Propagation Of Error Versus Standard Deviation
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The approach to uncertainty analysis that has been followed up to this point in the discussion has been what is called a top-down approach. Uncertainty components are estimated from direct repetitions of the measurement result. To contrast this with a propagation of
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error approach, consider the simple example where we estimate the area of a rectangle error propagation physics from replicate measurements of length and width. The area $$ area = length \cdot width $$ can be computed from each
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replicate. The standard deviation of the reported area is estimated directly from the replicates of area. Advantages of top-down approach This approach has the following advantages: proper treatment of covariances between measurements of length error propagation definition and width proper treatment of unsuspected sources of error that would emerge if measurements covered a range of operating conditions and a sufficiently long time period independence from propagation of error model Propagation of error approach combines estimates from individual auxiliary measurements The formal propagation of error approach is to compute: standard deviation from the length measurements standard deviation from the width measurements and combine the two into a standard error propagation excel deviation for area using the approximation for products of two variables (ignoring a possible covariance between length and width), $$ s_{area} = \sqrt{width^2 \cdot s_{length}^2 + length^2 \cdot s_{width}^2} $$ Exact formula Goodman (1960) derived an exact formula for the variance between two products. Given two random variables, \(x\) and \(y\) (correspond to width and length in the above approximate formula), the exact formula for the variance is: $$ V(\bar{x} \bar{y}) = \frac{1}{n} \left[ X^2 V(y) + Y^2 V(x) + 2XYE_{11} + 2X\frac{E_{12}}{n} + 2Y\frac{E_{21}}{n} + \frac{V(x) V(y)}{n} + \frac{Cov((\Delta x)^2, (\Delta y)^2) -E_{11}^2 }{n^2} \right] $$ with \(X = E(x)\) and \(Y = E(y)\) (corresponds to width and length, respectively, in the approximate formula) \(V(x)\) is the variance of \(x\) and \(V(y)\) is the variance \(y\) (corresponds to \(s^2\) for width and length, respectively, in the approximate formula) \( E_{ij} = {(\Delta x)^i, (\Delta y)^j}\) where \( \Delta x = x - X \) and \( \Delta y = y - Y \) \( Cov((\Delta x)^2, (\Delta y)^2) = E_{22} - V(x) V(y) \) To obtain the standard deviation, simply take the square root of the above formula. Also, an estimate of the statistic is obtained by substituting sample estimates for the corresponding popula
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data mining, and data visualization. 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 http://www.itl.nist.gov/div898/handbook/mpc/section5/mpc55.htm Error propagation SD vs SE up vote 7 down vote favorite I have 3 to 5 measures of a trait per individual in two different conditions (A and B). I'm plotting the average for each individual in each condition and I use the standard error (i.e., $SD/\sqrt{N}$, with $N$ = number of measurements) as error bars. Now I want to plot the difference between the average measure per individual in condition A and condition B. I http://stats.stackexchange.com/questions/70164/error-propagation-sd-vs-se know I can determine the propagated error doing: $$SD=\sqrt{SD_A^2+SD_B^2}$$ but how can I propagate standard errors (since I'm dealing with averages of measurements) instead of standard deviations? Does this make sense at all? standard-deviation standard-error error error-propagation share|improve this question edited Sep 16 '13 at 18:39 whuber♦ 146k18285546 asked Sep 16 '13 at 18:08 Ines 361 add a comment| 2 Answers 2 active oldest votes up vote 4 down vote You should simply treat your SE as SD, and use exactly the same error propagation formulas. Indeed, standard error of the mean is nothing else than standard deviation of your estimate of the mean, so the math does not change. In your particular case when you estimate SE of $C=A-B$ and you know $\sigma^2_A$, $\sigma^2_B$, $N_A$, and $N_B$, then $$\mathrm{SE}_C=\sqrt{\frac{\sigma^2_A}{N_A}+\frac{\sigma^2_B}{N_B}}.$$ Please note that another option that could potentially sound reasonable is incorrect: $$\mathrm{SE}_C \ne \sqrt{\frac{\sigma^2_A\sigma^2_B}{N_A+N_B}}.$$ To see why, imagine that $\sigma^2_A=\sigma^2_B=1$, but in one case you have a lot of observations and another case only one: $N_A=100, N_B=1$. The standard error of the mean of the first group is 0.1, and of the second it is 1. Now if you use the second (incorrect) formula, you would get approximately 0.14 as the joint standard error, which is far too small given that you second measurement is known $\pm 1$. The correct formula gives $\app
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 and math community on https://www.physicsforums.com/threads/error-propagation-with-averages-and-standard-deviation.608932/ the planet! Everyone who loves science is here! Error propagation with averages and standard deviation Page 1 of 2 1 2 Next > May 25, 2012 #1 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 webpages (e.g. UC physics or UMaryland physics) but have yet to find exactly error propagation 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 the mean ± SD of these rocks is: 21.6 ± 24.6 g. But now propagation of error 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 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 he
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