Question

In: Statistics and Probability

Guidelines for the Jolly Blue Giant Health Insurance Company say that the average hospitalization for a...

Guidelines for the Jolly Blue Giant Health Insurance Company say that the average hospitalization for a triple hernia operation should not exceed 30 hours. A diligent auditor studied records of 16 randomly chosen triple hernia operations at Hackmore Hospital, and found a mean hospital stay of 40 hours with a standard deviation of 20 hours. "Aha!" she cried, "the average stay exceeds the guideline." The p-value for a right-tailed test of her hypothesis is:

Solutions

Expert Solution

Ho :   µ =   30                  
Ha :   µ >   30       (Right tail test)          
                          
Level of Significance ,    α =    0.050                  
sample std dev ,    s =    20.0000                  
Sample Size ,   n =    16                  
Sample Mean,    x̅ =   40.0000                  
                          
degree of freedom=   DF=n-1=   15                  
                          
Standard Error , SE = s/√n =   20.0000   / √    16   =   5.0000      
t-test statistic= (x̅ - µ )/SE = (   40.000   -   30   ) /    5.0000   =   2.000
                          
critical t value, t* =        1.753   [Excel formula =t.inv(α/no. of tails,df) ]              
                          
p-Value   =   0.0320   [Excel formula =t.dist(t-stat,df) ]              


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