Question

A high school is examining whether or not a certain college admissions test prep course is helpful. To evaluate this, 15 students took the college admissions test. Afterwards, they went through the prep course and then took the admissions test again. Their before and after scores are shown below. With a significance level of 0.90, is the admissions test prep course effective?

Student Before After

1 27 29

2 28 29

3 30 31

4 32 31

5 16 20

6 25 27

7 27 27

8 25 26

9 27 30

10 23 28

11 25 26

12 24 24

13 22 25

14 31 32

15 25 25

Answer 1

H0: Null Hypothesis: 0 ( The admissions test prep course is not effective )

HA: Alternative Hypothesis: 0 ( The admissions test prep course is effective )

From the given data, values of d = Before - After are got as follows:

d = - 2, - 1, - 1, 1, - 4, - 2, 0, - 1, - 3, - 5, - 1, 0, - 3, - 1, 0

From d values, the following statistics are calculated:

n = 15

= - 1.5333

s_d = 1.6417

SE = s_d/

= 1.6417/

= 0.4239

Test Statistic is given by:
t = - 1.5333/0.4239

= - 3.6173

= 0.10

ndf = n - 1 = 15 - 1 = 14

From Table, critical value of t = - 1.3450

Since calculated value of t = - 3.6173 is less than critical value of t = - 1.3450, the difference is significant. Reject null hypothesis.

Conclusion: The data support the claim that the admissions test prep course is effective.