Question

Bastien, Inc. has been manufacturing small automobiles that have averaged 50 miles per gallon of gasoline in highway driving. The company has developed a more efficient engine for its small cars and now advertises that its new small cars average more than 50 miles per gallon in highway driving. An independent testing service road-tested 36 of the automobiles. The sample showed an average of 51.5 miles per gallon. The population standard deviation is 6 miles per gallon. The population standard deviation is 6 miles per gallon.

Required:
a. State the hypotheses.
b. What is the p-value associated with the sample results? With a 0.05 level of significance, test to determine whether or not the manufacturer's advertising campaign is legitimate.

Answer 1

Using the z-distribution, we have that:

a)

• $H_0: \mu = 50$

• $H_1: \mu > 50$

b) Since the p-value of the test is of 0.0668 > 0.05, there is not enough evidence to conclude that the manufacturer's advertising campaign is legitimate.

Item a:

At the null hypothesis, it is tested if the mean is still of 50 miles per gallon, that is:

$H_0: \mu = 50$

At the alternative hypothesis, it is tested if the mean has increased, that is:

$H_1: \mu > 50$

Item b:

We have the standard deviation for the population, hence, the z-distribution is used to solve this question.

The test statistic is given by:

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

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• $\sigma$ is the standard deviation of the sample.

• n is the sample size.

Hence, the values of the parameters are:

$\overline{x} = 51.5, \mu = 50, \sigma = 6, n = 36$

Hence, the value of the test statistic is:

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

$z = \frac{51.5 - 50}{\frac{6}{\sqrt{36}}}$

$z = 1.5$

Using a z-distribution calculator, with a right-tailed test, as we are testing if the mean is greater than a value, with z = 1.5, the p-value is of 0.0668.

Since the p-value of the test is of 0.0668 > 0.05, there is not enough evidence to conclude that the manufacturer's advertising campaign is legitimate.

You can learn more about the use of the z-distribution to test an hypothesis at https://brainly.com/question/16313918