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 random sample of 40 people, the mean body mass index (bmi) was 26.7 and the standard deviation was 6.07.
which distribution should be used to construct the confidence interval? choose the correct answer below.
a. use a t - distribution because the sample is random, the population is normal, and \\( \sigma \\) is unknown.
b. use a normal distribution because the sample is random, the population is normal, and \\( \sigma \\) is known.
c. use a t - distribution because the sample is random, \\( n\geq30 \\), and \\( \sigma \\) is unknown.
d. use a normal distribution because the sample is random, \\( n\geq30 \\), and \\( \sigma \\) is known
e. neither a normal distribution nor a t - distribution can be used because either the sample is not random, or \\( n < 30 \\), and the population is not known to be normal.
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: Determine the distribution

We are given a sample size \(n = 40\) (\(n\geq30\)), the sample mean \(\bar{x}=26.7\), and the sample standard deviation \(s = 6.07\) (population standard deviation \(\sigma\) is unknown). According to the Central Limit Theorem, when \(n\geq30\) and \(\sigma\) is unknown, we use the \(t -\)distribution.

Step2: Calculate the degrees of freedom and the \(t -\)value

The degrees of freedom is \(df=n - 1=40-1 = 39\). For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99=0.01\), and \(\frac{\alpha}{2}=0.005\). Using a \(t -\)table or a calculator, the \(t -\)value \(t_{\frac{\alpha}{2}}\) with \(df = 39\) is approximately \(t_{0.005}\approx 2.708\)

Step3: Calculate the margin of error \(E\)

The formula for the margin of error for a \(t -\)distribution is \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\). Substitute \(t_{\frac{\alpha}{2}} = 2.708\), \(s = 6.07\), and \(n = 40\) into the formula:

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Step4: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\). Substitute \(\bar{x}=26.7\) and \(E = 2.60\) into the formula:

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Answer:

A. The \(99\%\) confidence interval is \((24.10,29.30)\)