Question

use technology to construct the confidence intervals for the population variance $\sigma^{2}$ and the population standard deviation $\sigma$. assume the sample is taken from a normally distributed population.
$c = 0.90$, $s = 37$, $n = 17$
the confidence interval for the population variance is $(832.97, 2751.07)$.
(round to two decimal places as needed.)
the confidence interval for the population standard deviation is $(\square,\square)$.
(round to two decimal places as needed.)

Explanation:

Step1: Recall the relationship between variance and standard deviation

The standard deviation \(\sigma\) is the square - root of the variance \(\sigma^{2}\). If the confidence interval for the variance is \((a,b)\), then the confidence interval for the standard deviation is \((\sqrt{a},\sqrt{b})\).
We are given that the confidence interval for the variance \(\sigma^{2}\) is \((832.97,2751.07)\).

Step2: Calculate the square - roots

For the lower bound: \(\sqrt{832.97}\approx28.86\)
For the upper bound: \(\sqrt{2751.07}\approx52.45\)

Answer:

The confidence interval for the population standard deviation is \((28.86,52.45)\)