The guidelines for the Jolly Blue Giant health insurance company say that the average hospitalization for...

The 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 the records of 16 triple-hernia operations randomly selected at Hackmore Hospital and found an average hospital stay of 40 hours with a standard deviation of 20 hours. "Aha!" He exclaimed, "the average stay exceeds the standard." The p-value for this hypothesis test is

Select one: a. between .05 and .10

b. between .025 and .05

c. between .01 and .025

d. less than .01

Expert Solution

Ho : mu = 30

H1 : mu>30

we use t test since the sample is less than 30

t = xbar - mu / (s/√n)

40 -30 / (20/√16) =2

P value at t = 2 at d.f 15 is 0.031 so the appropriate p vaoue interval for rejection of null hypothesis is 0.05 to 0.10.

orchestra answered 2 years ago

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:

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