# What is the error of the mean of data that have uncertainty values attached to them?

Given a set of $$n$$ values, the error associated with their average will be

$$\text{standard deviation}/\sqrt{n}.$$

But if the values themselves have an uncertainty attached to them, such as $$100\pm 1,$$ $$110 \pm 1, \ldots$$ is the previous formula correct?

• for independent sources of noise (with variances $\sigma_1^2$ and $\sigma_2^2$) you could add the variances and then the error would be $N^{-1/2}\sqrt{\sigma_1^2+\sigma_2^2}$. May 9 at 21:19

One interpretation of your situation is that you have obtained a random sample $$Y_1, \ldots, Y_n$$ from a distribution with a finite variance, say $$\sigma^2,$$ but you have observed the values $$X_i = Y_i + E_i$$ where the variances of the "errors" $$E_i$$ are known. Let's refer to these error variances as $$\sigma_i^2.$$

One interpretation of the "error associated with the average" is that you are looking for the variance of the arithmetic mean of the $$X_i.$$ This mean is defined as

$$\bar X = \frac{1}{n}\sum_{i=1}^n X_i = \frac{1}{n}\sum_{i=1}^n \left(Y_i + E_i\right) = \bar Y + \bar E.$$

Assuming all the values $$Y_i$$ and all the errors $$E_j$$ are uncorrelated, the laws of variances say

$$\operatorname{Var}(\bar X) = \frac{1}{n^2}\sum_{i=1}^n (\operatorname{Var}(Y_i) + \operatorname{Var}(E_i)) = \frac{\sigma^2}{n} + \frac{1}{n^2}\sum_{i=1}^n \sigma_i^2.$$

Your formula for the "error" of $$\bar X$$ is the square root of the right hand side, neglecting the observational error variances $$\sigma_i^2.$$

At this point we're stuck, because we almost never assume we know $$\sigma^2.$$ Usually we are trying to use the observations $$X_i$$ to estimate it. But we can still say some things from this result, including

1. The error of $$\bar X$$ exceeds the error of $$\bar Y$$ (which is $$\sigma/\sqrt{n}$$).

2. Since you know or assume you know the $$\sigma_i^2,$$ if you can somehow obtain an estimate of $$\sigma^2,$$ then you can adjust it to estimate $$\operatorname{Var}(\bar X).$$

One subtlety is that with observations reported in the form "$$X_i \pm \tau_i,$$" as in the question, it is possible that the $$\tau_i$$ equal the $$\sigma_i,$$ or $$2\sigma_i,$$ or $$1.32 \sigma_i,$$ or $$1.645\sigma_i,$$ or maybe even something else--I have listed only the commonest conventions. You have to rely on your source of these data to tell you what it means.