Question

A major discount stock brokerage claims that customers calling in to make trades on the stock market are left on hold for an average of about 22 seconds. Suppose an executive is concerned about the satisfaction levels of the company’s clients and selects a random sample of 100 clients who phoned. The length of time that a client was left on hold was recorded. Assume that the sample mean was 23.5 second on hold. Suppose it is known that the standard deviation of all hold times is 8 seconds. Is there sufficient evidence to indicate that customers are left on hold for an average length of time greater than 22 seconds? Use a α = 0.05 level of significance and the P-value method.

Answer 1

Ho :   µ =   22                  
Ha :   µ >   22       (Right tail test)          
                          
Level of Significance ,    α =    0.05                  
population std dev ,    σ =    8.0000                  
Sample Size ,   n =    100                  
Sample Mean,    x̅ =   23.5000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   8.0000   / √    100   =   0.8000      
Z-test statistic= (x̅ - µ )/SE = (   23.500   -   22   ) /    0.8000   =   1.875
                          
  
p-Value   =   0.0304 [ Excel formula =NORMSDIST(z) ]              
Decision:   p-value<α, Reject null hypothesis                       
Conclusion: There is enough evidence to conclude that   customers are left on hold for an average length of time greater than 22 seconds