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.
select the correct choice below and, if necessary, fill in any answer boxes to complete your choice.
a. the 99% confidence interval is (□,□).
(round to two decimal places as needed.)
b. neither distribution can be used to construct the confidence interval

Explanation:

Step1: Calculate the sample mean

The formula for the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Sum up all the data values: \(1.6+3.7+6.3+3.6+4.5+6.7+6.3+4.4+3.3+2.5+1.3+2.7+4.8+5.1+3.4+3.2+6.3+3.8+3.4+6.5+4.9+3.9+2.5+7.4+4.8 = 100.5\)
\(n = 25\), so \(\bar{x}=\frac{100.5}{25}=4.02\)

Step2: Find the z - value for a 99% confidence interval

For a 99% confidence interval, the significance level \(\alpha=1 - 0.99 = 0.01\), and \(\alpha/2=0.005\).
Looking up in the standard normal distribution table, \(z_{\alpha/2}=z_{0.005} = 2.576\)

Step3: Calculate the margin of error

The formula for the margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)
Given \(\sigma = 1.21\), \(n = 25\)
\(E=2.576\times\frac{1.21}{\sqrt{25}}=2.576\times\frac{1.21}{5}=2.576\times0.242 = 0.623392\)

Step4: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
\(4.02-0.623392<\mu<4.02 + 0.623392\)
\(3.396608<\mu<4.643392\)

Answer:

A. The 99% confidence interval is \((3.40,4.64)\)