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 41 people, the mean body mass index (bmi) was 26.9 and the standard deviation was 6.14.
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, ( ngeq30 ), 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, the population is normal, and ( sigma ) is unknown.
d. use a normal distribution because the sample is random, ( ngeq30 ), 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

Since the sample is random, \(n = 41\geq30\), and the population standard deviation \(\sigma\) is unknown (we are given the sample standard deviation \(s = 6.14\)), we use the t - distribution.

Step2: Calculate the confidence interval

The formula for the confidence interval for the population mean when using the t - distribution is \(\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}\)
For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\). The degrees of freedom \(df=n - 1=41-1 = 40\).
From the t - distribution table, \(t_{0.005,40}\approx 2.704\)
We are given \(\bar{x}=26.9\), \(s = 6.14\), \(n = 41\)
The margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}=2.704\times\frac{6.14}{\sqrt{41}}\)
First, calculate \(\frac{6.14}{\sqrt{41}}\approx\frac{6.14}{6.403}\approx0.959\)
Then \(E = 2.704\times0.959\approx2.6\)
The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\)
\(26.9- 2.6<\mu<26.9 + 2.6\)

Answer:

The \(99\%\) confidence interval is \((24.3,29.5)\)