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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> ]

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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.

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    $\begingroup$ +1 Nice answer. Did you get the numerical result mentioned by the OP? $\endgroup$ – user10525 Jun 4 '12 at 11:55
  • $\begingroup$ @Procrastinator: I did, yes. I've added the numerical results to the answer now. Thanks for pointing that out :) $\endgroup$ – MånsT Jun 4 '12 at 12:00
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    $\begingroup$ 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. $\endgroup$ – Michael R. Chernick Jun 4 '12 at 12:18
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    $\begingroup$ (+1) This is a nice clear answer to the problem, as long as the normality assumption is reasonable. $\endgroup$ – Macro Jun 4 '12 at 12:28
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If your data are not normally distributed you can try bootstrap confidence intervals but my research has shown that you need larges samples to get the proper coverage (these intervals have advertised coverage only asymptotically). Also although you say that you know the measurements are unbiased and the actual mean parameter is 20 you can compute the sample estimate of bias and test to see if it is different from 0.

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  • $\begingroup$ +1 because everything in OP's reasoning is only valid when the data is normally distributed $\endgroup$ – Macro Jun 4 '12 at 12:27

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