Question

A sample is selected from a population with mean score of µ = 50. After a treatment is administered to the individuals in the sample, the mean is found to be M = 55 and the variance is s2 = 64.

A. Assume that the sample has n = 4 scores. Conduct a hypothesis test to evaluate the significance of the treatment effect and calculate Cohen’s d to measure the size of the treatment effect. Use a two-tailed test with ? = .05.

B. Assume that the sample has n = 16 scores. Conduct a hypothesis test to evaluate the significance of the treatment effect and calculate Cohen’s d to measure the size of the treatment effect. Use a two-tailed test with ? = .05. (Note: in order to receive full credit for part A & B of this question you must show all computational steps, the critical t value, the decision regarding the null hypothesis, and write the final conclusion with the t-test outcome reported in APA style (see example of how to write statistical conclusions on the PowerPoint slides).

C. Compare your answers to part A & B – have increasing the sample size affected the likelihood of rejecting the null hypothesis and the Cohen's d effect size?

Answer 1

a)

t= (Xbar - mu)/(sd/sqrt(n))

=(55 - 50)/(8/sqrt(4))

= 1.25

Also, and df = 4-1 = 3

So, using table of critical values for t-distribution the critical values are -3.182 and 3.182

Since, test statistics value < 2.571 Fail to reject Ho.

cohen's d= (xbar -mu)/s= 5/8

b)

n = 16

t = (55 - 50)/(8/sqrt(16)) = 2.5

df = n-1 = 15

t-critical = 2.13144 and -2.13144

since TS > critical value

we reject the null hypothesis

c)

as we increase sample size , the likelihood of rejecting null hyphesis increases

Also, the Cohen's d is same in both cases.