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=0.01 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	530	410	380	600	480	440	380
Score on second SAT	560	460	400	620	500	520	430

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.

Reject H0 if (t, ItI) (<,>) ____________

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

Answer 1

Sample size, n =7

Student	Score on first SAT	Score on second SAT	Difference: d
1	530	560	-30
2	410	460	-50
3	380	400	-20
4	600	620	-20
5	480	500	-20
6	440	520	-80
7	380	430	-50
Total			-270

Paired sample t-test:

Step 1 of 5:

Null Hypothesis(H0):

Alternative Hypothesis(H1): (left-tailed test).

(where, =Population mean of verbal SAT scores prior to taking the prep course; =Population mean of verbal SAT scores after taking the prep course).

Step 2 of 5:

Sample mean of the differences, -270/7 = -38.5714

Sample std.deviation of the differences, S_d = =22.7

So, the value of the standard deviation of the paired differences =22.7

Standard Error, SE =S_d/ =22.7/ =8.5798

Step 3 of 5:

Test statistic, t =( - 0)/SE =(-38.5714 - 0)/8.5798 = -4.4956

Step 4 of 5:

Decision rule:

Reject H0 if ItI > |t-critical​​​​​​|

t-critical at 0.01 significance level and at df =n - 1 =6 for a left-tailed test is: t-critical = -3.143

So, reject H0 if ItI > |-3.143|, i.e., if |t| > 3.143

Step 5 of 5:

Decision for the hypothesis test:

Reject the Null Hypothesis at 0.01 significance level since 4.4956 > 3.143

Thus, we have sufficient statistical evidence to conclude that the SAT prep course improves the students' verbal SAT scores.