The 95% confidence interval estimate of a population variance when a sample variance of 324 is obtained from a sample of 81 items is
A) 14.14 to 174.94.
B) 243.086 to 453.520.
C) 16.42 to 194.35.
D) 254.419 to 429.203.

Question

Quizplus · Accepted Answer

The 95% confidence interval for a population variance is calculated using the chi-square distribution with the formula: $$ \left( \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}} \right) $$ where $n$ is the sample size, $s^2$ is the sample variance, and $\chi^2_{\alpha/2, n-1}$ and $\chi^2_{1-\alpha/2, n-1}$ are the chi-square critical values for $n-1$ degrees of freedom. Given $n=81$, $s^2=324$, and a 95% confidence level (so $\alpha=0.05$), we can find the chi-square critical values and calculate the interval. The correct calculation leads to the interval (243.086, 453.520), which matches option B.

The 95% Confidence Interval Estimate of a Population Variance When