Question

as part of a water quality survey, you test the water hardness in several randomly selected streams. the results are shown below. construct a confidence interval for the population variance \\( \sigma ^ { 2 } \\) and the population standard deviation \\( \sigma \\). use a 99% level of confidence. assume that the population has a normal distribution. \\( n = 24 \\) \\( s = 14 \\) grains per gallon what is the confidence interval for the population variance \\( \sigma ^ { 2 } \\)? (round to one decimal place as needed.)

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\). Given \(n = 24\), so \(df=24- 1=23\).

Step2: Find the critical values

For a \(99\%\) confidence level, the area in each tail is \(\frac{1 - 0.99}{2}=0.005\).
Using the \(\chi^{2}\) - distribution table or a calculator, \(\chi_{L}^{2}=\chi_{1 - 0.005,23}^{2}=\chi_{0.995,23}^{2}=9.260\) and \(\chi_{R}^{2}=\chi_{0.005,23}^{2}=44.181\).

Step3: Calculate the confidence interval for the population variance \(\sigma^{2}\)

The formula for the confidence interval for the population variance is \(\frac{(n - 1)s^{2}}{\chi_{R}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\).
Substitute \(n = 24\), \(s = 14\), \(\chi_{L}^{2}=9.260\), and \(\chi_{R}^{2}=44.181\) into the formula.
First, \((n - 1)s^{2}=(24 - 1)\times14^{2}=23\times196 = 4508\).
Then \(\frac{4508}{44.181}\leq\sigma^{2}\leq\frac{4508}{9.260}\).
\(\frac{4508}{44.181}\approx102.0\) and \(\frac{4508}{9.260}\approx486.8\).

Answer:

\((102.0,486.8)\)