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.
(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
b. with 99% confidence, you can say that the
population variance is between
population variance is greater than
and
c. with 1% confidence, you can say that the
d. with 99% confidence, you can say that the
population variance is less than
population variance is between
6950743 and 30075903.
(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

For a \(99\%\) confidence interval of the population variance \(\sigma^{2}\) when the sample is from a normal distribution, the formula is \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}<\sigma^{2}<\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}\), where \(n = 27\), \(s = 3593\), \(s^{2}=3593^{2}=12909649\), \(\alpha=1 - 0.99 = 0.01\), \(\alpha/2=0.005\), \(1-\alpha/2 = 0.995\). The degrees of freedom \(df=n - 1=26\). From the \(\chi^{2}\) - distribution table, \(\chi_{0.005,26}^{2}=48.290\), \(\chi_{0.995,26}^{2}=11.524\).

Step2: Calculate the confidence interval for variance

Substitute the values into the formula:
\(\frac{(27 - 1)\times12909649}{48.290}<\sigma^{2}<\frac{(27 - 1)\times12909649}{11.524}\)
\(\frac{26\times12909649}{48.290}<\sigma^{2}<\frac{26\times12909649}{11.524}\)
\(\frac{335650874}{48.290}<\sigma^{2}<\frac{335650874}{11.524}\)
\(6950743<\sigma^{2}<30075903\)

Step3: Recall the formula for confidence interval of standard deviation

The confidence interval for the population standard deviation \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}}<\sigma<\sqrt{\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}}\)
Substitute the values: \(\sqrt{6950743}<\sigma<\sqrt{30075903}\)
\(2636<\sigma<5484\)

Answer:

(a) The \(99\%\) confidence interval for the population variance \(\sigma^{2}\) is \(6950743<\sigma^{2}<30075903\). Interpretation: With \(99\%\) confidence, you can say that the population variance is between \(6950743\) and \(30075903\) (Option D).
(b) The \(99\%\) confidence interval for the population standard deviation \(\sigma\) is \((2636,5484)\)