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=0.05α=0.05for 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	370370	400400	590590	570570	490490	460460	440440
Score on second SAT	400400	480480	610610	600600	530530	510510	480480 Copy Data 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.

Answer 1

Solution:

Here, we have to use paired t test for the difference between population means.

Step 1 of 5 :  

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: The verbal SAT score of the students do not improve after taking the course.

Alternative hypothesis: H_a: The verbal SAT score of the students improve after taking the course.

H₀: µ_d = 0 versus H_a: µ_d < 0

We assume d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course)

We are given level of significance = α = 0.05

The test statistic formula is given as below:

Test statistic for paired t test is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

Step 2 of 5:

From given data, we have

Dbar = -41.4286

S_d = 19.5180

n = 7

df = n – 1 = 6

α = 0.05

Lower critical value = -1.9432

(by using t-table)

Step 3 of 5:

Test statistic is given as below:

t = (-41.4286 – 0)/[ 19.5180/sqrt(7)]

t = -5.6158

Step 4 of 5:

We reject the null hypothesis if P-value < α = 0.05

Or

We reject the null hypothesis if test statistic value < Lower critical value = -1.9432

From given test statistic value, we have

P-value = 0.0007

(by using t-table)

P-value < α = 0.05

Test statistic < lower critical value

So, we reject the null hypothesis

Step 5 of 5:

There is sufficient evidence to conclude that the verbal SAT score of the students improve after taking the course.