Answer:
Cohen's d = 0.667
Explanation:
We are given the following in the question:
A paired sample t-test is conducted.
Sample size, n = 25
The mean difference between 25 pairs = 5.00
![M_1-M_2 = 5.00](https://img.qammunity.org/2021/formulas/mathematics/college/s3x2e9ibz0eicipgrjr9ogfytkptl58bcx.png)
where
are the means for paired t-test.
The standard deviation for the 25 difference scores = 7.50
![S.D_(Pooled) = 7.50](https://img.qammunity.org/2021/formulas/mathematics/college/da5jbl5arn4kf7gycuslbh7wzwdepcccsq.png)
Formula for Cohen's d:
![=(M_1-M_2)/(S.D_(Pooled))](https://img.qammunity.org/2021/formulas/mathematics/college/7mh3s1w8k43tpd506sv63ob5d40eybqonx.png)
Putting values, we get,
Cohen's d =
![=(5.00)/(7.50)\\\\=0.667](https://img.qammunity.org/2021/formulas/mathematics/college/hbft2q66jmpi8dsshpfl9jmoyhitkjdn02.png)
The value of Cohen's d coefficient is 0.667