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 1.38. the yards per carry of 25 randomly selected running backs are shown below. assume the yards per carry are normally distributed.
1.8 2.8 7.4 5.7 2.7 5.7 4.9 3.7 4.4 1.5 2.2 5.2 2.9
5.4 5.3 3.5 2.9 3.8 5.5 3.9 4.1 3.5 3.2 6.6 4.3
which distribution should be used to construct the confidence interval?
a. use a t - distribution because ( n<30 ) and ( sigma ) is known.
b. use a t - distribution because ( n<30 ) and ( sigma ) is unknown.
c. use a normal distribution because ( n<30 ), the data are normally distributed and ( sigma ) is unknown.
d. use a normal distribution because ( sigma ) is known and the data are normally distributed.
e. cannot use the standard normal distribution or the t - distribution because ( sigma ) 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 95% 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

First, sum up all the data values:

$$
LATEXBLOCK0
$$

The sample size \(n = 25\). The sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{100.5}{25}=4.02\)

Step2: Find the critical value

Since we use the normal distribution (\(\sigma = 1.38\) is known and data is normal), for a \(95\%\) confidence interval, the significance level \(\alpha=1 - 0.95=0.05\), and \(\alpha/2=0.025\). The critical value \(z_{\alpha/2}\) from the standard normal distribution table is \(z_{0.025}=1.96\)

Step3: Calculate the margin of error

The formula for the margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)
Substitute \(z_{\alpha/2}=1.96\), \(\sigma = 1.38\), \(n = 25\)

$$ E=1.96\times\frac{1.38}{\sqrt{25}}=1.96\times\frac{1.38}{5}=1.96\times0.276 = 0.54096 $$

Step4: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
Substitute \(\bar{x}=4.02\) and \(E = 0.54096\)

$$ 4.02-0.54096<\mu<4.02 + 0.54096 $$

$$ 3.48<\mu<4.56 $$

Answer:

A. The \(95\%\) confidence interval is \((3.48,4.56)\)