# Statistical uncertainty of weighted mean

by Brais
Tags: statistical, uncertainty, weighted
 P: 7 Hello! I am using physical data to do an analysis (~30k measurements). These measurements include energies, momenta, angles... of particles. I am calculating a value (call it v) at the end after a lengthy process, and if I introduce all the data into my program I did, the result is v±σ. If, however, I "bin" my events in energy and angle (say I made four bins in total), when I calculate "v", I get v1±σ1, v2±σ2, v3±σ3, v4±σ4. Then I combine these values into one using a weighted average (c stands for combined): $v_c = \sum(v_i/\sigma^2_i)/\sum(1/\sigma^2_i)$, and $\sigma_c = 1/\sqrt{\sum(1/\sigma^2_i)}$ (as can be seen here). When I do this, it turns out that $\sigma_c \simeq \sigma/2$. How can this be? I am using the same amount of statistics! Any reply or idea will be very welcome!! Thank you! Brais.
 P: 4,579 Hey Brais. The wiki article looks straight-forward, but perhaps you could just outline your calculations in a little more detail step by step to show the simplifications and assumptions you used.
 P: 7 Hi, thanks for your reply! I am calculating a fit. If I put all my data together I get an error that is higher than that of fitting different sets of points separately and then combining them with a weighted mean. I didn't do any simplification, just applied the expression seen in wikipedia. Brais
You are using a formula from that article that applies when you know the standard deviations of the distributions that are involved, but I'd guess that you don't. Your $\sigma_i$ are probably estimators of standard deviations that you computed from the sample. (The term "standard deviation" is ambiguous. It has at least 5 different meanings in statistics, depending on the context where it appears.)