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	370	380	450	500	360	400	360
Score on second SAT	420	480	500	580	400	460	400

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 H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5:

Make the decision for the hypothesis test.

Answer 1

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

using excel>addin>phstat>two sample test

we have

Paired t Test
Data
Hypothesized Mean Difference	0
Level of significance	0.1
Intermediate Calculations
Sample Size	7
DBar	-60.0000
Degrees of Freedom	6
SD	22.3607
Standard Error	8.4515
t Test Statistic	-7.0993
Lower-Tail Test
Lower Critical Value	-1.4398
p-Value	0.0002
Reject the null hypothesis

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

Ho:ud= 0

Ha:ud <0

Step 2 of 5:

the value of the standard deviation of the paired differences is 22.4

Step 3 of 5:

the value of the test statistic is -7.099

Step 4 of 5:

the decision rule for rejecting the null hypothesis is reject Ho if t <-1.440

Step 5 of 5:

Reject the null hypothesis , we will accept the calim that that the SAT prep course improves the students' verbal SAT scores