Question

A therapist was interested in determining whether patients experiences reduced anxiety following diaphragmatic breathing exercises. She includes 9 participants in her brief study. Each patient provides a rating for current anxiety, on a scale of 1 (least anxiety) to 10 (extreme anxiety). She then instructs them on a 45-minute diaphragmatic breathing exercise. Following the exercise, each patient again rates his/her anxiety on the same 1-10 scale. (Note that this is a repeated measures study because each patient/participant is measured twice, once before the treatment and once after the treatment.)

Patient	Before treatment	After treatment
A	8	7
B	7	5
C	6	6
D	7	6
E	9	7
F	8	5
G	5	4
H	9	4
I	7	4

1. Can the therapist conclude that there was a significant change in anxiety levels after treatment? Use a two-tailed test with α = .05. (Note that you will need to use – and clearly show – all 4 steps of hypothesis testing to answer this question.)

2. Compute the value of r² (percentage of variance accounted for) for these data.

3. Write a sentence showing how the outcome of the hypothesis test and the measure of effect size would appear in a research report (i.e., in APA format). Note that you can find an example of APA format for a repeated measures t-test both in your book, in the “In the Literature” section of chapter 11, and in the example I provided in this week’s resources. APA format is slightly different depending on the type of test used, so be sure to look for an example of a repeated-measures t-test!

Answer 1

The null hypothesis would be as followed:

H₀: D= 0

H₁: D≠ 0

The test-statistic would be as followed:

NOTE-

Treatement1= Before treatment

Treatment2= After treatment

M= sample mean

S= sample standard deviation

Now, test statistic is:

T-value= -4

So, by using t-table p-value would be as followed:

p-value= 0.00395

According to the decision theory, p-value < alpha-value (0.05)

So, we will reject the null hypothesis as the results are significant.

R²(coefficient of determinition)= 0.095238

Which shows that 9.524% of variability is explained in the data.