A university investigation was conducted to determine whethe

A university investigation was conducted to determine whether women and men complete medical school in significantly different amounts of time, on the average. Two independent random samples were selected and the following summary information concerning times to completion of medical school computed: Women Men Sample Size 90 100 Sample Mean 8.4 years 8.5 years Sample Standard Deviation 0.6 years 0.5 years Which test would be used?

Solution

This would be an independent samples z test, as both n values are greater than 30. [ANSWER]

As significance level is not given, we assume it is 0.05.

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    two   tailed test.      
Calculating the means of each group,              
              
X1 =    8.4          
X2 =    8.5          
              
Calculating the standard deviations of each group,              
              
s1 =    0.6          
s2 =    0.5          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    90          
n2 = sample size of group 2 =    100          

Thus,

sD =    0.080622577          
              
Thus, the z statistic will be              
              
z = [X1 - X2 - uD]/sD =    -1.240347346          
              
where uD = hypothesized difference =    0          
              
Now, the critical value for z is              
              
zcrit =    +/-   1.96
              
As |z| < 1.96,   WE FAIL TO REJECT THE NULL HYPOTHESIS.          
      
Thus, there is no significant evidence that women and men complete medical school in significantly different amounts of time, on the average. [CONCLUSION]