Propagation Of Error Average
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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 scientifically? The answer to this fairly common propagation of error division question depends on how the individual measurements are combined in the result. We will treat each
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case separately: Addition of measured quantities If you have measured values for the quantities X, Y, and Z, with uncertainties dX, dY, and dZ, and error propagation physics 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 can also serve as an upper bound for the error. Please note that the rule error propagation square root 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 m Multiplication of measured quantities In the same way as for sums and differences, we can also state the result for the
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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 rule for the uncertainty in c, dc = 0. Example: If an object is realeased from rest and is in free fall, and if you measure the velocity of this object at some point to be v = - 3.8+-0.3 m/s, how long has
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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 http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm question Anybody can answer 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 $x + y$ http://math.stackexchange.com/questions/123276/error-propagation-on-weighted-mean 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 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}
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