Question

To test the effectiveness of a studying program, participants are randomly assigned to either the control group or the treatment group (N=10 for each). At the end of the trials, the average test score increase in the control group is 3.5 points (s=0.7) and the treatment group increased 1.9 (s=1.1). Test the null hypothesis that there is no difference in test score improvement between the treatment and control groups (alpha = 0.05).

• For independent sample t-tests: mean values for each sample, the variances for each sample, estimated standard error of the difference in means, the t-ratio, degrees of freedom, the t-critical value, and your decision to reject or retain the null.

• For dependent sample t-tests: mean values for each sample, standard deviation for the difference between groups, standard error of the difference between the means, the t-ratio, degrees of freedom, the t-critical value, and your decision to reject or retain the null.

• For proportions: sample proportions, combined proportions, standard error of the difference, z-score, critical z-score, and your decision to reject or retain the null.

Answer 1

independent sample t-tests will be used

Ho :   µ1 - µ2 =   0              
Ha :   µ1-µ2 ╪   0              
                      
Level of Significance ,    α =    0.05              
                      
Sample #1   ---->   sample 1              
mean of sample 1,    x̅1=   3.50              
variance of sample 1,   s1² =    0.7² = 0.49
size of sample 1,    n1=   10              
                      
Sample #2   ---->   sample 2              
mean of sample 2,    x̅2=   1.90              
variance of sample 2,   s2² =    1.1² = 1.21   
size of sample 2,    n2=   10              
                      
difference in sample means =    x̅1-x̅2 =    3.5000   -   1.9   =   1.600
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.9220              
std error of difference, SE =    Sp*√(1/n1+1/n2) =    0.4123              

t-statistic = ((x̅1-x̅2)-µd)/SE = (   1.6000   -   0   ) /    0.41   =   3.881
Degree of freedom, DF=   n1+n2-2 =    18      
t-critical value , t* = ±2.1009   (excel formula =t.inv(α/2,df)  

since, test stat >2.1009 , reject the null hypothesis