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} \\)? (, ) (round to the nearest integer as needed.)

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 14\). So \(df=14-1 = 13\).

Step2: Find the critical values

For a \(99\%\) confidence level (\(\alpha=1 - 0.99=0.01\)), and \(\frac{\alpha}{2}=0.005\).
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.995,13}^{2}=3.565\) and \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.005,13}^{2}=29.819\).
The sample variance \(s^{2}=(3620)^{2}=13104400\).

Step3: Calculate the confidence interval for the population variance

The formula for the confidence interval for 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 the values: \(\frac{(14 - 1)\times13104400}{29.819}\leq\sigma^{2}\leq\frac{(14 - 1)\times13104400}{3.565}\).
First, \(\frac{13\times13104400}{29.819}=\frac{170357200}{29.819}\approx5713052\).
Second, \(\frac{13\times13104400}{3.565}=\frac{170357200}{3.565}\approx47786031\).

Answer:

\((5713052,47786031)\)