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: Find the degrees of freedom and critical values

The degrees of freedom \(df=n - 1\), where \(n = 27\), so \(df=27-1 = 26\).
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 \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.005,26}^{2}=53.672\).
The sample variance \(s^{2}=(3593)^{2}\).

Step2: Calculate the confidence interval for the population variance

The formula for the confidence interval of the population variance \(\sigma^{2}\) is \(\frac{(n - 1)s^{2}}{\chi_{R}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\).
Substitute \(n = 27\), \(s = 3593\), \(\chi_{L}^{2}=11.524\), \(\chi_{R}^{2}=53.672\) into the formula:
\(\frac{(27 - 1)\times(3593)^{2}}{53.672}\leq\sigma^{2}\leq\frac{(27 - 1)\times(3593)^{2}}{11.524}\)
First, calculate \((27 - 1)\times(3593)^{2}=26\times12909649 = 335650874\)
Then \(\frac{335650874}{53.672}\approx6253700\) and \(\frac{335650874}{11.524}\approx29126200\)

Answer:

\((6253700,29126200)\)