Question

1) Big Cat Mutual Insurance (BCMI) claims that in 2019, the average monthly premium paid for individual health coverage will be $177. The Nittany Lion, who just turned 26, was kicked off his parent’s insurance plan. He suspects that the average monthly cost for the individual plan will actually be higher. On the company’s Terms and Conditions webpage, they claim that monthly premiums are normally distributed with a standard deviation of $50. Let µ denote the true average monthly premium.

a) The Nittany Lion graduated from IE and wants to use his statistical background to test his suspicion. What are the appropriate null and alternative hypotheses he should use?

b) The Nittany Lion randomly asks 15 of his closest friends what their monthly premium is for BCMI’s insurance plan and he calculated xbar. Due to lack of jobs since the football season is over, he will not be able to purchase from BCMI if the average monthly premium exceeds $200. Using this information, calculate the probability of committing a Type I error for the test and depict the error probability graphically.

c) Using the test procedure of part (b), what is the probability of committing a Type II error if µ= 210? Depict the Type I error from part (b) and the Type II error graphically.

Answer 1

a) This will be a right tailed test

null hypothesis :$H_0:\mu =177$

alternate hypothesis : $H_a:\mu >177$

b)formula for converting to standard Z score $Z=\frac{X-\mu _0}{\frac{\sigma }{\sqrt{n}}}$

since he will not be able to purchase from BCMI if the average monthly premium exceeds $200

probability of committing a Type I error is $P(X>200)=P(Z>\frac{200-177}{\frac{50}{\sqrt{15}}})=P(Z>1.78)$

$P(Z>1.78)=1-P(Z<1.78)$

$P(Z<1.78)=0.9625$

$P(Z>1.78)=1-0.9625=0.0375$

c) This is a right tailed test.

We will fail to reject the null (commit a Type II error) if we get a Z statistic less than 1.78.

This 1.78 Z-critical value corresponds to some X critical value = 200

So I will incorrectly fail to reject the null as long as a draw a sample mean that less than 200. To complete the problem what I now need to do is compute the probability of drawing a sample mean less than 200 given µ = 210. Thus, the probability of a Type II error is given by

$P(X<200)=P(Z<\frac{200-210}{\frac{50}{\sqrt{15}}})=P(Z<0.77)=0.7794$