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I have measured $k_1$ and $k_2$ in two measurements and then I calculated $\Delta k_1$ and $\Delta k_2$. Now I want to calculate $k$ and $\Delta k$.

$k$ is just the mean of $k_1$ and $k_2$. I thought that I would need to square-sum the errors together, like so:

$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$

But if I measure $k_n$ $n$ times, $\Delta k$ would become greater and greater, not smaller. So I need to divide the whole root by some power of $n$, but I am not sure whether $1/n$ or $1/\sqrt n$. Which is it?

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More on measurements and errors: physics.stackexchange.com/q/23441/2451 and physics.stackexchange.com/q/23565/2451 –  Qmechanic Apr 12 '12 at 12:34
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2 Answers 2

up vote 7 down vote accepted

The formula you've specified

$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$

is the formula to obtain error of quantity $k$, as being dependent on $k_1$ and $k_2$ according to the following expression

$$ k = k_1 + k_2.$$

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

$$k = f(k_1, k_2, ...)$$

and use statistical expression

$$\Delta k = \sqrt{\sum_i \left(\frac{\partial f}{\partial k_i} \Delta k_i \right)^2}.$$

If

$$k = \frac{k_1 + k_2}{2}$$

then

$$ \Delta k = \frac{\sqrt{(\Delta k_1)^2 + (\Delta k_2)^2}}{2} $$

So the generalized answer might be: you have to divide with $n$ and not $\sqrt{n}$.

However, bare in your mind that the statistical expression above might be used when measured quantities are "independent" of each other. If $k_1$ and $k_2$ 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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Excellent answer, +1. –  Luboš Motl Apr 12 '12 at 11:57
    
Thanks a lot. I can finally comment everywhere ;-) –  Pygmalion Apr 12 '12 at 12:03
    
On the one hand, the error goes down with $1/\sqrt N$, but here the error goes down with $1/N$? How does that make sense? –  queueoverflow Apr 12 '12 at 15:34
    
You are talking about two different things. One thing is actually calculating standard deviation (or as you call it, error) from set of data, the other is calculating standard deviation of dependent quantity out of standard deviations of quantities that dependent quantity depends on. Standard deviation is by definition $\Delta x = \sqrt{\frac{1}{N} \sum_i \left(x_i-\bar{x} \right)}$, where $\bar{x} = \frac{\sum_i x_i}{N}$ –  Pygmalion Apr 12 '12 at 20:15
    
Sorry, I forgot the square in formula: $\Delta x = \sqrt{\frac{1}{N} \sum_i \left(x_i-\bar{x} \right)^2}$. To rephrase: in first step you calculate standard deviation from real data using this formula, and in second step you calculate standard deviation of one variable out of standard deviations of other variables using formulae in my answer. Hope this helps. –  Pygmalion Apr 12 '12 at 20:21
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$k$ is just the mean of $k_1$ and $k_2$

No, the best value of k is calculated using a weighted mean, weighting by the reciprocals of the squares of the respective individual uncertainty values. An accurate measurement must contribute more to the best value than an inaccurate measurement.

I thought that I would need to square-sum the errors together, like so:

$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$

No, instead $$ (\Delta k)^{-1} = \sqrt{(\Delta k_1)^{-2} + (\Delta k_2)^{-2}} $$

Intuitively, a very uncertain value must make little contribution. The uncertainty in k must always be less than or equal to the smallest of the individual uncertainties. Also, multiple, equally accurate measurements must decrease uncertainty.

see the example at the bottom of page 4 of this reference for the general case of n measurements:

http://www.physics.umd.edu/courses/Phys261/F06/ErrorPropagation.pdf

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