Question

according to a leasing firms reports, the mean number of miles driven annually in its leased cars is 12,700 miles with a standard deviation of 1060 miles. the company recently starting using new contracts which require customers to have the cars serviced at their own expense. the companys owner believes the mean number of miles driven annually under the new contracts, μ, is less than 12,700 miles. he takes a random sample of 9 cars under the new contracts. the cars in the sample had a mean of 11,805 annual miles driven. assume that the population is normally distributed. is there support for the claim, at the 0.05 level of significance, that the population mean number of miles driven annually by cars under the new contracts, is less than 12,700 miles? assume that the population standard deviation of miles driven annually was not affected by the change to the contracts. perform a one - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places, and round your responses as specified below. (if necessary, consult a list of formulas.) (a) state the null hypothesis h0 and the alternative hypothesis h1. h0:□ h1:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) □ (d) find the p - value. (round to three or more decimal places.) □ (e) can we support the claim that the population mean number of miles driven annually by cars under the new contracts is less than 12,700 miles? oyes ono

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is the status - quo, and the alternative hypothesis $H_1$ is the claim. So, $H_0:\mu = 12700$ and $H_1:\mu<12700$.

Step2: Determine test - statistic type

Since the population standard deviation $\sigma = 1060$ is known and the population is normally distributed, we use the z - test statistic.

Step3: Calculate the z - test statistic

The formula for the z - test statistic is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}=11805$, $\mu = 12700$, $\sigma = 1060$, and $n = 9$.
$z=\frac{11805 - 12700}{\frac{1060}{\sqrt{9}}}=\frac{-895}{\frac{1060}{3}}=\frac{-895\times3}{1060}\approx - 2.524$

Step4: Find the p - value

For a one - tailed z - test with $z\approx - 2.524$, we look up the value in the standard normal distribution table. The p - value is $P(Z < - 2.524)\approx0.006$

Step5: Make a decision

Since the p - value $0.006<0.05$ (the level of significance), we reject the null hypothesis.

Answer:

(a) $H_0:\mu = 12700$, $H_1:\mu<12700$
(b) z - test statistic
(c) $-2.524$
(d) $0.006$
(e) Yes