# Is $\sigma$ or $\sigma / \sqrt{N}$ is error of a measurement?

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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–  anna v Apr 11 '12 at 7:15
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

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 $1/\sqrt{N}$.
Okay, so it is $\sigma / \sqrt N$. Thanks for the detailed answer! –  queueoverflow Apr 12 '12 at 7:54