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.
With sample variance, find interval for population variance
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probability
statistics
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3Andy: You are putting questions on the site rapidly and without any e$x$planation a$b$out where you are stu$c$k, what you tried, what you know. This is not the way to pro$c$eed, see the *How to $a$sk* p$a$ge. – 2012-12-16
1 Answers
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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}. $ \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.