Question

a car company says that the mean gas mileage for its luxury sedan is at least 23 miles per gallon (mpg). you believe the claim is incorrect and find that a random sample of 7 cars has a mean gas mileage of 21 mpg and a standard deviation of 4 mpg. at a = 0.05, test the companys 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 σ is known. b. use a normal sampling distribution because the population is normal, and σ is known. c. use a t-sampling distribution because n < 30. d. use a normal sampling distribution because n > 30. e. use a normal sampling distribution because the population is normal, and σ is unknown. f. use a t-sampling distribution because the population is normal, and σ is unknown.

Explanation:

Step1: Recall sampling distribution rules

When the population is normally distributed, if the population standard deviation \(\sigma\) is unknown (here we are given the sample standard deviation \(s = 4\) and not \(\sigma\)), and the sample size \(n<30\) (here \(n = 7<30\)), we use the \(t -\)sampling distribution.

Step2: Analyze each option

• Option A: Incorrect because \(\sigma\) is unknown.
• Option B: Incorrect because \(\sigma\) is unknown.
• Option C: Incorrect, the reason \(n < 30\) alone is not sufficient when we consider the normality of the population and the fact that \(\sigma\) is unknown.
• Option D: Incorrect as \(n=7<30\).
• Option E: Incorrect, when \(\sigma\) is unknown we don't use the normal sampling distribution.
• Option F: Correct. The population is normal, \(\sigma\) is unknown (we use \(s\) instead), and \(n = 7<30\), so we use a \(t -\)sampling distribution.

Answer:

F. Use a \(t -\)sampling distribution because the population is normal, and \(\sigma\) is unknown.