Question

A sample is selected from a population with μ = 46, and a treatment is administered to the sample. After treatment, the sample mean is μ = 48 with a sample variance of σ² = 16. Based on this information, what is the value of Cohen’s d?

Answer 1

Answer:

0.5

Step-by-step explanation:

Solution:-

- The sample mean before treatment, μ1 = 46  

- The sample mean  after treatment, μ2 = 48

- The sample standard deviation σ = √16 = 4

- For the independent samples T-test, Cohen's d is determined by calculating the mean difference between your two groups, and then dividing the result by the pooled standard deviation.

                           Cohen's d = $\frac{u2 - u1}{sd_p_o_o_l_e_d}$

- Where, the pooled standard deviation (sd_pooled) is calculated using the formula:

                          $sd_p_o_o_l_e_d =\sqrt{\frac{SD_1^2 +SD_2^2}{2} }$

- Assuming that population standard deviation and sample standard deviation are same:

                          SD_1 = SD_2 =  σ = 4

- Then,

                           $sd_p_o_o_l_e_d =\sqrt{\frac{4^2 +4^2}{2} } = 4$

- The cohen's d can now be evaliated:

                          Cohen's d = $\frac{48 - 46}{4} = 0.5$