The upper 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) $( n - 1 ) s ^ { 2 } / \chi _ { .015 , n - 1 } ^ { 2 }$
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

Quizplus · Accepted Answer

The upper limit of a 95% confidence interval for the variance of a normal population is calculated using the formula $( n - 1 ) s ^ { 2 } / \chi ^ { 2 } _{.025 , n - 1}$ for the lower limit and $( n - 1 ) s ^ { 2 } / \chi ^ { 2 } _{.975 , n - 1}$ for the upper limit, where $\chi ^ { 2 } _{.975 , n - 1}$ is the chi-square value for 97.5% of the distribution with $n - 1$ degrees of freedom. Since we are looking for the upper limit, the correct formula is the one in option D.

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