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=0.05 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	1	2	3	4	5	6	7
Score on first SAT	500	370	430	420	530	450	470
Score on second SAT	540	410	4490	460	560	490	490

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.

Answer 1

Solution:

Here, we have to use paired t test.

H₀: µ_d = 0 versus H_a: µ_d < 0

This is a lower tailed test.

Step 2

S_d = 1521.3

Step 3

Test statistic for paired t test is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

From given data, we have

Dbar = -610

S_d = 1521.3262

n = 7

df = n – 1 = 6

t = (-610 – 0)/[ 1521.3262/sqrt(7)]

t = -1.0609

P-value = 0.1648

(by using t-table)

Step 4

We reject H₀ when P-value < 0.05

Step 5

P-value > α = 0.05

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that an SAT prep course improve the test score of students.