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 1.21. the yards per carry of 25 randomly selected running backs are shown below. assume the yards per carry are normally distributed.
1.6 3.7 6.3 3.6 4.5 6.7 6.3 4.4 3.3 2.5 1.3 2.7 4.8 5.1 3.4 3.2 6.3 3.8 3.4 6.5 4.9 3.9 2.5 7.4 4.8
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 normal distribution because n < 30, the data are normally distributed and σ is unknown.
c. use a t - distribution because n < 30 and σ is unknown.
d. use a t - distribution because n < 30 and σ is known.
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:

When constructing a confidence interval for the population mean, if the population standard deviation ($\sigma$) is known and the data is normally distributed (or the sample size $n$ is large enough for the Central Limit Theorem to apply), we use the normal distribution. Here, $n = 25$ (small sample, but $n<30$ is not a strict barrier if $\sigma$ is known and data is normal). The problem states that the population standard deviation ($\sigma = 1.21$) is known and the yards - per - carry are normally distributed.

Answer:

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