21 February, 21:26

# 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:

+4
1. 21 February, 21:50
0
To solve this problem, we make use of the t statistic. So what we have to do is to calculate for the t score in this case and using the standard distribution tables, we locate for the p value using the t score value.

The formula for t score is given as:

t = (x - μ) / (σ / sqrt (n))

Where,

x = statistical mean of the sample = 40

n=number of individuals in the sample = 16

s=standard deviation = 20

μ=Population Mean = 30

Substituting the given values:

t = (40 - 30) / (20 / sqrt 16)

t = 10 / 5

t = 2

Using the standard distribution tables at degrees of freedom = n - 1 = 15, the value of p to the right is:

p = 0.032