Question

A researcher wants to know whether a pharmaceutical intervention to reduce stress is working. N = 25 individuals are randomly assigned to receive the drug of interest and have a sample mean (M) of 76.4, whereas N = 29 individuals are randomly assigned to receive a placebo pill and have a sample mean of 83.0. Assume α = 0.05. The pooled variance (s²_pooled) is calculated to be 12.72. The standard deviation of the distribution for differences between means (s_difference)is calculated to be 0.819.

1. Which of the following test statistics is most appropriate to analyze these data?

a. z test

b. One-sample t test

c. Paired-samples t test

d. Independent samples t test

2. What is/are the critical value(s)?

3. What is the calculated test statistic?

4. Do you reject or fail to reject the null hypothesis?

5. Calculate the Cohen’s d measure of effect size.

Answer 1

a) Paired-samples t test(c).

*H0: μ1- μ2=0 VS   H1: μ1- μ2 ≠0

t= (xˉ1-​ xˉ2​)/√ s2(1/n1+1/n2) eq (1)

s2=12.72

n1=25 ,n2=29,

x-1=76.4 , x-2=83.0

substituting in eq(1) we get

t= 6.6/2.19

=3.013

*table value or t value :

df= n1+n2-2=25+29-2=52  

α = 0.05

P-value or table value =1.671 (taking df as 60 since 52 df is not there in the table )

1.671<3.013 which implies table value(p-value) < calculated value

therefore we reject the null hypothesis H0  ,there's difference in the mean value which implies μ1- μ2 ≠0 or μ1≠μ2

Cohen's d = (M2 - M1) ⁄ SDpooled

= (83 - 76.4) ⁄ 3.56 = 1.853933