A manufacturer of a new car claims the typical car will average 32 mpg of gasoline. An independent consumer group is skeptical of the claim and thinks the mean gas mileage is significantly different than the 32 claimed. A sample of 23 randomly selected cars produced a mean mpg of 30.15 with a standard deviation of 4.87. Assume that cars’ gas mileage is normally distributed in the population.(8 pts)

a. The manufactures want to determine if the mean mpg for the cars is significantly different from 32. Write the null and alternative hypotheses for this question.
b. Are the assumptions necessary for inference met?
c. Perform the appropriate test including the formula that you used, the calculated value of the t-statistic, and the t-critical you compared your t-statistic to using a significance level of .05. Then, state your conclusion to the test.
d. Now, suppose the researchers had wanted to find evidence specifically that the car obtained less than 32 mpg. Rewrite your null and alternative hypothesis, then repeat the test and state your conclusion.

a. The null hypothesis (H0) for this question is that the mean gas mileage (μ) for the cars is equal to 32. The alternative hypothesis (Ha) is that the mean gas mileage for the cars is significantly different from 32.

H0: μ = 32
Ha: μ ≠ 32

b. To determine if the assumptions necessary for inference are met, we need to check two things: the random sampling assumption and the normality assumption.

The question states that the sample of 23 cars was randomly selected, so the random sampling assumption is satisfied.

For the normality assumption, it is stated that the cars' gas mileage in the population is normally distributed. We can assume this is the case.

c. To perform the appropriate test, we will use a t-test. The formula for the t-statistic is given by:

t = (x̄ - μ) / (s / √n)

Where:
x̄ is the sample mean (30.15),
μ is the population mean (32),
s is the sample standard deviation (4.87), and
n is the sample size (23).

Substituting the given values into the formula, we get:

t = (30.15 - 32) / (4.87 / √23) ≈ -1.352

To compare this calculated t-statistic to the t-critical value, we need to determine the degrees of freedom (df). In this case, df = n - 1 = 23 - 1 = 22.

Using a significance level of 0.05, we can look up the t-critical value in a t-distribution table or use a statistical software. Assuming a two-tailed test, the t-critical value for df = 22 and α = 0.05 is approximately ±2.074.

Since |t| = 1.352 < 2.074, we fail to reject the null hypothesis (H0). There is not enough evidence to conclude that the mean gas mileage for the cars is significantly different from 32.

Therefore, we can conclude that the claim made by the manufacturer is reasonable based on the given data.

d. If the researchers wanted to find evidence specifically that the car obtained less than 32 mpg, the null and alternative hypotheses would be:

H0: μ ≥ 32 (mean gas mileage is greater than or equal to 32)
Ha: μ < 32 (mean gas mileage is less than 32)

To perform the test, we would calculate the t-statistic using the same formula as before:

t = (x̄ - μ) / (s / √n)

Substituting the given values into the formula, we get:

t = (30.15 - 32) / (4.87 / √23) ≈ -1.352

Since this is a one-tailed test (looking for evidence that the mean is less than 32), we compare the calculated t-statistic to the t-critical value for the given significance level. Using a significance level of 0.05 and the same degrees of freedom (df = 22), the t-critical value is approximately -1.717.

Since t = -1.352 > -1.717, we fail to reject the null hypothesis (H0). There is not enough evidence to conclude that the mean gas mileage for the cars is less than 32.

Therefore, based on the given data, we cannot conclude that the car obtains less than 32 mpg.