Question

The dean of the College of Business at the University of La Verne has observed for several years and found that the probability distribution of the salary of the alumni’s first job after graduation is normal. The college collected information from 121 alumni and finds that the mean of their salary is $60k. Assuming a 95% confidence level, please do the following

1. Suppose the dean believes that the average salary of the population should be about $58k per year, with a standard deviation of $2k. We need to conclude that the mean salary is less than what the dean has believed to be. (Hint: Hypothesis Testing and Confidence Interval - One sample; population standard deviation is known)

(a) What are the null and alternate hypotheses?

(b) What is the level of significance?

(c) What is the standard error?

(d) Decide on the test statistic and calculate the value of the test statistic (hint: write the equation and calculate the statistic)?

(e) What’s your decision regarding the hypothesis and interpret the result using test-score rejection region rule or p value rule.

Answer 1

A)

Ho :   µ =   58                  
Ha :   µ <   58       (Left tail test)          
                          
b)

Level of Significance ,    α =    0.05                  

population std dev ,    σ =    2.0000                  
Sample Size ,   n =    121                  
Sample Mean,    x̅ =   60.0000                  
                             
                          
c)

Standard Error , SE = σ/√n =   2.0000   / √    121   =   0.1818      

d)

Z-test statistic= (x̅ - µ )/SE = (   60.000   -   58   ) /    0.1818   =   11.00
                          

e)

  
p-Value   =   1.0000   [ Excel formula =NORMSDIST(z) ]  

          
Decision:   p-value>α, Do not reject null hypothesis

there is not enough evidence to say that the mean salary is less than what the dean has believed to be

                      

please revert back for doubt