Question

Use SPSS for this Application Exercise:
An educational psychologist believes that regular exercise is detrimental to academic achievement. Students are randomly selected from a local high school to participate in a semester long study. Then pairs of students with similar GPAs are randomly assigned to either a group that attends daily exercise classes or not. At the end of the study, each participant's GPA is recorded. What can be concluded with α = 0.05? Below are the data.

exercise	control
3.99 1.89 3.65 2.11 2.81 3.91 3.51 3.21 3.61	3.39 2.44 3.12 1.57 2.45 2.19 3.01 3.01 3.49

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- GPA semester academic achievement control exercise
Condition 2:
---Select--- GPA semester academic achievement control exercise

c) Input the appropriate value(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
p-value =  ; Decision:  ---Select--- Reject H0 Fail to reject H0

d) Using the SPSS results, compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

e) Make an interpretation based on the results.

The GPA of individuals doing regular exercise was significantly higher than those that do not.

The GPA of individuals doing regular exercise was significantly lower than those that do not.     

The GPA of individuals doing exercise did not significantly differ from those in the control group.

Answer 1

a) independent sample t test

b)

1 : exercise

2: control

c)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   3.188                  
standard deviation of sample 1,   s1 =    0.762                  
size of sample 1,    n1=   9                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   2.741                  
standard deviation of sample 2,   s2 =    0.625                  
size of sample 2,    n2=   9                  
                          
difference in sample means =    x̅1-x̅2 =    3.1878   -   2.7   =   0.45  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.6969                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.3285                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   0.4467   -   0   ) /    0.33   =   1.3596
                          
Degree of freedom, DF=   n1+n2-2 =    16                  
  
p-value =        0.0964 [excel function: =T.DIST.RT(t stat,df) ]              
Conclusion:     p-value>α , Fail to  reject null hypothesis   

d)

cohen's d =    |( x̅1-x̅2 )/Sp | =    0.641 (large)
      
r²=   0.104 (small)

e)

The GPA of individuals doing exercise did not significantly differ from those in the control group.