Error Propagation Average Calculation
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Community Forums > Mathematics > Set Theory, Logic, Probability, Statistics > Dismiss Notice Join Physics Forums Today! The friendliest, high quality science and math community on the planet! Everyone who loves science is here! Error propagation with averages and standard deviation Page 1 of error propagation equation calculator 2 1 2 Next > May 25, 2012 #1 rano I was wondering if someone how to calculate error propagation in excel could please help me understand a simple problem of error propagation going from multiple measurements with errors to an average incorporating these errors. I
How To Calculate Error Propagation Physics
have looked on several error propagation webpages (e.g. UC physics or UMaryland physics) but have yet to find exactly what I am looking for. I would like to illustrate my question with some example data. Suppose we want to know
Error Propagation Average Standard Deviation
the mean ± standard deviation (mean ± SD) of the mass of 3 rocks. We weigh these rocks on a balance and get: Rock 1: 50 g Rock 2: 10 g Rock 3: 5 g So we would say that the mean ± SD of these rocks is: 21.6 ± 24.6 g. But now let's say we weigh each rock 3 times each and now there is some error associated with the mass of each rock. Let's say that the error propagation mean mean ± SD of each rock mass is now: Rock 1: 50 ± 2 g Rock 2: 10 ± 1 g Rock 3: 5 ± 1 g How would we describe the mean ± SD of the three rocks now that there is some uncertainty in their masses? Would it still be 21.6 ± 24.6 g? Some error propagation websites suggest that it would be the square root of the sum of the absolute errors squared, divided by N (N=3 here). But in this case the mean ± SD would only be 21.6 ± 2.45 g, which is clearly too low. I think this should be a simple problem to analyze, but I have yet to find a clear description of the appropriate equations to use. If my question is not clear please let me know. Any insight would be very appreciated. rano, May 25, 2012 Phys.org - latest science and technology news stories on Phys.org •Game over? Computer beats human champ in ancient Chinese game •Simplifying solar cells with a new mix of materials •Imaged 'jets' reveal cerium's post-shock inner strength May 25, 2012 #2 viraltux rano said: ↑ I was wondering if someone could please help me understand a simple problem of error propagation going from multiple measurements with errors to an average incorporating these errors. I have looked on several error propagation webpages (e.g. UC physics or UMaryland physics) but have yet t
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Error Propagation Examples
more about hiring developers or posting ads with us Mathematics Questions Tags Users Badges Unanswered calculate error analysis Ask Question _ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related propagation of error division fields. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Error propagation on weighted https://www.physicsforums.com/threads/error-propagation-with-averages-and-standard-deviation.608932/ mean up vote 1 down vote favorite I understand that, if errors are random and independent, the addition (or difference) of two measured quantities, say $x$ and $y$, is equal to the quadratic sum of the two errors. In other words, the error of $x + y$ is given by $\sqrt{e_1^2 + e_2^2}$, where $e_1$ and $e_2$ and the errors of $x$ and $y$, respectively. However, I have not yet been able to find how to http://math.stackexchange.com/questions/123276/error-propagation-on-weighted-mean calculate the error of both the arithmetic mean and the weighted mean of the two measured quantities. How do errors propagate in these cases? statistics error-propagation share|cite|improve this question edited Mar 22 '12 at 17:02 Michael Hardy 158k15145350 asked Mar 22 '12 at 13:46 plok 10815 add a comment| 2 Answers 2 active oldest votes up vote 3 down vote accepted The first assertion assumes one takes mean squared errors, which in probabilistic terms translates into standard deviations. Now, probability says that the variance of two independent variables is the sum of the variances. Hence, if $z = x + y$ , $\sigma_z^2 = \sigma_x^2 + \sigma_y^2 $ and $$e_z = \sigma_z = \sqrt{\sigma_x^2 + \sigma_y^2} = \sqrt{e_x^2 + e_y^2} $$ Knowing this, and knowing that $Var(a X) = a^2 Var(X)$, if $z = a x + (1-a) y$ (weighted mean, if $ 0\le a \le1$) we get: $$\sigma_z^2 = a^2\sigma_x^2 + (1-a)^2\sigma_y^2 $$ $$e_z = \sqrt{a^2 e_x^2 + (1-a)^2 e_y^2} = a \sqrt{ e_x^2 + \left(\frac{1-a}{a}\right)^2 e_y^2} $$ In particular, if $a=1/2$ , then $e_z = \frac{1}{2}\sqrt{ e_x^2 + e_y^2} $ share|cite|improve this answer answered Mar 22 '12 at 15:05 leonbloy 29.8k63384 If I understand it correctly, and in a more general form, if we had $n$ independent variables with their corresponding weights, $c_{1}, c_{2}...c_{n}$, the error of their weighted mea
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