Question

***PLEASE TYPE ANSWERS***

Five students were tested before and after taking a class to improve their study habits. They were given articles to read which contained a known number of facts in each story. After the story each student listed as many facts as he/she could recall. The following data was recorded.

Before	10	12	14	16	12
Atter	15	14	17	17	20

	a.	What is the alternative hypothesis?
	b.	What is the null hypothesis?
	c.	What is your conclusion, using á = 0.052 tail?
	d.	What type error may you be making because of your conclusion in part c?   Please show all work.
	e.	What is the size of effect, using Cohen’s d?

Answer 1

Here, we have to use paired t test.

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

What is the null hypothesis?

Null hypothesis: H₀: There is no improvement in study habits after taking a class.

What is the null hypothesis?

Alternative hypothesis: H_a: There is an improvement in study habits after taking a class.

H₀: µ_d = 0 versus H_a: µ_d > 0

This is a right tailed test.

Test statistic for paired t test is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

From given data, we have

Dbar = 3.8

S_d = 2.7749

n = 5

df = n – 1 = 4

α = 0.052

t = (Dbar - µ_d)/[S_d/sqrt(n)]

t = (3.80 - 0)/[ 2.7749/sqrt(5)]

t = 3.0621

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

P-value = 0.0188

P-value < α = 0.052

So, we reject the null hypothesis

What is your conclusion, using á = 0.05_{2 tail}?

There is sufficient evidence to conclude that there is an improvement in study habits after taking a class.

What type error may you be making because of your conclusion in part c?

Answer: Type I error

What is the size of effect, using Cohen’s d?

d = Dbar/S_d

Dbar = 3.8

S_d = 2.7749

d = 3.8/2.7749

d = 1.369419