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?

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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?

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Answer:0.5Step-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 = [tex]\frac{u2 - u1}{sd_p_o_o_l_e_d}[/tex]- Where, the pooled standard deviation (sd_pooled) is calculated using the formula:                           [tex]sd_p_o_o_l_e_d =\sqrt{\frac{SD_1^2 +SD_2^2}{2} }[/tex]- Assuming that population standard deviation and sample standard deviation are same:                           SD_1 = SD_2 =  σ = 4- Then,                            [tex]sd_p_o_o_l_e_d =\sqrt{\frac{4^2 +4^2}{2} } = 4[/tex]- The cohen's d can now be evaliated:                           Cohen's d = [tex]\frac{48 - 46}{4} = 0.5[/tex]

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