Question

# An SAT prep course claims to improve the test score of students. The table below shows...

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=0.05α=0.05 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
1 540 560
2 570 610
3 460 500
4 410 510
5 530 550
6 590 620
7 480 530

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 H0H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5:

Make the decision for the hypothesis test.

Step 1

To Test :-

H0 :- µd = 0

H1 :- µd < 0

Step 2

S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √(4542.8571 / 7-1) = 27.5

Step 3

d̅ = Σ di/n = -300 / 7 = -42.8571

t = d̅ / ( S(d) / √(n) )
t = -42.8571 / ( 27.5162 / √(7) )
t = -4.121

Step 4

Reject null hypothesis if t < - t(α)
Critical value t(α) = t(0.05) = 1.9432
t < -1.9432

Step 5

t < - t(α) = -4.1208 < -1.9432
Result :- Reject null hypothesis

There is sufficient evidence to support the claim that  SAT prep course improves the student's verbal SAT scores.