Males use the Internet 1017 hours per week standard deviatio

Solution

Let u1 = mean of males
u2 = mean of females

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   <=   0  
Ha:   u1 - u2   >   0  
At level of significance =    0.05          
As we can see, this is a    right   tailed test.      
Calculating the means of each group,              
              
X1 =    10.17          
X2 =    9.08          
              
Calculating the standard deviations of each group,              
              
s1 =    11.71          
s2 =    12.26          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    118          
n2 = sample size of group 2 =    157          

Thus, df = n1 + n2 - 2 =    273          

Also, sD =    1.45583031          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    0.748713633          
              
where uD = hypothesized difference =    0          
              
Now, the critical value for t is              
              
tcrit =        1.650454303      
              
As |t| < 1.6504, we FAIL TO REJECT THE NULL HYPOTHESIS.          
          
Thus, there is no significant evidence that men use the Internet more hours than women. [CONCLUSION]