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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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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.

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