Nationwide graduates entering the actuarial field earn 40000

Nationwide graduates entering the actuarial field earn $40,000. A college placement officer feels that the mean salary is greater than this. She surveys n = 36 graduates entering the actuarial field and finds the average salary to be \\bar{x}x= $41,000. The population standard deviation is \\sigma= $3,000. Can her claim be supported at \\alpha= 0.05?

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   40000  
Ha:    u   >   40000  
              
As we can see, this is a    right   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 =    41000          
uo = hypothesized mean =    40000          
n = sample size =    36          
s = standard deviation =    3000          
              
Thus, z = (X - uo) * sqrt(n) / s =    2          
              
Also, the p value is              
              
p =    0.022750132          
              
As z > 1.6449, and P < 0.05, we   REJECT THE NULL HYPOTHESIS.          

Thus, there is significant evidence that the mean salary for the nationwide graduates entering the actuarial field is greater than $40,000. [CONCLUSION]