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.
\\( \mathrm { n } = 24 \\)
\\( \mathrm { s } = 14 \\) grains per gallon
what is the confidence interval for the population variance \\( \sigma ^ { 2 } \\)?
\\( ( 102.0,486.8 ) \\) (round to one decimal place as needed.)
interpret the results. select the correct choice below and fill in the answer box(es) to complete your choice.
(round to one decimal places as needed.)
\\( \bigcirc \\) a. with \\( 99 \\% \\) confidence, it can be said that the \\( \bigcirc \\) b. with \\( 99 \\% \\) confidence, it can be said that the
population variance is less than \\( \begin{array} { l } { square } end{array} \\). population variance is between \\( \square \\) and \\( \square \\).
\\( \bigcirc \\) c. with \\( 1 \\% \\) confidence, it can be said that the \\( \bigcirc \\) d. with \\( 1 \\% \\) confidence, it can be said that the
population variance is greater than \\( \square \\). population variance is between \\( \square \\) and \\( \square \\).

Explanation:

Step1: Recall the formula for confidence interval of population variance

The formula for the confidence interval of the population variance \(\sigma^{2}\) when the population is normally distributed is \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}<\sigma^{2}<\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}\), where \(n\) is the sample size, \(s\) is the sample standard deviation, \(\chi_{\alpha/2}^{2}\) and \(\chi_{1-\alpha/2}^{2}\) are the critical values from the chi - square distribution.

Given \(n = 24\), so the degrees of freedom \(df=n - 1=23\). For a \(99\%\) confidence interval, \(\alpha=1 - 0.99 = 0.01\), and \(\alpha/2=0.005\), \(1-\alpha/2 = 0.995\).

From the chi - square distribution table, \(\chi_{0.005,23}^{2}=44.181\) and \(\chi_{0.995,23}^{2}=9.260\), \(s = 14\), \(s^{2}=196\), \(n - 1=23\)

Step2: Calculate the lower and upper bounds of the confidence interval for \(\sigma^{2}\)

The lower bound: \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}=\frac{23\times196}{44.181}=\frac{4508}{44.181}\approx102.0\)
The upper bound: \(\frac{(n - 1)s^{2}}{\chi_{1 - \alpha/2}^{2}}=\frac{23\times196}{9.260}=\frac{4508}{9.260}\approx486.8\)

The interpretation of a confidence interval for the population variance: A \(99\%\) confidence interval for the population variance \(\sigma^{2}\) means that with \(99\%\) confidence, the population variance lies within the calculated interval.

Answer:

B. With \(99\%\) confidence, it can be said that the population variance is between \(102.0\) and \(486.8\)