Question

the waiting times (in minutes) of a random sample of 20 people at a bank have a sample standard deviation of 3.2 minutes. construct a confidence interval for the population variance $sigma^{2}$ and the population standard deviation $sigma$. use a 95% level of confidence. assume the sample is from a normally distributed population.
what is the confidence interval for the population variance $sigma^{2}$?
(5.9, 21.8) (round to one decimal place as needed.)
interpret the results. select the correct choice below and fill in the answer box(es) to complete your choice.
(round to one decimal place as needed.)
a. with 5% confidence, you can say that the population variance is greater than

b. with 95% confidence, you can say that the population variance is between and

c. with 5% confidence, you can say that the population variance is between and

d. with 95% confidence, you can say that the population variance is less than

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 \(n\) is the sample size, \(s\) is the sample standard deviation, \(\chi_{\alpha/2}^{2}\) and \(\chi_{1-\alpha/2}^{2}\) are the critical values from the chi - square distribution.
Given \(n = 20\), \(s=3.2\), \(\alpha=1 - 0.95 = 0.05\), so \(\alpha/2=0.025\) and \(1-\alpha/2 = 0.975\).
The degrees of freedom \(df=n - 1=19\).
From the chi - square distribution table, \(\chi_{0.025}^{2}(19)=32.852\) and \(\chi_{0.975}^{2}(19)=8.907\).
\((n - 1)s^{2}=(20 - 1)\times3.2^{2}=19\times10.24 = 194.56\)

Step2: Calculate the lower and upper bounds

Lower bound: \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}=\frac{194.56}{32.852}\approx5.9\)
Upper bound: \(\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}=\frac{194.56}{8.907}\approx21.8\)

Answer:

B. With \(95\%\) confidence, you can say that the population variance is between \(5.9\) and \(21.8\).