# Multiple measurements of the same quantity - combining uncertainties

I have a number of measurements of the same quantity (in this case, the speed of sound in a material). Each of these measurements has their own uncertainty.

$$v_{1} \pm \Delta v_{1}$$ $$v_{2} \pm \Delta v_{2}$$ $$v_{3} \pm \Delta v_{3}$$ $$\vdots$$ $$v_{N} \pm \Delta v_{N}$$

Since they're measurements of the same quantity, all the values of $v$ are roughly equal. I can, of course, calculate the mean:

$$v = \frac{\sum_{i=1}^N v_{i}}{N}$$

What would the uncertainty in $v$ be? In the limit that all the $\Delta v_i$ are small, then $\Delta v$ should be the standard deviation of the $v_i$. If the $\Delta v_i$ are large, then $\Delta v$ should be something like $\sqrt{\frac{\sum_i \Delta v_i^2}{N}}$, right?

So what is the formula for combining these uncertainties? I don't think it's the one given in this answer (though I may be wrong) because it doesn't look like it behaves like I'd expect in the above limits (specifically, if the $\Delta v_i$ are zero then that formula gives $\Delta v = 0$, not the standard deviation of the $v_i$).

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The answer from Pygmalion that you liked gives the correct (naive) treatment. Notice that in the average he has $N$ in the denominator, not $\sqrt{N}$. Also, you have not yet formed a standard deviation, which is not the same as a combined error (though in many cases there are useful relations between them). –  dmckee Mar 19 '13 at 15:41
Better than averaging, if each measurement has a different uncertainty, is to do a weighted mean, where the weights are $\frac{1}{\Delta\nu_1}$. –  user40838 Feb 18 '14 at 9:34

When you're combining measurements with different uncertainties, taking the mean is not the right thing to do. (Well, it's good enough if the uncertainties are almost the same.)

The right thing to do is chi-squared analysis, which gives a higher weight to the more accurate measurements. Here's how you do it:

$$\chi^2 = \sum \frac{(\text{observed value} - \text{true value})^2}{\text{(uncertainty associated with that observation)}^2}$$

You numerically choose the "true value" that minimizes $\chi^2$. That's your best guess.

Next, use the chi-square distribution to calculate the p-value (assuming the best guess is right). (Degrees of freedom is one less than the number of observations.) This will tell you whether your uncertainties were reasonable or whether you underestimated them. For example, if one measurement is $5.0 \pm 0.1$, and another measurement is $10.0 \pm 0.1$, then you probably underestimated your uncertainties.

IF you underestimated your uncertainties -- which is not unusual in practice -- then there are any number of reasonable procedures. I think the most standard procedure is to scale up all the uncertainties by the same factor until you get a reasonable $\chi^2$ p-value, say 0.5.

OK, now you have plausible measurement uncertainties, either because you did from the beginning or because you scaled them up. Next, you try varying the "true value" until the p-value dips down below, say, 5%. This procedure gives you lower bound and upper bound error bars on your final best-guess measurement.

I haven't done this in many years, sorry for any mis-remembering. I think it's discussed in Bevington&Robinson.

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This doesn't seem to answer the question. The OP hasn't yet gotten to the point of correctly forming a standard deviation. –  dmckee Mar 19 '13 at 15:42
@dmckee -- Thanks, I changed "standard deviation" to "uncertainty" in the formula. That should be less confusing. –  Steve B Mar 19 '13 at 17:31
The problem is the chi-squared is a figure of merit for choosing between possible fits or explanations. The OP isn't that far along the analysis chain yet. –  dmckee Mar 19 '13 at 18:00

You seem to be mixing up several concepts here.

In particular you are interested in the error on a mean, refer to the error on a sum (which you need on the way to the error on a mean), and talk about the standard deviation.

(all work here in the naive version assuming zero correlation.)

• Error of a sum of uncertain quantities: $X = \sum_i x_i$ and $\Delta X = \sqrt{ \sum_i \left(\Delta x_i\right)^2 }$

• Error of the mean of uncertain quantities: Divide the sum by the number of measurements. The number of measurements is definite, you are not at all unsure how many figures you have processed so this is just division. $$\bar{x} = \frac{X}{N} = \frac{\sum_i x_i}{N} \quad ,$$ and $$\Delta \bar{x} = \frac{\Delta X}{N} = \frac{\sqrt{\sum_i \left( \Delta x_i\right)^2}}{N} \quad.$$ (Note that later on you will encounter the phrase "error of the mean" in the context of large samples. That's different.)

If you individual $\Delta x_i$s vary considerably it is better to use the error weighted mean.

• Standard Deviation: This is a figure expressing the dispersion of you $x_i$s, and is figured without reference to your $\Delta x_i$s. Generally represented with $\sigma$ and we call $\sigma^2$ the "variance". $$\sigma^2 = \frac{1}{N} \sum_i \left( x_i - \bar{x} \right)^2 \quad ,$$ with a minor correction that if you have to get $\bar{x}$ from the same list (which you do) you use $$\sigma^2 = \frac{1}{N-1} \sum_i \left( x_i - \bar{x} \right)^2 \quad .$$

If you have estimated your $\Delta x_i$s correctly then there should be a relationship between the standard deviation and the error on the mean, but that is for another day.

You complain in your question that the error of the mean does not go to the standard deviation in the limit that $\Delta x_i = 0$, but that is because they represent different concepts. It is possible to have experiment where the individual measurements are drawn from a broad distribution but are known very well (high standard deviation, but low individual uncertanties) or where you re-measure the same underling quantity with poor instruments (zero standard deviation, but large $\Delta x_i$s). In a lot of ways the cases can be treated with the same math, but they are different.

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If you're really sure that some measurements are much more accurate then others, then taking the mean of all the measurements is the wrong thing to do. You would want to give extra weight to the more accurate measurements. Granted, the OP asked about the mean. But still, he/she should be guided towards a better analysis method. –  Steve B Mar 19 '13 at 17:38
@Steve Yes. That would be physics.stackexchange.com/questions/55983/… for instance, but to me this feels like a question related to a first lab. YMMV and all that. –  dmckee Mar 19 '13 at 17:49

## protected by Qmechanic♦Feb 18 '14 at 10:05

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