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
( 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.)
a. with 99% confidence, it can be said that the
b. with 99% confidence, it can be said that the
population variance is less than
population variance is between 102.0 and 486.8
c. with 1% confidence, it can be said that the
d. with 1% confidence, it can be said that the
population variance is greater than
population variance is between and
what is the confidence interval for the population standard deviation \\( \sigma \\)?
( ) (round to one decimal place 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 \((\sqrt{a},\sqrt{b})\).

Step2: Calculate the square - roots of the endpoints of the variance confidence interval

Given the confidence interval for the variance \(\sigma^{2}\) is \((102.0,486.8)\).
For the lower bound of the standard deviation: \(\sqrt{102.0}\approx10.1\)
For the upper bound of the standard deviation: \(\sqrt{486.8}\approx22.1\)

Answer:

\((10.1,22.1)\)