Question

1. Exhibit 10-1
   Salary information regarding two independent random samples of male and female employees of a large company is shown below.
   

   	Male	Female
   Sample size	64	36
   Sample mean salary (in $1000s)	44	41
   Population variance	128	72

   Refer to Exhibit 10-1. At 95% confidence, we have enough evidence to conclude that the  _____.

   	a.	We fail to reject the null hypothesis; we conclude that the average average salary of males is at least as much as females.
   	b.	We reject the null hypothesis; we conclude that the average average salary of males is significantly lower than females.
   	c.	We fail to reject the null hypothesis; we conclude salaries of males and females are equal.
   	d.	We reject the null hypothesis; we conclude that the average average salary of males is significantly greater than females.

Answer 1

difference in sample means = x̅1 - x̅2 =    44   -   41   =   3
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    2.0000          
                  
Z-statistic = ((x̅1 - x̅2)-µd)/SE =    3   /   2.0000   =   1.5000
                  
  
p-value =        0.1336 [excel formula =2*NORMSDIST(z)]      

p-value>α , Do not reject null hypothesis

We fail to reject the null hypothesis; we conclude salaries of males and females are equal.