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=0.01 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 1 2 3 4 5 6 7
Score on first SAT 360 530 580 480 550 460 520
Score on second SAT 400 570 610 550 580 510 550

Step 1 of 5: State the null and alternative hypotheses for the test. Ho: μd (=,≠,<,>,≤,≥) 0 Ha: μd (=,≠,<,>,≤,≥) 0

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

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. Reject Ho if (t, I t I) (<,>) _____

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

 Pre Post Difference 360 400 -40 530 570 -40 580 610 -30 480 550 -70 550 580 -30 460 510 -50 520 550 -30

1).

Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd < 0

2).

∑d = -290

∑d² = 13300

n = 7

Mean , x̅d = Ʃd/n = -290/7 = -41.429

Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] = √[(13300-(-290)²/7)/(7-1)] = 14.6385 = 14.64

3)

Test statistic:

t = (x̅d)/(sd/√n) = (-41.4286)/(14.6385/√7) = -7.488

4)

df = n-1 = 6

Critical value, t_c = T.INV(0.01, 6) = -3.143

Reject Ho if t < -3.143

5)

Decision:

Reject the null hypothesis

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