The lower limit of a 95% confidence interval for the variance $\sigma ^ { 2 }$ of a normal population using a sample of size n and variance value $s ^ { 2 }$ is given by:
A) $( n - 1 ) s ^ { 2 } / \chi _ { .05,_ { n - 1 } } ^ { 2 }$
B) $( n - 1 ) s ^ { 2 } / \chi _ { .025 , n - 1 } ^ { 2 }$
C) $( n - 1 ) s ^ { 2 } / \chi _ { .95, { n - 1 } } ^ { 2 }$
D) $( n - 1 ) s ^ { 2 } / \chi ^ { 2 } _{.975 , n - 1}$
E) None of the above answers are correct.

Question

The lower limit of a 95% confidence interval for the variance $\sigma ^ { 2 }$ of a normal population using a sample of size n and variance value $s ^ { 2 }$ is given by:
A) $( n - 1 ) s ^ { 2 } / \chi _ { .05,_ { n - 1 } } ^ { 2 }$
B) $( n - 1 ) s ^ { 2 } / \chi _ { .025 , n - 1 } ^ { 2 }$
C) $( n - 1 ) s ^ { 2 } / \chi _ { .95,  { n - 1 } } ^ { 2 }$
D) $( n - 1 ) s ^ { 2 } / \chi ^ { 2 } _{.975 , n - 1}$
E) None of the above answers are correct.

Quizplus · Accepted Answer

For a 95% confidence interval for the variance, the lower limit is calculated using the chi-square distribution with $\alpha/2$ in the denominator, which is $\chi^2_{.025, n-1}$.

The Lower Limit of a 95% Confidence Interval for the Variance