A car company says that the mean gas mileage for its luxury sedan is at least 24 miles per gallon ( $\mathrm{mpg}$ ). You believe the claim is incorrect and find that a random sample of 8 cars has a mean gas mileage of $21 \mathrm{mpg}$ and a standard deviation of $5 \mathrm{mpg}$. At $\alpha=0.05$, test the company's claim. Assume the population is normally distributed. Click here to view the t-distribution table. Click here to view page 1 of the normal table. Click here to view pace 2 of the normal table.

Step 1 ：State the null hypothesis and the alternative hypothesis. Null hypothesis: \(\mu \geq 24\). Alternative hypothesis: \(\mu < 24\).

Step 2 ：Formulate an analysis plan. The significance level is defined as \(\alpha = 0.05\). A one-sample t-test will be used to test the hypotheses.

Step 3 ：Analyze sample data. Using the formula for the t-score: \(t = \frac{{\bar{x} - \mu}}{{s/\sqrt{n}}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the standard deviation of the sample and \(n\) is the sample size. Substituting the given values: \(t = \frac{{21 - 24}}{{5/\sqrt{8}}} = -2.121\).

Step 4 ：Interpret the results. The null hypothesis will be rejected if the calculated t-score is less than the critical t-value with degrees of freedom \(df = n - 1 = 8 - 1 = 7\). From the t-distribution table, the critical t-value for a one-tailed test at the 0.05 significance level with 7 degrees of freedom is approximately 1.895. Since the calculated t-score (-2.121) is less than the critical t-value (1.895), the null hypothesis is rejected. Therefore, \(\boxed{\text{The mean gas mileage for the car company's luxury sedan is less than 24 mpg.}}\)