Question

A consumer advocacy group is doing a large study on car rental practices. Among other things, the consumer group would like to do a statistical test regarding the mean monthly mileage, H, of cars rented in the US, this year. The consumer group has reason to believe that the mean monthly mileage of cars rented in the U.S. this year is less than last year's mean, which was 2750 miles. The group chooses a random sample of 75 cars rented in the U.S. this year and records their monthly mileages Suppose that the population of monthly mileage of cars rented in the U.S. this year has a standard deviation of 750 miles and that the group performs its hypothesis test using the 0.05 level of significance. Based on this information, answer the questions below. Carry your intermediate computations to at least four decimal places, and round your responses as indicated. (If necessary, consult a list of formulas.) Huis What are the null and alternative hypotheses that the group should use for the test? less than less than or equal to greater than er not to to Assuring that the actual value of us 2507 miles, what is the probability that the group rejects the null hypothesis? Round your response to at least two decimal places 0 What is the probability that the group rejects the null wypothesis when, in fact, it is true? Round your response to at least two decimal places The probability of committing Type Iterror in the second test is greater Suppose that the group decides to perform another Masticates using the same population, the same null and alternatives and the same samples but for this second test the group ses a wonificance vel of 0.01 instead of a significance level of 0.05 Assuming that the actual value of wis 2507 miles, how does the probability that the group commits Type error in the second tout compare to the probably that the group commits a Type Il error in the original sest? The probability of committing a Type Il error in the second The probabilities of committing a Type Iterror are equal

Answer 1

sample size = n = 75

H₀ : $\mu$ = 2750 miles

H_a : $\mu$ < 2750 miles

Population standard deviation = $\sigma$ = 750 miles

standard error = se=  $\sigma$/sqrt(n) = 750/sqrt(75) = 86.60 miles

Here we will reject the null hypothesis if $\bar x$ <2750 + NORMSINV(0.05) * se

$\bar x$ <  2750 - 1.645 * 86.60

$\bar x$ < 2607.55 miles

Now here true value is = 2507 miles

HO

so here P(Reject the null hypothesis) = P($\bar x$ < 2607.55 miles ; = 2507 miles ; se = 86.80 miles)

HO

z = (2607.55 - 2507)/86.80

z = 1.161

P(Reject the null hypothesis) = P(Z < 1.161) = 0.8772

Probability that the group will reject the null hypothesis when even it is true,then it would be significance level = 0.05

Here significance level is reduced to 0.01 from 0.05, so that would decrease the rejection region, so that would also increase the probability of type II error. So here in this case, type II error willl increase with respect to previous case.