Error Propagation Average Value
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or more quantities, each with their individual uncertainties, and then combine the information from these quantities in order to come up with a final result of our experiment. How can you state your answer for the combined result of these measurements and their uncertainties error propagation average standard deviation scientifically? The answer to this fairly common question depends on how the individual measurements are error propagation weighted average combined in the result. We will treat each case separately: Addition of measured quantities If you have measured values for the quantities
Error Propagation Mean Value
X, Y, and Z, with uncertainties dX, dY, and dZ, and your final result, R, is the sum or difference of these quantities, then the uncertainty dR is: Here the upper equation is an approximation that
How To Find Error Propagation
can also serve as an upper bound for the error. Please note that the rule is the same for addition and subtraction of quantities. Example: Suppose we have measured the starting position as x1 = 9.3+-0.2 m and the finishing position as x2 = 14.4+-0.3 m. Then the displacement is: Dx = x2-x1 = 14.4 m - 9.3 m = 5.1 m and the error in the displacement is: (0.22 + 0.32)1/2 m = 0.36 propagation of error division m Multiplication of measured quantities In the same way as for sums and differences, we can also state the result for the case of multiplication and division: Again the upper line is an approximation and the lower line is the exact result for independent random uncertainties in the individual variables. And again please note that for the purpose of error calculation there is no difference between multiplication and division. Example: We have measured a displacement of x = 5.1+-0.4 m during a time of t = 0.4+-0.1 s. What is the average velocity and the error in the average velocity? v = x / t = 5.1 m / 0.4 s = 12.75 m/s and the uncertainty in the velocity is: dv = |v| [ (dx/x)2 + (dt/t)2 ]1/2 = 12.75 m/s [(0.4/5.1)2 + (0.1/0.4)2]1/2 = 3.34 m/s Multiplication with a constant What if you have measured the uncertainty in an observable X, and you need to multiply it with a constant that is known exactly? What is the error then? This is easy: just multiply the error in X with the absolute value of the constant, and this will give you the error in R: If you compare this to the above rule for multiplication of two quantities, you see that this is just the special case of that
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Error Propagation Formula Physics
developers or posting ads with us Cross Validated Questions Tags Users Badges Unanswered Ask Question _ error propagation square root Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Join them; it error propagation calculator 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 Average over two variables: Why do standard error of mean and http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm error propagation differ and what does that mean? up vote 3 down vote favorite I'm doing an experiment with a cryostat to determine the critical temperature for lead. To avoid asymmetries, I determine the critical temperature both through heating (going from 2 K to 10 K) and cooling (10 K -> 2 K). Now I have two values, that differ slighty and I average them. So a measurement of (6.942 $\pm$ 0.020) K and (6.959 $\pm$ 0.019) K gives me http://stats.stackexchange.com/questions/71419/average-over-two-variables-why-do-standard-error-of-mean-and-error-propagation an average of 6.951 K. Now the question is: what is the error of that average? One way to do it would be to calculate the variance of this sample (containing two points), take the square root and divide by $\sqrt{2}$. This gives me an SEM of 0.0085 K, which is too low for my opinion (where does this precision come from?) The other way is to say the the mean is a function of two variables, $\bar{T} = \frac{T_1 + T_2}{2}$, therefore by error propagation the error is $\Delta T = \frac12\sqrt{(\Delta T_1)^2+(\Delta T_2)^2}$, and that gives me a much more rational value of 0.014. I see how those values differ in terms of numbers, but which one is correct when talking about the correct estimate for the standard deviation? mean standard-error measurement-error error-propagation share|improve this question edited Sep 29 '13 at 21:32 gung 74.1k19160309 asked Sep 29 '13 at 21:05 Wojciech Morawiec 1164 @COOLSerdash That's actually another point I have thought about: The numbers after the $\pm$ denote the error of the thermometer, as given by the manufacturer. My interpretation of that was always that the manufacturer did a lot of measurements with a calibrated source and calculated the 'descriptive' variance of those values, therefore saving me the fuss of doing several measurements. –Wojciech Morawiec Sep 29 '13 at 21:46 2 Understanding systematic errors is the key. Let $\mu$ be the critical temperature (CT).
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 http://math.stackexchange.com/questions/123276/error-propagation-on-weighted-mean 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 The best answers are error propagation 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 $\sqrt{e_1^2 + e_2^2}$, where $e_1$ and $e_2$ and the error propagation average 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 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: