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I have five values for the volume of sodium hydroxide needed to neutralise a fixed quantity of hydrochloric acid, each trial with an uncertainty of 0.05 mL. If I take the average of these five values, in what way would the uncertainty propagate? I can think of three of the following ways, but don't understand which would be correct:

  1. Because we can see the calculation of an average as the sum of the values for the five trials (with their uncertainties), and then a division by 5, we sum up the uncertainty to get 0.25 mL and then divide it by 5 to get the uncertainty of the average to be 0.05 mL.

  2. Because we can see the calculation of an average as the sum of the values for the five trials (with their uncertainties), and then a division by 5 (which has no uncertainty), we sum up the uncertainty to get 0.25 mL and then divide it by 5 to get the uncertainty of the average to be 0.25 mL.

  3. Range divided by two is often another way to calculate the uncertainty of an average value, which for my case results in a value different to the above two for the uncertainty of the average.

Which one, and why, would be the correct one to state next to the average value (with a plus-minus sign), as an uncertainty of the average value?

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If we assume that each of the trials had a normally distributed standard error of 0.05 mL, and that all of these errors were independent of each other, then the correct way to combine the errors is via quadrature. The idea here is that the average $\bar{x}$ will depend on each of the measurements $x_1, x_2, \dots$ by $$ \bar{x} = \frac{1}{5} (x_1 + x_2 + x_3 + x_4 + x_5). $$ If each $x_i$ has an uncertainty $\Delta x_i$, then the general formula for the uncertainty in the derived quantity $\bar{x}$ will be $$ (\Delta \bar{x})^2 = \sum_i \left( \frac{\partial \bar{x}}{\partial x_i} \Delta x_i \right)^2. $$ In the present case, all of the $\Delta x_i$ values are the same, and this works out to be $$ \Delta \bar{x} = \sqrt{ 5 \cdot \left(\frac{1}{5} \Delta x \right)^2} = \frac{\Delta x}{\sqrt{5}} \approx 0.0223 \, \text{mL}. $$ Note that this implies that as as you take more and more repeated measurements, you should get a better and better idea of what the "real" value is, since $\Delta \bar{x}$ is decreasing with time.

That said, there are some caveats to this technique. In particular, it makes a pretty strong (if standard) assumption about your measurement technique: namely, that you're equally likely to get a value that's too high or too low, and that your errors are distributed in some sort of bell curve about some "true" value. The method described here still holds (I think) under certain relaxations of the latter assumption. The former assumption, however, is pretty necessary for this logic to work; and it's easy to think of a situation where it might not hold. (In particular, "last-digit" measurement error might tend to be a bit too high or too low depending on many factors.) Proper error analysis requires a lot of careful thought; there's no one-size-fits-all approach to it.

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  • $\begingroup$ Thanks @MichaelSeifert, however, I do not think the 5 in the denominator enters the formula for the uncertainty of the average. The first 5 in front of the bracket is just the number of elements averaged over - a consequence of the summation operator. Then the second 5 in the denominater just before dX in your last equation is just the partial derivative. In other words, if you had summed instead of averaged (dropped the division by 5 in your first equation) you would get the same uncertainty as for the average. Or am I missing something? $\endgroup$ Oct 20, 2019 at 10:07
  • $\begingroup$ Thanks again @MichaelSeifert but I am basically wondering what, if anything happens to the 1/5 in your first equation for the average? Does that at all enter the calculation of the uncertainty of the derived quantity? $\endgroup$ Oct 21, 2019 at 18:14
  • $\begingroup$ Ah I now see, it was actually quite simple to understand. Feel free to delete my comments as they might confuse other users. Thanks for the patience! $\endgroup$ Oct 25, 2019 at 12:57
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You don't say how the 0.05 ml error arises - and that is crucial.

If it is a random error in estimating the volume, then I think the best estimate of the error is to calculate the mean and then calculate the uncertainty in that mean through standard error propagation, which assumes the measurements are independent of each other and that the error is pseudo-Gaussian. In which case the error in the mean is $0.05/\sqrt{5}$ ml and the more measurements you make, the smaller the uncertainty becomes.

On the other hand if the uncertainty is dominated by some systematic error - for instance you are using the same measuring cylinder for each measurement and the scale could be out by up to 0.05 ml. Then the uncertainty in your mean is at least 0.05 ml. I say "at least", because there then could be some additional measurement uncertainty that you have not taken account of. This could be estimated by looking at the standard deviation (of the population) of the 5 measurements.

This latter point would also be a good way of estimating the total uncertainty in the case where you thought there were no systematic errors or you were uncertain about your uncertainties! You take the standard deviation of the population to be an estimate of the true uncertainty in each measurement and then divide that by $\sqrt{5}$ to estimate the standard error in the mean.

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Either options 1 or 3 would be valid options, but perhaps one is more suitable than the other, depending on the details.

We can decide which one of option 1 and 2 is correct by understanding where the rules for propagating error come from.

Mathematically, the rules for calculating the uncertainty in a derived or calculated value based on measured values comes from a Taylor expansion. If $f(x_1, x_2, ...)$ is a function of the measured values $x_1, x_2, ...$, the the uncertainty on $f$ is

$\delta f^2 = \left(\frac{df}{dx_1}\right)^2\delta x_1^2 + \left(\frac{df}{dx_2}\right)^2\delta x_2^2+ ...$

In your case, $f = \frac{1}{5}\Sigma_{i=1}^5 x_i$,

and, $\frac{df}{dx_i} = \frac{1}{5}$,

therefore $\delta f^2 = \frac{1}{5^2}\Sigma_{i=1}^5 \delta x_i^2$.

Because all of your $\delta x_i$'s are equal, we can simplify this to

$\delta f^2 = \frac{1}{5} \delta x^2$,

or $\delta f = \frac{1}{\sqrt{5}} \delta x$,

However, sometimes you may not know all sources of variation in your measurement, and this makes it hard to estimate $\delta x_i$. In this case, and if you have enough measurements, you can use the data itself to estimate the uncertainty. This could be (among other things)

  • Standard deviation
  • Standard error
  • Range/2
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