I have read the difference between precision and accuracy. As normally (mistakenly) they are considered to be the same, but in actual they are two different things.

My question is regarding the accuracy and precision of a measurement system. Lets say if i am measuring current across a resistor, which has a voltage tolerance of ±0.25%, and it is connected to a micro-controller having an ADC which is 0.5% accurate (including all the offset,gain and INL errors), my total accuracy with which I'm measuring the current is ±0.75%, i.e i will have a total absolute error of ±0.75%.

I have the following question:

1) If my system is ±0.75% accurate, can i say that its 1.50% accurate (i.e without the "±" sign) ?

2) For the above described system, can i say that if i take 20 different readings, and try to observe the precision of my system, will it be ±0.75% (maximum) as well ? For this i made a rough sketch attached herewith, showing the accuracy of ±0.75% by a red circle. Some readings are taken which (should always) lie inside this circle. So by looking at these readings, i think that we can say that our system's maximum precision will be ±0.75% as well. Am i right ?

enter image description here

3) I asked about precision in 2), am i right to say --> ± 0.75% i.e can precision be denoted by ± ? (like for accuracy ± is used.)

4) If precision can be denoted by ±, then just like 1), can i also say that the precision for this system is 1.50% ? (i.e without the "±" sign) ?

NOTE: The reason i asked 1) and 4) , is because normally i see things written like, this thing is x% precise or x% accurate, i.e with no ± sign with it. So without going in its detail, one should think of it as ±x% or ±0.5x% ? I mean what is the norm here ? I hope i made my self clear.

Please help me clear my doubts. Your helpful suggestions and comments would be appreciated.

Thank you.

  • \$\begingroup\$ You have made a mistake at the very beginning. You can not simply 'add accuracies'. In fact it is \$u_I=\sqrt{(\frac{\partial I}{\partial V} u_v)^2+(\frac{\partial I}{\partial R} u_R)^2}\$. (u denotes uncertainty) \$\endgroup\$
    – venny
    Commented Dec 6, 2015 at 13:24
  • \$\begingroup\$ i can add them simply for worst case scenarios, as my answer will be always greater than or equal to yours, as you have averaged the value. So i am considering the worst case here. i simply added the errors being offered by the resistor and adc to get the worst case error \$\endgroup\$
    – yiipmann
    Commented Dec 6, 2015 at 13:37
  • \$\begingroup\$ If I throw three darts at the bullseye and they all hit treble-20, how would you express the accuracy and precision? I try again and they all hit treble-3. How, if at all, do the results of the second attempt differ from the first? \$\endgroup\$
    – Chu
    Commented Dec 6, 2015 at 17:37

1 Answer 1

  1. Generally if an accuracy is given as 1.5%, it is assumed that this means ±1.5% unless otherwise stated. Thus you should not substitute one for the other.
  2. Precision can be better than accuracy. In your example, however, since the readings are near the edge of the circle and spread around, the precision is not much better than the accuracy. If the readings were closer together then the precision would be better than the accuracy.
  3. Precision could be expressed using ± meaning that the readings will cluster around the average reading within that percentage.
  4. Again, if you express precision without the ± sign, then the ± will generally be assumed. It is better to use the ± sign and avoid any ambiguity.
  • \$\begingroup\$ Yes you are right, precision can be better, but i was asking about the worst case, that is why i showed the readings far from each other. I got all my answers from your post. Thanks alot! \$\endgroup\$
    – yiipmann
    Commented Dec 6, 2015 at 14:08

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