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To combine independent measures of the same quantity, the value is obtained by weighting each measurment by its inverse error, and the total error is obtained by combining the individual errors in inverse quadrature:
This means that more measurements produces a smaller error, decreasing by ~1/sqrt(n), as you would expect.
But it sounds like maybe each of your measurements is not actually of the same quantity, but rather contributes an nth of the total. In that case you do add each part to get the whole, and add the errors in quadrature:
It also sounds like you may not believe that your n measurements are actually independent. Why is that? In any case, you can test whether the measurements are independent by computing the corrleation coefficient between them and seeing whether it is non-zero at a significant level.
![x = \left[ \frac{x_1}{\sigma_1} + \frac{x_2}{\sigma_2} +
\cdots \frac{x_n}{\sigma_n} \right] \left[ \frac{1}{\sigma_1} +
\frac{1}{\sigma_2} + \cdots + \frac{1}{\sigma_n} \right]^{-1} \qquad
\frac{1}{\sigma^2} = \frac{1}{\sigma_1^2} +
\frac{1}{\sigma_2^2} + \cdots + \frac{1}{\sigma_n^2}](/~talkmath/tex/img/13e8d0051c3ce1f75f8cb329bbbc2113-1.gif "x = \left[ \frac{x_1}{\sigma_1} + \frac{x_2}{\sigma_2} +
\cdots \frac{x_n}{\sigma_n} \right] \left[ \frac{1}{\sigma_1} +
\frac{1}{\sigma_2} + \cdots + \frac{1}{\sigma_n} \right]^{-1} \qquad
\frac{1}{\sigma^2} = \frac{1}{\sigma_1^2} +
\frac{1}{\sigma_2^2} + \cdots + \frac{1}{\sigma_n^2}")
Originally Posted by ichbin
