Propagation Of Error When Taking An Average
Contents |
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,
Propagation Of Error Division
high quality science and math community on the planet! Everyone who loves science is average uncertainty here! Error propagation with averages and standard deviation Page 1 of 2 1 2 Next > May 25, 2012 #1
Error Propagation Calculator
rano 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 physics 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 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: multiplying uncertainties 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 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 o
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 about Stack Overflow the company Business Learn
Error Propagation Square Root
more about hiring developers or posting ads with us Mathematics Questions Tags Users Badges error propagation chemistry Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in
Error Propagation Inverse
related 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 https://www.physicsforums.com/threads/error-propagation-with-averages-and-standard-deviation.608932/ 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 $\sqrt{e_1^2 + e_2^2}$, where $e_1$ and $e_2$ and the errors of $x$ and $y$, respectively. However, I have not yet been able to find http://math.stackexchange.com/questions/123276/error-propagation-on-weighted-mean 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}\sqrt{ e_x^2 + e_y^2} $ share|cite|improve this answer answered Mar 22 '12 at 15:05 leonbloy 29.9k63384 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
be down. Please try the request again. Your cache administrator is webmaster. Generated Mon, 24 Oct 2016 19:59:55 GMT by s_wx1157 (squid/3.5.20)