Diane suspects that car garages chage women more than they d

Diane suspects that car garages chage women more than they do men. She took her car for service to 12 shops. She was given average estimate of $85 with standard deviation of $28. Her friend Steve took her car to 9 shops. He was given average estimate of $65 with standard deviation of $21. At 0.05 level of significance, assuming normal populations with equal standard deviation, solve Diane\'s suspicions.

Solution

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 =    85          
X2 =    65          
              
Calculating the standard deviations of each group,              
              
s1 =    28          
s2 =    21          
              
Thus, the pooled standard deviation is given by              
              
S = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]               
              
As n1 =    12   , n2 =    9  
              
Then              
              
S =    25.28989813          
              
Thus, the standard error of the difference is              
              
Sd = S sqrt (1/n1 + 1/n2) =    11.15179686          
              
As ud = the hypothesized difference between means =    0   , then      
              
t = [X1 - X2 - ud]/Sd =    1.793432956          
              
Getting the critical value using table/technology,              
df = n1 + n2 - 2 =    19          
tcrit =    +   1.729132812      
              
Getting the p value using technology,              
              
p =    0.044418528          
              
Thus, as we see, comparing t and tcrit (or, comparing p and the significance level) we   REJECT THE NULL HYPOTHESIS.          

Thus, there is significant evidence that car garages charge women more than they do men. [conclusion]