What is the 90% confidence interval for the variance of exam scores for 28 algebra students, if the standard deviation of their last exam was 12.7?
A) $108.6

Question

What is the 90% confidence interval for the variance of exam scores for 28 algebra students, if the standard deviation of their last exam was 12.7? 
A) $108.6 < \sigma ^ { 2 } < 269.6$
B) $10.4 < \sigma ^ { 2 } < 16.4$
C) $123.7 < \sigma ^ { 2 } < 312.7$
D) $122.8 < \sigma ^ { 2 } < 316.5$

Quizplus · Accepted Answer

To find the 90% confidence interval for the variance, use the chi-square distribution with the formula: $\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}$, where $n$ is the sample size, $s^2$ is the sample variance, and $\chi^2$ values are from the chi-square distribution table for $n-1$ degrees of freedom at the specified confidence level. Given $n=28$ and $s=12.7$, thus $s^2 = 161.29$. Plugging these into the formula with the appropriate chi-square values for 27 degrees of freedom at 90% confidence gives the interval $108.6 < \sigma^2 < 269.6$.

What Is the 90% Confidence Interval for the Variance of Exam