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I am trying to find the %Accuracy, in which I used this equation: %Accuracy = 100-%Error.

So far, I have faced two problems with this equation:

  1. When the Exact Value is Zero, the fraction can’t be used -> Solved by adding the same value to both Exact and Measured, to avoid the null denominator

  2. When the Measured value is twice bigger or smaller, the %Accuracy value will be in negative values, In which I don't see the meaning behind it.

For example:

if: Exact = 20; Measured = 25; %Accuracy = 75%;

But, when: Exact = 20; Measured = 45; %Accuracy = -25%;

What is the meaning of -25%? and How to change the range of [0-100], to take the values that are outside of it accurately?

Teco
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    According to the equation you wrote, for the second scenario, $%Error = \dfrac{1}{20} \times 100 = 5%$ and hence accuracy is $95%$. – Aniruddha Deshmukh Aug 22 '18 at 08:34
  • Thanks for your answer, it was a mistake of mine. The question is edited now, as the thing I am looking for is how to know the meaning of measured numbers that are twice bigger than the exact measurement. Can you help me in answering this question?

    Appreciated :)

     – Teco Aug 22 '18 at 08:52

1 Answers1

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For your problem 1, if the exact answer is zero, then a miss is as good as a mile. That is, any measured value other than zero is equally wrong. For your problem 2, there is a better way. If x and y are two non-zero real numbers, then define

%Error := 100 * min(1, 2 * abs(x - y) / (abs(x) + abs(y))), while if only one number is zero the %Error := 100%, and if both are zero then %Error := 0%. With this definition %Error is always between 0% and 100%. Now your definition of %Accuracy = 100-%Error behaves as you would expect.

Alternatively, you can adjust your original equation to %Error =100 * min(1, |Exact - Measured| / Exact).

Somos
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