Question 32, 6.4.23-T
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The waiting times (in minutes) of a random sample of 22 people at a bank have a sample standard deviation of 4.6 minutes. Construct a confidence interval for the population variance $\sigma^{2}$ and the population standard deviation $\sigma$. Use a $90 \%$ level of confidence. Assume the sample is from a normally distributed population.

What is the confidence interval for the population variance $\sigma^{2}$ ?
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Step 1 ：Given values are: sample size \(n = 22\), sample standard deviation \(s = 4.6\), and level of significance \(\alpha = 0.10\).

Step 2 ：First, calculate the sample variance \(s^2 = s^2 = 21.16\).

Step 3 ：Next, calculate the critical values from the Chi-Square distribution. The lower critical value \(\chi^2_{lower} = 11.59\) and the upper critical value \(\chi^2_{upper} = 32.67\).

Step 4 ：Substitute these values into the formula to get the confidence interval for the population variance. The lower confidence interval \(CI_{lower} = 13.60\) and the upper confidence interval \(CI_{upper} = 38.34\).

Step 5 ：The confidence interval for the population variance \(\sigma^2\) is between 13.6 and 38.3. This means that we are 90% confident that the true population variance lies within this interval.

Step 6 ：Final Answer: The confidence interval for the population variance \(\sigma^2\) is \(\boxed{[13.6, 38.3]}\).