Question

as part of a survey, a marketing representative asks a random sample of 21 business owners how much they would be willing to pay for a website for their company. she finds that the sample standard deviation is $553. assume the sample is taken from a normally distributed population. construct 99% confidence intervals for (a) the population variance $\sigma^{2}$ and (b) the population standard deviation $\sigma$. interpret the results.
(a) the confidence interval for the population variance is (6,940,763, 30,515,503) (round to the nearest integer as needed.)
interpret the results. select the correct choice below and fill in the answer box(es) to complete your choice. (round to the nearest integer as needed.)
a. with 1% confidence, you can say that the population variance is between □ and □.
b. with 99% confidence, you can say that the population variance is greater than □.
c. with 1% confidence, you can say that the population variance is less than □.
d. with 99% confidence, you can say that the population variance is between 6940763 and 30515503.
(b) the confidence interval for the population standard deviation is (□,□) (round to the nearest integer as needed.)

Explanation:

Step1: Recall the formula for confidence interval of variance

The formula for the confidence interval of the population variance \(\sigma^{2}\) when the sample is from a normal population 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, and \(\chi_{\alpha/2}^{2}\) and \(\chi_{1 - \alpha/2}^{2}\) are the critical values of the chi - square distribution.

Given \(n = 21\), \(s=3103\), and the confidence level \(C = 0.99\), then \(\alpha=1 - 0.99 = 0.01\), and \(\alpha/2=0.005\), \(1-\alpha/2 = 0.995\). The degrees of freedom \(df=n - 1=20\).

From the chi - square distribution table, \(\chi_{0.005,20}^{2}=39.997\) and \(\chi_{0.995,20}^{2}=7.434\)

Step2: Calculate the confidence interval for variance

First, calculate \((n - 1)s^{2}=(21 - 1)\times(3103)^{2}=20\times9628609 = 192572180\)

Then, \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}=\frac{192572180}{39.997}\approx4814657\approx4814657\) (rounded to the nearest integer)

\(\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}=\frac{192572180}{7.434}\approx25904248\approx25904248\) (rounded to the nearest integer)

Step3: Recall the formula for confidence interval of standard deviation

The formula for the confidence interval of the population standard deviation \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}}<\sigma<\sqrt{\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}}\)

We know that \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}\approx4814657\) and \(\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}\approx25904248\)

\(\sqrt{4814657}\approx2194\) and \(\sqrt{25904248}\approx5090\)

Answer:

(a) For the interpretation of the variance confidence interval, the correct choice is D. With \(99\%\) confidence, you can say that the population variance is between \(4814657\) and \(25904248\)

(b) The confidence interval for the population standard deviation is \((2194,5090)\)