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Calculating Percentage Difference confusion Watch

1. I always thought that percentage difference was calculated like this: ((X1-X2)/average of X) * 100

But on the mark scheme of a past AQA EMPA, one of the questions is:

The theoretical separation of the Moiré fringes when α = 2°, shows D = 0.0859 mm.
Calculate the percentage difference between this value and the student’s spreadsheet ( spreadsheet shows, D=0.0855)
result for D when α = 2°.

The mark scheme shows:

((0.0859-0.0855)/0.0859) × 100 (working must show 0.0859 in denominator, or 0/2)
= 0.466% or 0.47% only ! (ie 0.5% is worth 1 max)

Using an average of D would still give the same answer to such S.F but it clearly says 0 marks unless denominator is 0.0859. (average of D should be 0.0857)?

Anyone can clarify this for me?

Also when doing percentage difference, in the numerator, does it matter which value you put in first - basically is the percentage difference always meant to be positive (bigger value - smaller value) or does it not matter (so if you use smaller value and get negative percentage difference, just take away the negative sign)?

Thanks for any help
In this case the % difference would always be positive. It's simply a number showing how far the measured value differs from the "true" value".
And that is also why the true value is in the denominator. The % difference is given, in this case, with reference to the true (theoretical) value and measures how far away from it the student's results are.
Imagine you had a true value of 50 for some quantity and 3 students measured values of 45, 49 and 54
If you were to give the % difference between the students' results and the true value, does it not seem logical that the denominator would be the true value in each case?
Anyway, whether or not it seems right or logical, this is what the exam board is asking for and expects.
3. Thanks for your reply, so do you only do ((X1-X2)/Average X)) when you are comparing the percentage difference of 2 values that you measured yourself?
Thanks for your reply, so do you only do ((X1-X2)/Average X)) when you are comparing the percentage difference of 2 values that you measured yourself?
Yes, more or less. In this (your) case the two values are just any two values with neither one being more important than the other.
Using the true value in the denominator is really just showing that it is the important value from which you are measuring the difference of all the others.
5. Thanks for the help. For some reason it won't let me rep you.

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Updated: April 16, 2013
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