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 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. 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: Determine the distribution

Since the population standard deviation \(\sigma = 1.26\) is known, and the sample size \(n=25<30\) but the data is normally distributed. We use the standard normal distribution (z - distribution). The formula for the confidence interval for the population mean is \(\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\).
First, calculate the sample mean \(\bar{x}\).
\(\bar{x}=\frac{2.5 + 5.8+5.7+3.7+7.4+4.8+6.3+4.5+7.2+2.3+1.4+6.4+2.9+5.7+2.9+3.6+5.2+5.3+5.6+6.2+5.4+3.6+5.4+4.7+6.1}{25}\)
\(\bar{x}=\frac{125}{25}=5\)
For a 95% confidence interval, \(\alpha = 1 - 0.95=0.05\), and \(\alpha/2=0.025\). The \(z\) - value \(z_{\alpha/2}=z_{0.025}\). From the standard normal table, \(z_{0.025} = 1.96\).

Step2: Calculate the margin of error

The margin of error \(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)
Substitute \(\sigma = 1.26\), \(n = 25\), and \(z_{\alpha/2}=1.96\) into the formula.
\(E=1.96\times\frac{1.26}{\sqrt{25}}=1.96\times\frac{1.26}{5}=1.96\times0.252 = 0.49392\)

Step3: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
Substitute \(\bar{x}=5\) and \(E = 0.49392\)
\(5-0.49392<\mu<5 + 0.49392\)
\(4.51<\mu<5.49\) (rounded to two decimal places)

Answer:

A. The 95% confidence interval is \((4.51,5.49)\)