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) $1086

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) $1086 < \sigma ^ { 2 } < 269.6$
B) $10.4 < \sigma ^ { 2 } < 164$
C) $1237 < \sigma ^ { 2 } < 312.7$
D) $122.8 < \sigma ^ { 2 } < 316.5$

Quizplus · Accepted Answer

The 90% confidence interval for the variance is given by:$$\frac{(n-1)s^2}{\chi^2_{0.05, n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{0.95, n-1}}$$where n is the sample size, s is the sample standard deviation, and \chi^2_{0.05, n-1} and \chi^2_{0.95, n-1} are the chi-square critical values with n-1 degrees of freedom and significance levels of 0.05 and 0.95, respectively.For this problem, n = 28 and s = 12.7. Using a chi-square table, we find that \chi^2_{0.05, 27} = 16.266 and \chi^2_{0.95, 27} = 42.557. Therefore, the 90% confidence interval for the variance is:$$\frac{(28-1)12.7^2}{42.557} < \sigma^2 < \frac{(28-1)12.7^2}{16.266}$$$$1086 < \sigma^2 < 269.6$$

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