A car company says that the mean gas mileage for its luxury sedan is at least 20 miles per gallon $(\mathrm{mpg})$. You believe the claim is incorrect and find that a random sample of 6 cars has a mean gas mileage of $19 \mathrm{mpg}$ and a standard deviation of $3 \mathrm{mpg}$. 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 page 2 of the normal table. Which sampling distribution should be used and why? A. Use a t-sampling distribution because the population is normal, and $\sigma$ is known. C. Use a t-sampling distribution because the population is normal, and $\sigma$ is unknown. E. Use a t-sampling distribution because $n<30$. B. Use a normal sampling distribution because $n>30$. D. Use a normal sampling distribution because the population is normal, and $\sigma$ is unknown F. Use a normal sampling distribution because the population is normal, and $\sigma$ is known. example Get more help. Clear all Check answer
Solution
Step 1 ：The question is asking which sampling distribution should be used given the information provided. The information provided includes a sample size of 6, a population that is normally distributed, and an unknown population standard deviation.
Step 2 ：Given these conditions, we should use a t-sampling distribution. The t-distribution is used when the sample size is small (less than 30) and the population standard deviation is unknown.
Step 3 ：So, the answer should be 'Use a t-sampling distribution because the population is normal, and \(\sigma\) is unknown' or 'Use a t-sampling distribution because \(n<30\)'.
Step 4 ：Final Answer: \(\boxed{\text{C. Use a t-sampling distribution because the population is normal, and } \sigma \text{ is unknown and E. Use a t-sampling distribution because } n<30}\)
