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:

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

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq