How to obtain the standard error of measurements that already have error bars?

Question

Suppose I make three measurements: 9, 10 and 11. Let's say there is no uncertainty associated with these values because they are counts. Numbers of children in a class say.

I want to know the average - 10 - and I want to know the error of this value, so I take the standard deviation (1) and get the standard error from this - 0.577. I would report the average number of children per class as 10 +/- 0.577.

But suppose they were values with uncertainty already associated. Let's say those numbers were actually repeated measurements of the weight of something, in kg, and the machine I use to measure them is quite poor so they have large uncertainty values to begin with - 9 +/- 5, 10 +/- 4, and 11 +/- 7.

If I take the average of these measurements now, how would I express the error? The standard error of 0.577 seems misleading, as it doesn't take into account how very uncertain the initial measurements were. Should I just forget about standard error, and use the rules for adding errors? Should I only use standard error of the starting values have no uncertainty associated?

Answer 1

Let's say you need to know the value of $y$ and for some reason you cannot measure it directly but you can measure $x_1$,$x_2$...$x_n$ and you can express $y$ as a function of these variables. Now you want to calculate $y=f\left(x_1,...,x_n\right)$.

For each $i$ you get $$x_i=\bar{x}_i \pm \sigma_i.$$ Then the average value of $y$ is obviously $$\bar y= f\left(\bar x_1,...,\bar x_n\right)$$

and the uncertainty is $$\sigma_y=\sqrt{ \left(\sum_{i=1}^n \left. \frac{\partial y}{\partial x_i}\right|_{x_1=\bar x_1,..., x_n = \bar x_n}\right) \cdot \sigma_i^2}.$$

Answer 2

This is a very important topic in physics, usually named as data reconciliation. So, you want to minimize $$S=\frac12\sum_{i=1}^n\left(\frac {y_i-Y}{\sigma_i}\right)^2$$ Differentiate with respect to $Y$ to get $$S'=-\sum_{i=1}^n \frac{y_i}{\sigma_i}+Y\sum_{i=1}^n \frac{1}{\sigma_i^2}$$ Since you want the minimum, set $$S'=0\implies Y=\frac{\sum_{i=1}^n \frac{y_i}{\sigma_i} }{\sum_{i=1}^n \frac{1}{\sigma_i^2}}$$ and $$\sigma^2_Y=\frac 1 n\sum_{i=1}^n \sigma_i^2$$ So, for your case $$Y=\frac{7902}{803}\approx 9.84 \qquad \text{ and}\qquad \sigma_Y=\sqrt{30}\approx 5.48$$