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I know that generally the uncertainty in the mean of a sample should be equal to:

$\frac{V_{max} - V_{min}}{2} $

where $V_{max}$ is the maximum value and $V_{min}$ the minimum value of the sample of data. However, what if each value has its own uncertainty? For example, I have to values:

$R1 = 12.8 \pm 0.2$ m

$R2 = 13.6 \pm 0.4$ m

The mean would be $13.2$ m, but what about the uncertainty? Will be it be the range $1.4/2$ or will it be the combined uncertainty of each measurment?

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  • $\begingroup$ (Vmax - Vmin)/2 would be very sensitive to outliers. Shouldn't you use just the standard deviation of the mean, i.e. the standard deviation divided by the square root of the number of observations? $\endgroup$
    – Nils L.
    Commented Jul 21, 2021 at 8:34

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If you have two uncorrelated quantities $x$ and $y$ with uncertainties $\delta x$ and $\delta y$, then their sum $z=x+y$ has uncertainty

$$\delta z = \sqrt{(\delta x)^2 + (\delta y)^2}$$

The average would then have uncertainty $$\frac{\delta z}{2} = \frac{\sqrt{(\delta x)^2 + (\delta y)^2}}{2}$$

Intuitively, one might imagine that

$$\delta z = \delta x + \delta y$$

However, this overestimates the uncertainty in $z$. If $x$ and $y$ are uncorrelated, then it is very unlikely that their errors would constructively add in this way. It is of course possible that $x$ and $y$ are correlated, but then more complicated analysis is required.

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  • $\begingroup$ Could you provide a reason (or a reference to a reputable source) to why that is the case? $\endgroup$
    – user63248
    Commented Mar 13, 2018 at 20:29
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    $\begingroup$ The reason is that measured quantities are typically assumed to correspond to normally distributed random variables, and the uncertainty is the standard deviation. Adding two such random variables results in a random variable with standard deviation given by the above formula. This can be found in essentially any reference on experimental techniques, such as this one. $\endgroup$
    – J. Murray
    Commented Mar 13, 2018 at 21:05
  • $\begingroup$ What about two quantities that measure the same thing but are incompatible with each other? Increase the uncertainty in both symmetrically until they are compatible and then perform error propagation as indicated? $\endgroup$
    – Nils L.
    Commented Jul 21, 2021 at 8:31
  • $\begingroup$ @NilsL. I’m not sure what you mean by that. If you’re saying that the two quantities you’re averaging aren’t the same to within your uncertainty estimates, then that isn’t relevant - it also happens to be true for the quantities in the OP. $\endgroup$
    – J. Murray
    Commented Jul 21, 2021 at 10:58
  • $\begingroup$ @J.Murray Correct, I mean that the "errorbars overlap", i.e. that the values are metrologically compatible for some common confidence level like 95 or 99%. I disagree that this is should not be relevant, since missing compatibility indicates that the uncertainty was underestimated for at least one quantity. The combined uncertainty can then not be trustworthy. Imagine the following: If x and y are incompatible and their uncertainties are very small, you can end up with a result for z that is incompatible with both x and y. How would that make sense? $\endgroup$
    – Nils L.
    Commented Jul 22, 2021 at 21:56

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