As part of a survey, a marketing representative asks a random sample of 27 business owners how much they would willing to pay for a website for their company. She finds that the sample standard deviation is \( \$ 3525 \). 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 ( \( \square, \square \) ). (Round to the nearest integer as needed.)

To construct a \( 99\% \) confidence interval for the population variance \( \sigma^2 \) using the sample standard deviation, we can use the Chi-Square distribution. The formula for the confidence interval for the population variance is given by: \[ \left( \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}} \right) \] Where: - \( n \) is the sample size, - \( s \) is the sample standard deviation, - \( \chi^2_{\alpha/2, n-1} \) is the critical value from the Chi-Square distribution for \( \alpha/2 \) and \( n-1 \) degrees of freedom. Given: - Sample size \( n = 27 \) - Sample standard deviation \( s = 3525 \) - Degrees of freedom \( df = n - 1 = 26 \) - Confidence level \( 99\% \) implies \( \alpha = 0.01 \), so \( \alpha/2 = 0.005 \) Now, we need to find the critical values from the Chi-Square distribution for \( df = 26 \): - \( \chi^2_{0.005, 26} \) (upper critical value) - \( \chi^2_{0.995, 26} \) (lower critical value) Using a Chi-Square table or calculator: - \( \chi^2_{0.005, 26} \approx 47.245 \) - \( \chi^2_{0.995, 26} \approx 10.645 \) Now we can calculate the confidence interval for the population variance: 1. Calculate \( s^2 \): \[ s^2 = 3525^2 = 12425725 \] 2. Calculate the lower limit: \[ \text{Lower limit} = \frac{(n-1)s^2}{\chi^2_{0.995, 26}} = \frac{26 \times 12425725}{47.245} \approx \frac{323870950}{47.245} \approx 6851955.56 \] 3. Calculate the upper limit: \[ \text{Upper limit} = \frac{(n-1)s^2}{\chi^2_{0.005, 26}} = \frac{26 \times 12425725}{10.645} \approx \frac{323870950}{10.645} \approx 30415663.56 \] Thus, the \( 99\% \) confidence interval for the population variance \( \sigma^2 \) is approximately: \[ (6851956, 30415664) \] ### Interpretation: This means we are \( 99\% \) confident that the true population variance of the amount business owners are willing to pay for a website lies between approximately \( 6,851,956 \) and \( 30,415,664 \) dollars squared. ### (b) Confidence interval for the population standard deviation \( \sigma \): To find the confidence interval for the population standard deviation \( \sigma \), we take the square root of the variance limits: 1. Lower limit for \( \sigma \): \[ \sqrt{6851956} \approx 2616.36 \] 2. Upper limit for \( \sigma \): \[ \sqrt{30415664} \approx 5520.36 \] Thus, the \( 99\% \) confidence interval for the population standard deviation \( \sigma \) is approximately: \[ (2616, 5520) \] ### Interpretation: We are \( 99\% \) confident that the true population standard deviation of the amount business owners are willing to pay for a website lies between approximately \( 2,616 \) and \( 5,520 \) dollars.