A random sample of the starting salaries of 35 randomly sele

A random sample of the starting salaries of 35 randomly selected graduates with bachelor’s degrees last year gave sample mean and standard deviation $41,202 and $7,621, respectively. Test whether the data provide sufficient evidence, at the 5% level of significance, to conclude that the mean starting salary of all graduates last year is less than the mean of all graduates two years before, $43,589.

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   43589  
Ha:    u   <   43589  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical z, as alpha =    0.05   ,      
alpha =    0.05          
zcrit =    -   1.644853627      
              
Getting the test statistic, as              
              
X = sample mean =    41202          
uo = hypothesized mean =    43589          
n = sample size =    35          
s = standard deviation =    7621          
              
Thus, z = (X - uo) * sqrt(n) / s =    -1.85299599          
              
Also, the p value is              
              
p =    0.031941467          
              
Comparing |z| < 1.645, (or, p < 0.05), we   REJECT THE NULL HYPOTHESIS.      

Thus, there is sufficient evidence, at the 5% level of significance, to conclude that the mean starting salary of all graduates last year is less than the mean of all graduates two years before, $43,589. [CONCLUSION]