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.

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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.

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varialectio t1_jadehk0 wrote

1 - Yes. The relative uncertainty must stay the same so X ± 1 per 10 seconds is the same as X * 6 ± 6 per minute.

2 - Averaging multiple results reduces the error by a factor of 1/✓n. Root 5 is a little over 2 so averaging five measurements would roughly halve the uncertainty.

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saywherefore t1_jadf9gk wrote

The easiest way to see the effect of uncertainties in a calculation is to redo the sums with the extreme values of the range.

So let’s say your average value was 20cm^3 per 10 seconds. The limits are:

20 - 1 = 19cm^3 per 10 seconds = 19 x 6 = 114 cm^3 per minute

20 + 1 = 21cm^3 per 10 seconds etc

This also helps to see how uncertainties combine, by considering all the limits that affect the final result in the same direction, although as another commenter alluded to, you don’t necessarily need to combine all your uncertainties linearly.

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