Propagate Error Through Average
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Community Forums > Mathematics > Set Theory, Logic, Probability, Statistics > We've just passed 300 Insights! View them here! What a resource! Dismiss Notice Dismiss Notice Join Physics Forums Today! The friendliest, high quality science error propagation calculator and math community on the planet! Everyone who loves science is here! Error propagation error propagation physics with averages and standard deviation Page 1 of 2 1 2 Next > May 25, 2012 #1 rano I was wondering error propagation square root 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
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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 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 error propagation inverse 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 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 ce
"change" in the value of that quantity. Results are is obtained by mathematical operations on the data, and small changes in any data error propagation definition quantity can affect the value of a result. We say that "errors in
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the data propagate through the calculations to produce error in the result." 3.2 MAXIMUM ERROR We first consider how data
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errors propagate through calculations to affect error limits (or maximum error) of results. It's easiest to first consider determinate errors, which have explicit sign. This leads to useful rules for error propagation. Then we'll https://www.physicsforums.com/threads/error-propagation-with-averages-and-standard-deviation.608932/ modify and extend the rules to other error measures and also to indeterminate errors. The underlying mathematics is that of "finite differences," an algebra for dealing with numbers which have relatively small variations imposed upon them. The finite differences we are interested in are variations from "true values" caused by experimental errors. Consider a result, R, calculated from the sum of two data quantities A and B. https://www.lhup.edu/~dsimanek/scenario/errorman/propagat.htm For this discussion we'll use ΔA and ΔB to represent the errors in A and B respectively. The data quantities are written to show the errors explicitly: [3-1] A + ΔA and B + ΔB We allow the possibility that ΔA and ΔB may be either positive or negative, the signs being "in" the symbols "ΔA" and "ΔB." The result of adding A and B is expressed by the equation: R = A + B. When errors are explicitly included, it is written: (A + ΔA) + (B + ΔB) = (A + B) + (Δa + δb) So the result, with its error ΔR explicitly shown in the form R + ΔR, is: R + ΔR = (A + B) + (Δa + Δb) [3-2] The error in R is: ΔR = ΔA + ΔB. We conclude that the error in the sum of two quantities is the sum of the errors in those quantities. You can easily work out the case where the result is calculated from the difference of two quantities. In that case the error in the result is the difference in the errors. Summarizing: Sum and difference rule. When two quantities ar
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more http://math.stackexchange.com/questions/123276/error-propagation-on-weighted-mean about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Mathematics Questions Tags Users Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer error propagation The best answers are voted up and rise to the top Error propagation on weighted 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 $\sqrt{e_1^2 + e_2^2}$, propagate error through 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 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 158k16145350 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
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