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=0.1 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	1	2	3	4	5	6	7
Score on first SAT	400	420	510	530	380	440	460
Score on second SAT	440	490	560	560	410	510	500

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 H 0 . Round the numerical portion of your answer to three decimal places.

Step 5 of 5: Make the decision for the hypothesis test.

Answer 1

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

Null hypothesis: H₀: The SAT prep course do not improves the students' verbal SAT scores.

Alternative hypothesis: H_a: The SAT prep course improves the students' verbal SAT scores.

H₀: µ_d = 0 versus H_a: µ_d < 0

This is lower tailed test.

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

The calculation table for mean and standard deviation is given as below:

No.	Score on first SAT	Score on second SAT	Di	(Di - DBar)^2
1	400	440	-40	51.02040816
2	420	490	-70	522.4489796
3	510	560	-50	8.163265306
4	530	560	-30	293.877551
5	380	410	-30	293.877551
6	440	510	-70	522.4489796
7	460	500	-40	51.02040816
Total			-330	1742.857143

n = 7

Mean = Dbar = ∑D/n = -330/7 = -47.1429

SD = sqrt[∑(Di - DBar)^2)/(n – 1)]

SD = sqrt(1742.857143/6)

SD = 17.0434

Standard deviation = S_d = 17.0

Step 3 of 5: Compute the value of the test statistic.

Test statistic for paired t test is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

t = (-47.1429 – 0)/[ 17.0434/sqrt(7)]

t = -7.3183

Test statistic = -7.318

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis H 0 .

n = 7

df = n – 1 = 6

α = 0.1

Critical t value = -1.4398

(by using t-table)

Decision rule: Reject the null hypothesis H₀ when test statistic t < -1.440

Step 5 of 5: Make the decision for the hypothesis test.

We have

Test statistic = -7.318

Critical t value = -1.4398

Test statistic < Critical t value

So, we reject the null hypothesis

There is sufficient evidence to conclude that The SAT prep course improves the students' verbal SAT scores.