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 $3593. 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 (round to the nearest integer as needed.)

Explanation:

Step1: Determine the degrees of freedom and critical values

The degrees of freedom \(df=n - 1=27-1 = 26\).
For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\frac{\alpha}{2}=0.005\).
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.995,26}^{2}=11.524\) and \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.005,26}^{2}=53.672\)

Step2: Calculate the confidence interval for the population variance \(\sigma^{2}\)

The formula for the confidence interval of the population variance is \(\frac{(n - 1)s^{2}}{\chi_{R}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\)
Given \(s = 3593\), then \(s^{2}=3593^{2}=12909649\)
\(\frac{(27 - 1)\times12909649}{53.672}\leq\sigma^{2}\leq\frac{(27 - 1)\times12909649}{11.524}\)
\(\frac{26\times12909649}{53.672}\leq\sigma^{2}\leq\frac{26\times12909649}{11.524}\)
\(\frac{335650874}{53.672}\leq\sigma^{2}\leq\frac{335650874}{11.524}\)
\(6253741\leq\sigma^{2}\leq29126250\)

Answer:

\((6253741,29126250)\)