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I have to find the uncertainty of a quantity $Q$ doing two mean values. For example for a set of parameters I measure ten times $Q$, I obtain a mean value $Q_1$ and variance ${\rm Var}(Q_1)$. Then for a different set of parameters I measure ten times $Q$ and obtain $Q_2$ and ${\rm Var}(Q_2)$ etc. At the end I compute the mean value which is the sum of $Q_i$ but these 10 from one set of parameter to the other are correlated so I don't know how to compute the variance.

Put differently, I don't know how to compute the variance if I have two averages.

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Related physics.stackexchange.com/questions/23643 –  Pygmalion Apr 26 '12 at 8:50
    
Question (v3) possibly more at home at stats.stackexchange.com ? –  Qmechanic Apr 26 '12 at 9:29
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1 Answer

If measurements of $Q_1$ and $Q_2$ were "independent" of each other, you could use expression

$$Q = \frac{Q_1 + Q_2}{2}$$

Generally, to obtain experimental error of a dependent quantity (and the expression stated in your question), you start with the expression for dependent quantity

$$Q = f(Q_1, Q_2, ...)$$

and use statistical expression

$$\sigma(Q) = \sqrt{\sum_i \left(\frac{\partial f}{\partial Q_i} \sigma(Q_i) \right)^2}.$$

If

$$Q = \frac{Q_1 + Q_2}{2}$$

then

$$ \sigma(Q) = \frac{\sqrt{(\sigma(Q_1))^2 + (\sigma(Q_2))^2}}{2} $$

If $Q_1$ and $Q_2$ are the same quantity measured in two measurements, this is not exactly true, so the exact statistical expression is much more complicated.

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thank you but here I have strong correlations I think so I should consider also covariance, isn't it? –  PanAkry Apr 26 '12 at 8:57
    
I am not quite sure what you mean by correlations? You are measuring the same quantity, are you? I guess it should be independent of the parameters. Let me put it in this way: if parameters influence precision of the measurements then you can establish the proper average value and variance by first setting up "weights" for individual measurements. –  Pygmalion Apr 26 '12 at 9:02
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