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This question already has an answer here:

Suppose I have taken 5 measurements, each with an associated error, let's call these values $\eta_i\pm\sigma_i$, where $i$ is an integer from 1 to 5.

I would like to find the mean and standard deviation of these quantities, $\bar{\eta}\pm\bar{\sigma}$.

The mean can be calculated simply: $\bar{\eta}=\sum_{i=1}^{5}\eta_i$.

How is the standard deviation $\bar{\sigma}$ calculated, including the contribution from the deviation of $\eta_i$ about $\bar{\eta}$, and also the contribution from the errors $\sigma_i$?

EDIT To give some more context, the $\eta_i$ are a function of the number of counts $N$ of a Poisson distributed variable. Due to the high number of counts, $\sigma_i$ is a function of $\sqrt{N}$, which is the standard deviation of the count number.

Thus, I definitely know the errors $\sigma_i$. However, I would like to combine my repeated measurements to find a mean and deviation.

EDIT#2 My reading into the topic thus far has led me to this page on propagating errors (https://en.wikipedia.org/wiki/Propagation_of_uncertainty).

The example formulas tab has an example for the function $f=aA+bB$, which should be applicable in my case where I would like to find the mean. For my purposes (finding the mean) $a=1, b=1$. Using this entry in the table, $\sigma_f=\sqrt{\sigma_A^2+\sigma_B^2+2\sigma_{AB}}$, where $\sigma_{AB}$ is the covariance.

If I were to apply this reasoning to my problem, I should get:

$\bar{\sigma}=\sqrt{\sum_{i=1}^5\sigma_i^2 + 2cov}$, where $cov$ is the covariance.

I am rather unfamiliar with the covariance, but I think that in my example, it is equal to the variance of my 5 different $\eta_i$.

In other words, $cov=\frac{\sum_{i=1}^5(\eta_i-\bar{\eta})^2}{5}$

My question, is this interpretation right for the propagation of my errors, and if not, where has my reasoning gone wrong?

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marked as duplicate by whuber Apr 16 '18 at 15:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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If you know the errors, then you can eliminate them and it's hard to see why you would not. In the usual case where you know that there may be error but don't know what it is, then taking the mean and standard deviation in the usual way is one way to estimate error.

In fact, that's one of the first uses of this sort of analysis: To figure out the size of the Earth, several different measurements were taken and averaged.

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    $\begingroup$ Assuming I do know the errors, it seems like I shouldn't just ignore them. To use your analogy of the size of the Earth, 3 measurements of $6000\pm500$km, when averaged, doesn't have a standard deviation of 0, which is what happens when they are eliminated as you suggest. $\endgroup$ – ocelto Apr 15 '18 at 21:46
  • $\begingroup$ But you don't know the errors. If you knew the errors, you would incorporate them into the estimate and would have error free estimates. You never get a measurement of $6000 \pm 500$ or anything like that. You estimate the errors from the set of measurements. $\endgroup$ – Peter Flom Apr 15 '18 at 21:54
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    $\begingroup$ I have added an edit with more context. The gist is that I definitely do know the errors in this case. $\endgroup$ – ocelto Apr 15 '18 at 22:07

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