Answer:
The Cohen's D is given by this formula:
![D = \frac{\bar X_A -\bar X_B}{s_p}](https://tex.z-dn.net/?f=%20D%20%3D%20%5Cfrac%7B%5Cbar%20X_A%20-%5Cbar%20X_B%7D%7Bs_p%7D)
Where
represent the deviation pooled and we know from the problem that:
represent the pooled variance
So then the pooled deviation would be:
![s_p = \sqrt{4}= 2](https://tex.z-dn.net/?f=%20s_p%20%3D%20%5Csqrt%7B4%7D%3D%202)
And the difference of the two samples is
, and replacing we got:
![D = \frac{1}{2}= 0.5](https://tex.z-dn.net/?f=%20D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%3D%200.5)
And since the value for D obtained is 0.5 we can consider this as a medium effect.
Step-by-step explanation:
Previous concepts
Cohen’s D is a an statistical measure in order to analyze effect size for a given condition compared to other. For example can be used if we can check if one method A has a better effect than another method B in a specific situation.
Solution to the problem
The Cohen's D is given by this formula:
![D = \frac{\bar X_A -\bar X_B}{s_p}](https://tex.z-dn.net/?f=%20D%20%3D%20%5Cfrac%7B%5Cbar%20X_A%20-%5Cbar%20X_B%7D%7Bs_p%7D)
Where
represent the deviation pooled and we know from the problem that:
represent the pooled variance
So then the pooled deviation would be:
![s_p = \sqrt{4}= 2](https://tex.z-dn.net/?f=%20s_p%20%3D%20%5Csqrt%7B4%7D%3D%202)
And the difference of the two samples is
, and replacing we got:
![D = \frac{1}{2}= 0.5](https://tex.z-dn.net/?f=%20D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%3D%200.5)
And since the value for D obtained is 0.5 we can consider this as a medium effect.