Question

A car dealer recommends that transmissions be serviced at 30,000 miles. To see whether her customers are adhering to this recommendation, the dealer selects a random sample of 40 customers and finds that the average mileage of the automobiles serviced is 30,456. The standard deviation of the population is 1684 miles. By finding the P-­value, determine whether the owners are having their transmissions serviced at 30,000 miles. Use α = 0.10. Are the owners having their transmissions serviced at 30,000 miles?

Answer 1

Answer:

No, the owners are not having their transmissions serviced at 30,000 miles.

Step-by-step explanation:

We are given that a car dealer recommends that transmissions be serviced at 30,000 miles.

The car dealer selects a random sample of 40 customers and finds that the average mileage of the automobiles serviced is 30,456. The standard deviation of the population is 1684 miles.

Let $\mu$ = true average mileage of the automobiles serviced.

So, Null Hypothesis, $H_0$ : $\mu$ = 30,000 miles      {means that the owners are having their transmissions serviced at 30,000 miles}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 30,000 miles      {means that the owners are having their transmissions serviced at different than 30,000 miles}

The test statistics that will be used here is One-sample z-test statistics because we know about population standard deviation;

                             T.S.  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~  N(0,1)

where, $\bar X$ = sample average mileage serviced = 30,456 miles  

            $\sigma$ = population standard deviation = 1684 miles

            n = sample of customers = 40

So, the test statistics =  $\frac{30,456-30,000}{\frac{1684}{\sqrt{40} } }$  

                                     =  1.71

The value of z-statistics is 1.71.

Also, the P-value of the test statistics is given by;

                    P-value = P(Z > 1.71) = 1 - P(Z $\leq$ 1.71)

                                  = 1 - 0.9564 = 0.0436

For the two-tailed test, the P-value is calculated as = 2 $\times$ 0.0436 = 0.0872.

Since the P-value of our test statistics is less than the level of significance as 0.0872 < 0.10, so we have sufficient evidence to reject our null hypothesis as the test statistics will fall in the rejection region.

Therefore, we conclude that the owners are having their transmissions serviced at different than 30,000 miles.