Question

the annual earnings of 14 randomly selected computer software engineers have a sample standard deviation of $3620. assume the sample is from a normally distributed population. construct a confidence interval for the population variance \\( \sigma^{2} \\) and the population standard deviation \\( \sigma \\). use a 99% level of confidence. interpret the results.
what is the confidence interval for the population variance \\( \sigma^{2} \\)?
(5713052, 47786031) (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
5713052 and 47786031.
what is the confidence interval for the population standard deviation \\( \sigma \\)?
(\square.\square) (round to the nearest integer as needed.)

Explanation:

Step1: Recall the formula for confidence interval of population variance

The formula for the confidence interval of the population variance \(\sigma^{2}\) is \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}<\sigma^{2}<\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}\), where \(n = 14\), \(s=3620\), \(\alpha=1 - 0.99=0.01\), \(\alpha/2 = 0.005\), \(1-\alpha/2=0.995\). The degrees of freedom \(df=n - 1=13\).
From the \(\chi^{2}\) - distribution table, \(\chi_{\alpha/2}^{2}=\chi_{0.005}^{2}=29.819\) and \(\chi_{1-\alpha/2}^{2}=\chi_{0.995}^{2}=3.565\).

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

First, calculate \((n - 1)s^{2}=(14 - 1)\times3620^{2}=13\times13104400 = 170357200\).
The lower bound for \(\sigma^{2}\) is \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}=\frac{170357200}{29.819}\approx5713052\).
The upper bound for \(\sigma^{2}\) is \(\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}=\frac{170357200}{3.565}\approx47786031\).

Step3: Calculate the confidence interval for \(\sigma\)

Since \(\sigma=\sqrt{\sigma^{2}}\), the lower bound for \(\sigma\) is \(\sqrt{5713052}\approx2390\) and the upper bound for \(\sigma\) is \(\sqrt{47786031}\approx6913\).

Answer:

The confidence interval for the population variance \(\sigma^{2}\) is \((5713052,47786031)\).
The confidence interval for the population standard deviation \(\sigma\) is \((2390,6913)\).
For the interpretation, the correct choice is D.