Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

I vaguely remember seeing this question here before . . . . . .

$$ \frac{(n-1)S^2}{\sigma^2} = \frac{1}{\sigma^2}\sum_{i=1}^n (X_i - \bar X)^2 \sim \chi^2_{n-1},\text{ where }\bar X = \frac{X_1+\cdots+X_n}{n}. $$

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.)

There you have it.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.