Calculating the total uncertainty of the final product

Question

I have performed a large number of calculations in order to determine the ethanol content of wine. These included titrations using burettes, mass measurements using analytical balances, and others. I know the uncertainties of the apparatus.

I am given a task to find the ethanol content of wine (done) but also to calculate the total uncertainty by propagation of errors. As I understand it, the total uncertainty is given by

$$\delta F=\sqrt{\left(\delta q\right)^2+\left(\delta w\right)^2+...}$$

But - does this include the standard error? That is, to calculate the total uncertainty of the final value, will I need to find the standard error for all the sets of titrations (eg - for titration 1, a set of 3 values, find the standard error taking into account all results, even non-concordant ones, and do this for titration 2), and treat those values as fractional uncertainties? Should I add this value to the fractional uncertainties of the burette measurements (ie its own measurement uncertainty, not the standard errors from the readings), and to the measurement uncertainties of the other apparatus?

Finally - does the operation which I apply to the values that the uncertainties effect change the equation for the total uncertainty? Or do I just use the same equation as above?

Answer 1

General approach

\begin{equation} \sigma_{F}^2 = \sum \left( \frac{\partial F}{\partial x_i} \right)^2 \mathrm{d}x_i^2 \end{equation}

Here we neglect all correlation terms formed by squaring what is essentially the first order Taylor expansion with respect to the variables about their uncertainty. $x_i$ is a variable or parameter which contains uncertainty and $F$ is the function which depends on the variables and parameters $x_i$. $\mathrm{d}x_i$ is the uncertainty in $x_i,$ i.e., you measure the mass of your solution 10 times and it is 1 kg ± 0.1 kg. Your $\mathrm{d}x_i$ for mass is 0.1 kg. If you want you can think of $\mathrm{d}x_i$ as $\Delta x_i$ instead.

For a given calculation $f$ you take its derivative $F$ wrt each variable/parameter, hence you know its rate of change, and $\mathrm{d}x$ is the known/measured uncertainty of that variable/parameters, i.e., how much the function might be out for a given error in a specific parameter.

So, in your original expression you only accounted for the uncertainty in the parameters, but not how their error would propagate into each and every function/equation/calculation they are used. If a function is linear in a given $x_i$ then at most it will be out by the uncertainties magnitude multiplies by some constant in the equation. However, it may be that the function contains $x_i^{10}$ in which case the error in $x_i$ will be to the 10th power which could be huge! So, use the above equation, with the appropriate derivatives to "propagate" error.

Some background derivation

The change in a function due to a variable/parameter can follow from its differentiation

\begin{equation} \mathrm{d}f = \left( \frac{\partial f}{\partial x} \right) \mathrm{d}x + \left( \frac{\partial f}{\partial y} \right) \mathrm{d}y \end{equation}

It is possible that the error due to variable $x$ is equal but opposite to $y,$ and the errors cancel. However, when we square it, which gives us

\begin{equation} \sigma_f^2 = \left( \frac{\partial f}{\partial x} \right)^2 \mathrm{d}x^2 + \left( \frac{\partial f}{\partial y} \right)^2 \mathrm{d}y^2 + 2\left( \frac{\partial f}{\partial x} \right)\left( \frac{\partial f}{\partial y} \right) \mathrm{d}xy, \end{equation}

this way errors don't cancel. Also, we almost always assume the correlation term $\mathrm{d}xy$ is negligible. Neglecting the correlation terms, we find ourselves left with the top equation.

More info can be found here: Wikipedia uncertainty propagation. Also, don't forget to donate to Wikipedia. :)