Question

as part of a survey, a marketing representative asks a random sample of 27 business owners how much they would be willing to pay for a website for their company. she finds that the sample standard deviation is $5500. 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 (6950743, 30075903).
(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.)
o a. with 1% confidence, you can say that the
population variance is between
and.
o b. with 99% confidence, you can say that the
population variance is greater than

o c. with 1% confidence, you can say that the
population variance is less than

o d. with 99% confidence, you can say that the
population variance is between
and

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 of the chi - square distribution.

Given \(n = 27\), \(s=3500\), so \(s^{2}=(3500)^{2}=12250000\), and the confidence level \(c = 0.99\), then \(\alpha=1 - c=0.01\), \(\alpha/2 = 0.005\), \(1-\alpha/2=0.995\)

The degrees of freedom \(df=n - 1=27 - 1 = 26\)

From the chi - square distribution table, \(\chi_{\alpha/2}^{2}=\chi_{0.005}^{2}=45.642\), \(\chi_{1-\alpha/2}^{2}=\chi_{0.995}^{2}=11.524\)

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

The lower bound: \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}=\frac{26\times12250000}{45.642}\approx6950743\)

The upper bound: \(\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}=\frac{26\times12250000}{11.524}\approx30075903\)

Answer:

D. With \(99\%\) confidence, you can say that the population variance is between \(6950743\) and \(30075903\)