
2 530 580
3 520 560
4 530 610
5 360 390
6 380 400
7 550 610

Step 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 Null Hypothesis)

Answer 1

H₀: = 0

H₁: < 0

= (-50 + (-50) + (-40) + (-80) + (-30) + (-20) + (-60))/7 = -47.143

s_d = sqrt(((-50 + 47.143)^2 + (-50 + 47.143)^2 + (-40 + 47.143)^2 + (-80 + 47.143)^2 + (-30 + 47.143)^2 + (-20 + 47.143)^2 + (-60 + 47.143)^2)/6) = 19.8

The test statistic t = ( - D)/(s_d/)

                             = (-47.143 - 0)/(19.8/)

                             = -6.299

At alpha = 0.1, the critical value is t_0.1,6 = -1.440

Reject H₀, if t < -1.440

Since the test statistic value is less than the critical value(-6.299 < -1.440), so we should reject the null hypothesis.