A company is interested in assessing how satisfied their emp

A company is interested in assessing how satisfied their employees are with their jobs after a job redesign effort. Previously, the company has collected scores on a popular measure of job satisfaction, the Minnesota Satisfaction Questionnaire (Short-Form) and obtained population-based parameters of = 60 and an = 20. Six months after the job redesign effort, the company collected data on a sample of 25 employees and obtained a mean job satisfaction score of M = 72. What is the probability of obtaining a sample mean greater than 72? In other words, out of all possible sample means that could be drawn from the sampling distribution, what proportion of them will be greater than 72?

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    72      
u = mean =    60      
n = sample size =    25      
s = standard deviation =    20      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    3      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3   ) =    0.001349898 [answer]