Question

An automobile manufacturer has given its jeep a 26.5 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this jeep since it is believed that the jeep performs over the manufacturer's MPG rating. After testing 240 jeeps, they found a mean MPG of 26.7. Assume the population standard deviation is known to be 2.5. Is there sufficient evidence at the 0.05 level to support the testing firm's claim? Step 1 of 6: State the null and alternative hypotheses.

Answer 1

Answer:

The null hypothesis is $H_0: \mu = 26.5$

The alternate hypothesis is $H_a: \mu > 26.5$

The pvalue of the test is 0.1075 > 0.05, which means that there is not sufficient evidence at the 0.05 level to support the testing firm's claim.

Step-by-step explanation:

An automobile manufacturer has given its jeep a 26.5 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this jeep since it is believed that the jeep performs over the manufacturer's MPG rating.

At the null hypothesis, we test that the mean is 26.5 MPG, that is:

$H_0: \mu = 26.5$

At the alternate hypothesis, we test that the mean is above 26.5 MPG, that is:

$H_a: \mu > 26.5$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

26.5 is tested at the null hypothesis:

This means that $\mu = 26.5$

After testing 240 jeeps, they found a mean MPG of 26.7. Assume the population standard deviation is known to be 2.5.

This means that $n = 240, \mu = 26.7, \sigma = 2.5$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{26.7 - 26.5}{\frac{2.5}{\sqrt{240}}}$

$z = 1.24$

Pvalue of the test and decision:

The pvalue of the test is the probability of finding a mean above 26.7, which is 1 subtracted by the pvalue of z = 1.24.

Looking at the z-table, z = 1.24 has a pvalue of 0.8925

1 - 0.8925 = 0.1075

The pvalue of the test is 0.1075 > 0.05, which means that there is not sufficient evidence at the 0.05 level to support the testing firm's claim.