Question 32, 6.4.23-T
HW Score: $86.46\mathrm{\%},27.67$ of 32 points
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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 $90\mathrm{\%}$ 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 ${\chi }_{lower}^2=11.59$ and the upper critical value ${\chi }_{upper}^2=32.67$.

Step 4 ：Substitute these values into the formula to get the confidence interval for the population variance. The lower confidence interval $C{I}_{lower}=13.60$ and the upper confidence interval $C{I}_{upper}=38.34$.

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 $\boxed{{\displaystyle [13.6,38.3]}}$.