# With sample variance, find interval for population variance

The variance of in a random sample of 16 is 2.5. What is a 95% confidence interval for the variance of the population, assuming the population is normally distributed.

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Andy: You are putting questions on the site rapidly and without any explanation about where you are stuck, what you tried, what you know. This is not the way to proceed, see the How to ask page. – Did Dec 16 '12 at 18:57

$$\frac{(n-1)S^2}{\sigma^2} = \frac{1}{\sigma^2}\sum_{i=1}^n (X_i - \bar X)^2 \sim \chi^2_{n-1},\text{ where }\bar X = \frac{X_1+\cdots+X_n}{n}.$$
So find $A,B$, such that $\Pr(\chi_{n-1}^2 >B) = \Pr(\chi^2_{n-1}>A) = 0.05/2$.
Then $\Pr(A < \chi^2_{n-1}<B) = 0.95$. $$\Pr\left( A < \frac{(n-1)S^2}{\sigma^2} < B \right) = 0.95.$$ $$\Pr\left( \frac{(n-1)S^2}{B} < \sigma^2 < \frac{(n-1)S^2}{A} \right) =0.95.$$ (I'll let you fill in the details of algebra, etc.)