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 1.24. the yards per carry of 25 randomly selected running backs are shown below. assume the yards per carry are normally distributed.
2.3 6.2 3.4 5.4 4.2 2.8 3.3 3.8 5.7 2.3 1.7 3.5 5.9 6.8 3.3 5.3 3.6 5.9 6.9 5.4 5.1 2.9 4.3 4.6 4.7
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

First, sum up all the data values:

$$
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The sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 25\). So \(\bar{x}=\frac{107.5}{25}=4.3\)

Step2: Determine the critical value

Since the population standard deviation \(\sigma=1.24\) is known and the data is normally distributed (given), we use the standard - normal distribution (\(z\) - distribution). For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99 = 0.01\), and \(\alpha/2=0.005\). The critical value \(z_{\alpha/2}\) is the \(z\) - score such that \(P(Z>z_{\alpha/2}) = 0.005\). From the standard - normal table, \(z_{\alpha/2}=z_{0.005} = 2.576\)

Step3: Calculate the margin of error

The margin of error \(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). Substitute \(\sigma = 1.24\), \(n = 25\), and \(z_{\alpha/2}=2.576\)

$$ E=2.576\times\frac{1.24}{\sqrt{25}}=2.576\times\frac{1.24}{5}=2.576\times0.248 = 0.64 $$

Step4: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\). Substitute \(\bar{x}=4.3\) and \(E = 0.64\)

$$ 4.3-0.64<\mu<4.3 + 0.64 $$

$$ 3.66<\mu<4.94 $$

Answer:

A. The \(99\%\) confidence interval is \((3.66,4.94)\)