Question

use the standard normal distribution or the t - distribution to construct a 99% confidence interval for the population mean. justify your decision. if neither distribution can be used, explain why. interpret the results.
in a recent season, the population standard deviation of the yards per carry for all running backs was 1.24. the yards per carry of 25 randomly selected running backs are shown below. assume the yards per carry are normally distributed.
2.3 6.2 3.4 5.4 4.2 2.8 3.3 3.8 5.7 2.3 1.7 3.5 5.9
6.8 3.3 5.3 3.6 5.9 6.9 5.4 5.1 2.9 4.3 4.6 4.7
which distribution should be used to construct the confidence interval?
○ a. use a t - distribution because ( n<30 ) and ( sigma ) is unknown.
○ b. use a normal distribution because ( n<30 ), the data are normally distributed and ( sigma ) is unknown.
○ c. use a normal distribution because ( sigma ) is known and the data are normally distributed.
○ d. use a t - distribution because ( n<30 ) and ( sigma ) is known.
○ e. cannot use the standard normal distribution or the t - distribution because ( sigma ) is unknown, ( n<30 ), and the data are not normally distributed.

Explanation:

Step1: Recall the conditions for using normal and t - distributions

• For a confidence interval of the population mean \(\mu\):
• If the population standard deviation \(\sigma\) is known and the population is normally distributed (or \(n\geq30\) by the Central Limit Theorem), we use the standard normal distribution \(z\).
• If the population standard deviation \(\sigma\) is unknown and we estimate it with the sample standard deviation \(s\), and the population is normally distributed (or \(n\geq30\)), we use the \(t\) - distribution.

Step2: Analyze the given information

• We are given that \(n = 25<30\), but the population standard deviation \(\sigma=1.24\) (known), and the data (yards per carry) are normally distributed.

Answer:

C. Use a normal distribution because \(\sigma\) is known and the data are normally distributed.