Error propagation question

  • Thread starter dipluso
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Main Question or Discussion Point

Hi,

I've been trying to reproduce the output of an analytical machine here at work by doing the calculation myself in Excel but I can't get the error to match. Perhaps I am propagating the error incorrectly...

The calculation is z = 2^(x-y)

The values are:
x = 24.96
y = 25.98
and the std. devs are:
deltax=0.265016
deltay=0.085049

I figured that the error in z should be:
delta z = z * sqrt((2*deltax/x)^2 + (2*deltay/y)^2)
which would give 0.010958

However, the software/machine reports the error as 0.07
I am not 100% sure how the software is calculating the error but it says it uses the standard deviations.

Am I making a mistake in my formula for error propagation? (I haven't done this in a while, frankly).

Any advice much appreciated.

Thanks,
-Alex
 

Answers and Replies

1,222
2
Remember that whatever variation is present in x and y will be exponentiated by the time it appears in Z. I am not sure about your error analysis method, but I prefer to go back to basics and do a simulation:

Code:
x = 24.96;
y = 25.98;

sx = 0.265016;
sy = 0.085049;

(* this simulates a million measurements of X and Y
 by drawing from a normal distribution with the
 mean and standard deviation you have given *)

distX = RandomReal[NormalDistribution[x, sx], 1000000];
distY = RandomReal[NormalDistribution[y, sy], 1000000];

distZ = 2^(distX - distY);

StandardDeviation[distZ]

0.0978848
I get a somewhat different answer from the instrument you are reading, so either the measurements are not normally distributed (i.e. the instrument is automatically correcting for a systematic bias) or else the standard deviations are not yet representative of the true population.

I attached a bitmap graphic of the Histograms for all three variables in this simulation, although for visual clarity I reduced the size of the simulation from 1 million measurements to only 1 thousand. The main feature to notice is that the Z distribution is slightly skewed towards larger values, likely as a result of exponentiating x - y. Hopefully someone more knowledgeable about statistics can explain how to properly quantify the error in the Z distribution.
 

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Last edited:
nicksauce
Science Advisor
Homework Helper
1,272
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Well in my opinion the error should be given by: (this is the error formula I've always been taught)

[tex]\sigma^2_z = (\frac{\partial z}{\partial x})^2\sigma^2_x + (\frac{\partial z}{\partial y})^2\sigma^2_y}[/tex]

which gives an error of 0.095135
 
4
0
Thanks to you both.
I have to say that you both give very reasonable answers - and indeed they report similar errors. I don't know why the value reported by the machine's software is different (mine was just plain wrong). Indeed it may be that it takes so other invisible factor into account. I'll have to ask the manufacturer.
Thanks again,
-A.
 

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