Submitted by slackslackliner t3_11dlvyu in askscience
Lets say you are measuring O2 production of enzyme under various conditions. You measure the volume produced per 10 seconds. You would like to display this as a per min rate, so you multiple the recorded value by 6.
Your equipment, gas syringe, has a measurement uncertainty of ± 1cm^(3).
Does this mean the final value should have an uncertainty of ± 6cm^(3).
Does this change if I have 5 final values and average them? Does the average still have an uncertainty of ± 6cm^(3)?
Thanks for any help on this simple question, my googling wasn't helping. I am sorry but I really should know the answer, but I just don't.
Redingold t1_jadkevs wrote
Generally this is not a straightforward task, however if the uncertainties on the variables and the correlations between the variables are small, you can use a formula:
For a function f(x, y, z...), where x, y, z... have uncertainties σ*x, σy, σz..., the uncertainty of the function σf* is approximately sqrt((∂f/∂x)^(2)σ*x^2 + (∂f/∂y)^(2)σy^2 + (∂f/∂z)^(2)σz*^2 + ...).
You can see for simple cases that this produces sensible results. For example, for the case of multiplying a value by 6 to convert between volume per 10 seconds and volume per minute, we'd have f(x) = 6x. This gives us ∂f/∂x = 6, so σ*f* = sqrt(6^(2)σ*x^(2)), or σf* = 6σ*x*. Multiplying a value by 6, therefore, increases the uncertainty on it by 6 as well.
For another example, consider adding two variables, so f(x, y) = x + y. Then, ∂f/∂x and ∂f/∂y both equal 1, so σ*f* = sqrt(σ*x^(2) + σy^(2)). This is only an approximation based on assuming x and y are uncorrelated, if they are correlated then this isn't quite accurate (if they're correlated then σf* = sqrt(σ*x^(2) + σy^(2) + 2σxy*)), but the nice thing about this formula is you can use it in a lot of different situations.