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need help with the problem below. Thank you!!

A national survey showed that Puregold cold cuts were priced, on the average, at $5.20 per pound. Supposed a national survey of 23 retail outlets was taken and the price of Puregold cold cuts was ascertained. If the figures represent these prices, what is a 90% confidence interval for the population variance of these prices? Assume prices are normally distributed in the population.

Prices in dollars per pound 5.18 5.17 5.05 5.22 5.22 5.22 5.15 5.19 5.08 5.19 5.25 5.28 5.26 5.21 5.19 5.19 5.20 5.23 5.24 5.30 5.14 5.19 5.33

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Thanks for your help, Michael Hardy! – user59117 Mar 19 '13 at 18:32

Let $\sigma^2$ be the population variance and let $$ S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i -\bar X)^2 $$ be the sample variance. Then $(n-1)S^2/\sigma^2$ has a chi-square distribution with $n-1$ degrees of freedom. In your example, $n$ is $23$.

Go to a table for the chi-square distribution with $22$ degrees of freedom, or to whichever software you use for that, and find numbers $A$ and $B$ such that the probability that such a random variable is less than $A$ is half of $5\%$ and the probability that such a random variable is more than $B$ is half of $5\%$. You've got $$ \Pr\left( A < \frac{(n-1)S^2}{\sigma^2} < B \right) = 0.95. $$ Therefore $$ \Pr\left(\frac{(n-1)S^2}{B} <\sigma^2< \frac{(n-1)S^2}{A} \right) = 0.95. $$ That's a confidence interval for $\sigma^2$.

(I'll let you crunch the numbers to find $(23-1)S^2$.)

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