So if I have made three measurements, measuring the same physical quantity $d$:
$a \pm \Delta a$
$b \pm\Delta b$
$c \pm\Delta c$
and I want to take the average of $a, b$ and $c$ to get the physical quantity $d$, what is the error associated with it?
I have seen the following formula:
$\frac{\Delta d}{d} = \frac{\sqrt{(\Delta a)^2 + (\Delta b)^2 + (\Delta c)^2}}{a+b+c}$
However, what I don't understand about this formula is that if $\Delta a = \Delta b = \Delta c = 0$, then the error is 0 but I would expect that the error is non-zero and should equal be equal to the standard deviation of the three values $a,b$ and $c$ when the average of them is found.
Any help with this is much appreciated.
Best Answer
Step one is find the mean. The mean is not the average of $(a, b, c)$ because you have to weigh the more accurate measurements higher. That is done as follows:
$$ d \equiv \langle d\rangle= \frac{\frac{a}{\Delta a^2}+\frac{b}{\Delta b^2}+\frac{c}{\Delta c^2}}{\frac{1}{\Delta a^2}+\frac{1}{\Delta b^2}+\frac{1}{\Delta c^2}}$$
Now if any $\Delta = 0$, this formula fails. If only one error is zero, then its associated value is the correct answer, and the other measurements are garbage. If two values are zero, and their associated measurements disagree, then you don't understand your uncertainties, at all. If all three are zero, then set them to 1--as your error analysis isn't working.
In order to calculate the standard deviation, you start with computing the second moment, which is the weighted average of the squares:
$$ d_2 \equiv \langle d^2\rangle= \frac{\frac{a^2}{\Delta a^2}+\frac{b^2}{\Delta b^2}+\frac{c^2}{\Delta c^2}}{\frac{1}{\Delta a^2}+\frac{1}{\Delta b^2}+\frac{1}{\Delta c^2}}$$
From here, the weighted standard deviation, $\sigma$, is computed in the usual fashion:
$$ \sigma^2 = d_2 - d^2 $$
For unweighted data, the error on the mean (the standard error), is computed from the standard deviation via:
$$ stderr = \frac{\sigma}{\sqrt N} $$
where $N$ is the number of measurements. However, you don't have $N=3$ measurements, because they are weighted. For instance, if $\Delta a = \Delta b = 1$ and $\Delta C = 1,000,000$, that 3rd measurement is useless: you have 2 good measurements. You can verify with the above formulae that the measurement of $c$ is worthless.
So, you have to replace $N$ with the effective number of measurements:
$$ N_{eff}= \frac{(\frac{1}{\Delta a^2}+\frac{1}{\Delta b^2}+\frac{1}{\Delta c^2})^2}{(\frac{1}{\Delta a^2})^2+(\frac{1}{\Delta b^2})^2+(\frac{1}{\Delta c^2})^2}$$
You can verify that if all errors are equal, this reduced to $3$.