# Interval estimation of $\sigma^2$ with the reliability of $95\%$

I got these values from the measurement by a telescope: 20.1, 20.2, 19.9, 20, 20.5, 20.5, 20, 19.8, 19.9, 20. I know that the actual distance is 20 km and the error of the measurement is not affected by the systematic error. What is the accuracy of the telescope? [Answer is: ${σ}^2$=0.061]

Now I am supposed to calculate ${σ}^2$ with a reliability of 95% using the interval estimation. So, ${σ}^2$ should be:
$\frac{n* s^{2} }{ \chi^{2}_{0.975} \big(n-1\big) } ; \frac{n* s^{2} }{ \chi^{2}_{0.025} \big(n-1\big) }$

After using $n = 10$, $k = n-1 = 9$ and $s^{2} = 0.061$ I get this:

$\frac{10*0.061}{ \chi^{2}_{0.975} \big(9\big) }; \frac{10*0.061}{ \chi^{2}_{0.025} \big(9\big) }$

Is it right? In the numerator I should use n = 10 or n-1 = 9? In the denumerator I should use n-1 = 9 or n = 10 ?

When I'm looking for the values in the table of $\chi^2$ quantiles, should I look for $0.95$ and $0.05$ or $0.975$ and $0.025$? When they say they need the reliability of $95\%$, I assume I will use $1- \frac{ \alpha }{2}$ and $\frac{ \alpha }{2}$. Since my $\alpha$ is $0.05$, I think $0.975$ and $0.025$ are the right values.

[ Answer is: $σ$ = <0.173 , 0.433> ]

-

Short answer: Use $n$ when you know the actual distance. Use $n-1$ when you don't know the actual distance.

Long answer: Assume that $X_1,\ldots,X_n$ are independent normal random variables with mean $\mu=20$ and unknown variance $\sigma^2$.

By the scale and location properties of the normal distribution, this means that $Y_i=\frac{X_i-20}{\sigma}$ are standard normal, i.e. $N(0,1)$, random variables.

By the definition of the $\chi^2$ distribution, the sum of $n$ independent squared $N(0,1)$ variables is $\chi^2(n)$. This means that $\sum_{i=1}^n Y_i^2$ is $\chi^2(n)$ distributed.

Your estimator of $\sigma^2$ is $$s^2=\frac{1}{n}\sum_{i=1}^n(X_i-20)^2=\frac{\sigma^2}{n}\sum_{i=1}^n Y_i^2.$$

Thus $\frac{ns^2}{\sigma^2}=\sum_{i=1}^n Y_i$ is $\chi^2(n)$ distributed. Hence, using the definition of quantiles, $$P\Big(\chi^2_{\alpha/2}(n)\leq \frac{ns^2}{\sigma^2} \leq \chi^2_{1-\alpha/2}(n)\Big)=1-\alpha.$$

After some algebra, we find that this also means that $P\Big(\frac{ns^2}{\chi^2_{1-\alpha/2}(n)}\leq\sigma^2 \leq \frac{ns^2}{\chi^2_{\alpha/2}(n)}\Big)=1-\alpha$. Consequently, your $1-\alpha$ confidence interval for $\sigma^2$ is

$$\Big(\frac{ns^2}{\chi^2_{1-\alpha/2}(n)}, \frac{ns^2}{\chi^2_{\alpha/2}(n)}\Big).$$

As pointed out by Procrastinator, you obtain the corresponding interval for $\sigma$ by taking the square roots of the interval limits. This works because $$P\Big(\sqrt{\frac{ns^2}{\chi^2_{1-\alpha/2}(n)}}\leq\sigma \leq \sqrt{\frac{ns^2}{\chi^2_{\alpha/2}(n)}}\Big)=P\Big(\frac{ns^2}{\chi^2_{1-\alpha/2}(n)}\leq\sigma^2 \leq \frac{ns^2}{\chi^2_{\alpha/2}(n)}\Big)=1-\alpha.$$

Plugging in your values, the interval for $\sigma^2$ is $(0.02978054, 0.18786730 )$. Thus the interval for $\sigma$ becomes $(0.1725704, 0.4334366)\approx (0.173,0.433)$.

If $\mu$ had been unknown, you would have used the estimator $\hat{\sigma}^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar{X})^2$ instead. It can be shown that $\hat{\sigma}^2$ is $\chi^2(n-1)$ distributed. Using the same derivation as above, we then find that the confidence interval becomes $\Big(\frac{(n-1)\hat{\sigma}^2}{\chi^2_{1-\alpha/2}(n-1)}, \frac{(n-1)\hat{\sigma}^2}{\chi^2_{\alpha/2}(n-1)}\Big)$ instead.

-
+1 Nice answer. Did you get the numerical result mentioned by the OP? –  user10525 Jun 4 '12 at 11:55
@Procrastinator: I did, yes. I've added the numerical results to the answer now. Thanks for pointing that out :) –  MånsT Jun 4 '12 at 12:00
I think the important point in MansT's answer is that the chi square distribution used to get the confidence interval depends on the observations having a normal distribution. –  Michael Chernick Jun 4 '12 at 12:18
(+1) This is a nice clear answer to the problem, as long as the normality assumption is reasonable. –  Macro Jun 4 '12 at 12:28