Question

Administrators want to know if test anxiety is impacted by the number of college years completed. After completing their freshman year, a random sample of students was selected and given the College Test Anxiety Questionnaire (CTAQ); higher scores indicate more test anxiety. After completing their junior year they were again tested. What can the administrators conclude with α = 0.05?

freshman	junior
2.1 7.5 3.2 6.3 5.5 5.2 4.6 5.2	5.9 7.2 7.4 6.8 8.5 6.2 7.3 5.2

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

b)
Condition 1:
---Select--- junior OR CTAQ OR test anxiety OR number of college years OR freshman
Condition 2:
---Select---junior OR CTAQ OR test anxiety OR number of college years OR freshman

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 OR 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 OR trivial effect OR small effect OR medium effect OR large effect
r² = _____ ;   ---Select--- na OR trivial effect OR small effect OR medium effect OR large effect

e) Make an interpretation based on the results.

A) Students showed significantly less anxiety in their junior year as opposed to their freshman year.

B) Students showed significantly more anxiety in their junior year as opposed to their freshman year.    

C) Students showed no significant anxiety difference between their junior and freshman year.

Answer 1

a) Independent-Samples t-test

b)

condition 1 : freshman

condition 2: junior

c)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   4.950                  
standard deviation of sample 1,   s1 =    1.6928                  
size of sample 1,    n1=   8                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   6.813                  
standard deviation of sample 2,   s2 =    1.0274                  
size of sample 2,    n2=   8                  
                          
difference in sample means =    x̅1-x̅2 =    4.9500   -   6.8   =   -1.863  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.4002                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.7001                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -1.8625   -   0   ) /    0.70   =   -2.6603
                          
Degree of freedom, DF=   n1+n2-2 =    14                  
p-value =        0.0093   [ excel function: =T.DIST(t stat,df) ]               
Conclusion:     p-value <α , Reject null hypothesis                      

d)

effect size,          
cohen's d =    |( x̅1-x̅2 )/Sp | =    1.330  

e)

B) Students showed significantly more anxiety in their junior year as opposed to their freshman year.