Question

A social researcher is interested in understanding the effect of college education on wages. The workers in one group have earned an associate’s degree while members of the other group hold at least a bachelor’s degree. He would like to run a hypothesis test with α = .10 to see if those with a bachelor’s degree have significantly higher hourly wages than those with an associate’s degree.

Bachelor’s degree			Associate’s degree
Participant	Hourly wage		Participant	Hourly wage
1	11.25		11	10.25
2	12.25		12	11.25
3	11.60		13	9.10
4	9.80		14	10.00
5	9.40		15	9.70
6	12.60		16	11.00
7	11.80		17	10.10
8	12.70		18	9.40
9	11.90		19	9.20
10	14.10		20	10.50

k.Determine the critical t value(s) for this hypothesis test based on the degree of freedom, from (d), and the preset alpha level. (1 point total)

l.Compare the calculated t statistic with the critical t value by stating which is more “extreme”, and then make a decision about the hypothesis test by stating clearly “reject” or “fail to reject” the null hypothesis. (1 point total: .5 for comparison, .5 for decision)

m.Calculate the pooled standard deviation for the populations and then use it to calculate the standardized effect size of this test. (2 point total: 1 for pooled standard deviation, 1 for effect size. Deduct .5 if a result is wrong but the process is correct.)

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.1                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   11.740                  
standard deviation of sample 1,   s1 =    1.3751                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   10.050                  
standard deviation of sample 2,   s2 =    0.7272                  
size of sample 2,    n2=   10                  
                          
difference in sample means =    x̅1-x̅2 =    11.7400   -   10.1   =   1.69  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.1000                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.4919                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   1.6900   -   0   ) /    0.49   =   3.435
                       
Degree of freedom, DF=   n1+n2-2 =    18                  
t-critical value , t* =        1.734   (excel formula =t.inv(α/2,df)              
Decision:   | t-stat | > | critical value |, so, Reject Ho                      
p-value =        0.0030   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis                     

effect size,      
cohen's d =    |( x̅1-x̅2 )/Sp | =    1.536