Question

A researcher would like to determine if relaxation training will affect the number of headaches for chronic headache sufferers. For a week prior to training, each participant records the number of headaches suffered. Participants then receive relaxation training, and for the week following training the number of headaches is again measured. The data are as follows

Pre-Training Post-Training

6 2

9 4

4 4

5 3

3 4

9 8

7 6

6 2

4 0

7 2

Formulate the hypotheses.

Determine the critical value(s).

Calculate the test statistic.

Calculate and evaluate Cohen’s d and r2.

What decision can you make and what does it indicate?

What type of error could you be making and what would that mean?

Report your results with an APA phrase. Redo as a one-tail test, expecting an decrease in number of headaches. What is different?

Answer 1

Here we have paired data so paired t test should be used. Following table shows the calculations:

Pre training	Post training	d	(d-mean)^2
6	2	4	2.25
9	4	5	6.25
4	4	0	6.25
5	3	2	0.25
3	4	-1	12.25
9	8	1	2.25
7	6	1	2.25
6	2	4	2.25
4	0	4	2.25
7	2	5	6.25
Total		25	42.5

Sample size n = 10 Sample sum of difference: d = 25 Sample mean of differences: d=2^ = 25/10 = 2.5 Sample standard deviation of differences: (d-d) - =sqrt(42.5/(10-1)) SdVn-1 =2.1731

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Hypotheses are: H:Hq = 0 HI, * o Level of significance: a =0.05 Test is two tailed. Degree of freedom: df =n-1=9 Test Statistics: t d -12 = 12.5-0)/(2.1731/sqrt(10)) = 3.638 Critical value: Rejection Region: 2.26 If |t| >2.26, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value=0.005 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: Excel function for p-value: ETINV(0.05,9) ETDIST(3.638,9,2) =IDIS

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Cohen's d effect size is

$d=\frac{2t}{\sqrt{df}}=\frac{2\cdot 3.638}{\sqrt{9}}=2.43$

It is a large effect size.

And effect size r2 is

$r^{2}=\frac{t^{2}}{t^{2}+df}=\frac{3.638^{2}}{3.638^{2}+9}=0.60$

It is a medium to large effect size.

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The APA style reuslts are:

t(9) = 3.638, p-value < 0.05

Since we reject the null hypothesis so type II error is possible. It means that researcher incorrectly conldue that trainingis not effective while actually it is.

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Hypotheses are: H:Hg = 0 H:Ha > 0 Level of significance: a =0.05 Test is one tailed (right tailed). Degree of freedom: df =n-1=9 Test Statistics: a-la t== = (2.5-0)/(2.1731/sqrt(10)) = 3.638 Critical value: Rejection Region: 1.833 Ift > 1.833, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value=0.0027 Decision: Since p-value is less than level to significance so we reject the null hypothesis Excel function for critical value: Excel function for p-value: TINV(0.1,9) =TDIST(3.638,9,1)

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