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I have roughly 39 data points that are values for % error. I have been researching and found that it is not correct to simply take the average of the % errors. Is this correct? If so, what is the correct way to take the average for % error values? All of the data values are weighted the same. Thanks!

m.car
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2 Answers2

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You certainly can average the percent error values. That is a well defined operation. As Dilbert says, you can multiply them, too. Whether or not it expresses what you want can be very subtle. You are probably remembering the fact that averaging the percentage errors will not give the same result as dividing the average error by the average true value. If you explain carefully what you want to express, it will lead to the correct way to compute it.

Ross Millikan
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  • I want to say, our average percent error for all data sets is XX. – m.car Jul 10 '17 at 15:06
  • Say you have two samples, one with true value 10 and measured at 9 for 10% error and one with true value 100 and measured at 99 for 1% error. Do you want to say the average percentage error is 5.5%? The absolute error is 1 in both cases, so you could say the average percent error is 1 out of the average value of 55 for about 1.8%. If the true values are close together the naive averaging of percent errors will be fine. If not, you need to think about what you are trying to do. – Ross Millikan Jul 10 '17 at 15:25
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I assume that you have given the errors in one direction and not the absolute error.

Let $r_i$ the deviation from the real data point in decimal notation.

Then the average deviation is

$$d=\sqrt[n]{(1+r_1)\cdot (1+r_2)\cdot (1+r_3)\cdot \ldots\cdot (1+r_n)}-1$$

$r_i$ can be negative and positive. For instance, if the real value is 120 and the deviation is $-12$, then $r_1=\frac{-12}{120}=-0.1$ and consequently $1+r_1=0.9$.

To get the value in percentage you have to multiply $d$ by $100\%$.

callculus42
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  • I am using the absolute error... Actual-theoretical/theoreticalX100 – m.car Jul 10 '17 at 15:07
  • The forumula you have posted in you comment is not for the absolute error. It is the same formula as I have used in the answer.$ \frac{108-120}{120}=\frac{-12}{120}=-0.1$ – callculus42 Jul 10 '17 at 15:10
  • If you have given the percentage deviations then you have to transform them in decimal notation. In general $-x%=-0.0x$. Numerical example: $-3.2%=-0.032$ – callculus42 Jul 10 '17 at 16:09