Question

Question 9 (1 point)
Twelve students who were not satisfied with their ACT scores particiapted in an online 10-hour training program. The ACT scores before and after the training for the 12 students are given below:

Student Before After
1 23 27
2 25 26
3 27 31
4 30 32
5 24 26
6 25 24
7 27 31
8 26 28
9 28 30
10 22 25
11 20 24
12 29 32
Test a claim that the program is effective in improving a student’ ACT score.

What is the p-value?

Question 9 options:

Essentially 0

0.0325

0.0478

1.000

Answer 1

What is the p-value?

Essentially 0

Solution:

Here, we have to use paired t test.

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

Null hypothesis: H0: the program is not effective in improving a student’ ACT score.

Alternative hypothesis: Ha: the program is effective in improving a student’ ACT score.

H0: µd = 0 versus Ha: µd > 0

This is a right tailed test.

We take difference as after minus before.

Test statistic for paired t test is given as below:

t = (Dbar - µd)/[Sd/sqrt(n)]

From given data, we have

µd =0

Dbar = 2.5

Sd = 1.5076

n = 12

df = n – 1 = 11

α = 0.05

t = (Dbar - µd)/[Sd/sqrt(n)]

t = (2.5 - 0)/[ 1.5076/sqrt(12)]

t = 5.7446

The p-value by using t-table is given as below:

P-value = 0.0001

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the program is effective in improving a student’ ACT score.

Note:

If you take difference as before minus after, then you will get p-value equal to 1.000.

Answer will be 1.000 in case you take difference as before minus after.

Answer is depends on how you take difference, however, question doesn’t tell about how to take difference.