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I wonder whether $\sigma$ or $\sigma / \sqrt{N}$ is error of a measurement. When I measure, say $0, 1, -1, 1, -1$, I have a $\sigma = 1$. I just measure $0, 1, -1$, I also have $\sigma = 1$.

But in the former case, I had more measurements, so the error should be smaller. So would the error be 1 or rather $1/\sqrt{5}$ and $1/\sqrt{3}$ respectively?

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Okay, but that does not really answer my question. –  queueoverflow Apr 11 '12 at 7:40
    
If your data are "0, 1, -1, 1, -1", then it sounds like you have some kind of quantization effect that might change your analysis. –  nibot Apr 11 '12 at 13:10
    
I just made that up since it has an obvious mean and an easy residual. –  queueoverflow Apr 12 '12 at 7:00
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up vote 4 down vote accepted

You're measuring the quantity $X$ and you got results $+1,0,-1$ and perhaps $+1,-1$ again. Assuming that your systematic error is zero, these numbers are randomly generated around the right value you want to know.

That's why you want to estimate the right value as the average of the results you obtained. That's $$ \overline{X}= \frac{(-1)+0+(+1)}{3} = \frac{(-1)+0+(+1)+(-1)+(+1)}{5} = 0$$ So there's no doubt about the mean value. It's zero in both cases. However, you also want to know the error of $\overline{X}$. That's calculated as the square root of the expectation value of $(X-\overline{X})^2$. Because $\overline{X}=0$, we just have the expectation value of $X^2$ in this case which is $$ \frac{(-1)^2+0^2+(+1)^2}{9} = \frac{2}{9} $$ in the case of three measurements or $$ \frac{(-1)^2+0^2+(+1)^2+(-1)^2+(+1)^2}{25} = \frac{4}{25} $$ There are no mixed terms because the individual deviations are independent.

So the errors are $\sqrt{2}/3$ and $2/5$, respectively. Note that you had an error in the numerator as well. Your results $1/\sqrt{3}$ or $1/\sqrt{5}$ would occur if there were 3 or 5 terms in the numerator equal to 1 i.e. 3 or 5 measurements equal to $\pm 1$, respectively. But one of the measurements was, in both cases, equal to zero which reduces the variance and reduces the error from your incorrect $\sqrt{3/9}$ or $\sqrt{5/25}$ to $\sqrt{2/9}$ and $\sqrt{4/25}$, respectively.

Yes, if you repeat the measurement many times, the statistical error will go down as $1/\sqrt{N}$. The proof is de facto contained in the simple calculation above. Because we're computing the average which has $1/N$ in it, this produces $1/N^2$ when squared and isn't quite compensated by the numerator which is the sum of $N$ terms so it goes just like $N$. So the expectation value of $(\Delta X)^2$ goes like $1/N$ and $\sigma$ therefore goes like $1/\sqrt{N}$.

Again, I should emphasize that there can be errors that are fixed for your gadget – the gadget persistently produces a result that is off by some unknown but always the same deviation. Such errors are known as the systematic errors and they can't be diminished by repeating the measurement many times. The total error of your measurement is composed both of statistical errors and systematic errors. You usually want to repeat the experiments a sufficient number of times so that the statistical error drops to the level of the systematic error or slightly below it; it doesn't make much sense to repeat the experiment too many times again because it won't reduce the total error, now dominated by the systematic error.

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Okay, so it is $\sigma / \sqrt N$. Thanks for the detailed answer! –  queueoverflow Apr 12 '12 at 7:54
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