Question

use the standard normal distribution or the t - distribution to construct a 95% 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 1.26. the yards per carry of 25 randomly selected running backs are shown below. assume the yards per carry are normally distributed 2.5 5.8 5.7 3.7 7.4 4.8 6.3 4.5 7.2 2.3 1.4 6.4 2.9 5.7 2.9 3.6 5.2 5.3 5.6 6.2 5.4 3.6 5.4 4.7 6.1 which distribution should be used to construct the confidence interval? a. use a normal distribution because σ is known and the data are normally distributed. b. use a t - distribution because n < 30 and σ is unknown. c. use a t - distribution because n < 30 and σ is known. d. use a normal distribution because n < 30, the data are normally distributed and σ is unknown. e. cannot use the standard normal distribution or the t - distribution because σ is unknown, n < 30, and the data are not normally distributed.

Explanation:

Step1: Recall the conditions for normal and t - distributions

• For a normal distribution ($z$-distribution): We use it when the population standard deviation $\sigma$ is known and either the sample size $n\geq30$ or the population is normally distributed.
• For a t - distribution: We use it when the population standard deviation $\sigma$ is unknown and either the sample size $n\geq30$ or the population is normally distributed.

Step2: Analyze the given information

We are given that the population standard deviation $\sigma = 1.26$ (so $\sigma$ is known), the sample size $n = 25<30$, and the yards per carry are normally distributed.

Answer:

A. Use a normal distribution because $\sigma$ is known and the data are normally distributed.