Question

Instructions:

• Show all work, including every calculation and every formula, with proper symbols. Formulas must be those used in the required textbook for the course.
• Use the t-table in your book to find the critical t value(s).
• Calculators are fine; computer software of any kind is NOT. (You are welcome to use SPSS to check your work, but all values generated/reported below must come from hand calculations or the t table in your book.)

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

Ho:D = 0

H1:D ≠ 0

The test-statistic would be as followed

Treatment 1	Treatment 2	Diff (T2 - T1)	Dev (Diff - M)	Sq . Dev
8	7	-1	1	1
7	5	-2	0	0
6	6	0	2	4
7	6	-1	1	1
9	7	-2	0	0
8	5	-3	-1	1
5	4	-1	1	1
9	4	-5	-3	9
7	4	-3	-1	1

M : -2

S : 18

Note:

Treatment1 - Before Treatment

Treatment2 - After Treatment

M - Sample mean

S - Standard deviation

Difference scores calculations:

Mean: -2

= 0

= = = 2.25

= = = 0.25

= = = 0.5

T-value Calcultion:

t = = =-4

T-value = -4

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

p-value = 0.00395

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

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

(Coefficient of determination) = 0.095238

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