Question

Guidelines for the Jolly Blue Giant Health Insurance Company say that the average hospitalization for a triple hernia operation should not exceed 30 hours. Without knowing the population distribution or the population standard deviation, a diligent auditor studied records of 64 randomly chosen triple hernia operations at Hackmore Hospital. She found a mean hospital stay of 40 hours with a sample standard deviation of 24 hours. "Aha!" she cried, "The average stay exceeds the guideline." At α = .005, can she reject her null hypothesis?

Select one:

a. Yes, because her t_crit = 1.690 and her t_stat = 2.5

b. Yes, because her t_crit = 2.66 and her t_stat = 3.33

c. No, because her t_crit = 2.15 and her t_stat = 1.78

d. No, because z­­_crit = 1.96 and her z­_stat = 2.5

Answer 1

Solution

The null and alternative hypothesis is ,

H0 :   < 30

Ha : > 30

= 40

s = 24

n = 64

degrees of freedom = n - 1 = 64 - 1 = 63

Test statistic = t =

= ( - ) / s / n

= (40 - 30) / 24 / 64

Test statistic = t = 3.33

This is the right  tailed test

= 0.005  

P( t > t) = 0.005

= 1 - P( t < t) = 0.005

= P( t < t) = 1 -0.005

= P( t < t) = 0.995

=  P( t < 2.66) = 0.995

= critical value = 2.66

reject the null hypothesis, because test statistic > critical value

correct option is = b)