Recently I did an experiment on isolating casein and now that I am processing the results, I feel my method of uncertainty calculation is not feasible. The casein was isolated 3 times and the results are as follows: 2.386, 2.628 and 2.164 grams. The scale had an uncertainty of 3 decimal places. Furthermore, the casein was isolated from 100 mL of milk, measured using a measuring cylinder with an uncertainty of 1 mL. The mean uncertainty was found using the following formula:

$$Δx_\mathrm{avg} = \frac{Δx}{\sqrt{N}} = \frac{R}{2\sqrt{N}}$$

Hence, my calculations were:

$$\mathrm{Range} = 2.628 - 2.164 = 0.464$$

Thus the uncertainty for the average would be:

$$\frac{0.464}{2\sqrt{3}} = 0.133945262 = 0.2$$

This would result in a final value for the average amount of casein in the isolated milk of $\pu{2.4 ± 0.2 g}$.

However, this error seems unreasonable considering that my measurements were to 3 decimal places. Furthermore, how would I add the uncertainty of the 100 mL of milk to this value? Would I simply add the uncertainties?

  • 4
    $\begingroup$ You need to do two different things here: Error propagation and statistical comparison. If your statistical error is larger than expected from the error propagation, then your assumptions for possible errors are incomplete or wrong. $\endgroup$
    – Karl
    May 11, 2019 at 22:29
  • 3
    $\begingroup$ Resolution of scales has minimal influence on the statistical error introduced by the whole process. You can see the results differ at the first decimal place. $\endgroup$
    – Poutnik
    May 12, 2019 at 3:03

1 Answer 1


If we have 2 variables with absolute uncertainties(+) $A \pm \sigma_A$ and $B \pm \sigma_B$

(+) expressed as standard deviations, but applicable to confidence intervals as well.

Then $C = A \pm B$ has uncertainty

$$\sigma_C = \sqrt{{\sigma_A}^2 + {\sigma_B}^2}$$

For $C=A/B$ or $C=A\cdot B$,

$$\sigma_C = C \cdot \sqrt{\left(\frac {\sigma_A}{A}\right)^2+ \left(\frac {\sigma_B}{B}\right)^2 }$$

Let $A$ is content of cassein in 100 ml of milk.
Let $B$ is the actual volume of used milk.
Let $C=A\cdot B$ is the mass of cassein in the volume of milk used in analysis.

Then $\sigma_C$ is the estimation of the error of the amount of cassein (known as the result error)
$\sigma_A$ is the estimation of the error of the cassein mass in 100 ml of milk (unknown)
$\sigma_B$ is the estimation of the error of the milk volume (1 ml )

The error introduced by scales can be neglected, or introduced as additive variable $D=0$ with absolute error estimation $\sigma_D=0.001~ \text{g}$.

Another, the biggest factor is the used analytical method with its bias $E$ and error $\sigma_E$.

$$\begin{align} C &= D + E + A\cdot B \\ \sigma_{AB} &= A\cdot B \cdot \sqrt{\left(\frac {\sigma_A}{A}\right)^2+ \left(\frac {\sigma_B}{B}\right)^2 }\\ \sigma_C &= \sqrt{{\sigma_D}^2 + {\sigma_E}^2 + {\sigma_{AB}}^2}\\ \sigma_{AB} &= 2.4 \cdot \sqrt{\left(\frac {\sigma_A}{2.4}\right)^2+ \left(\frac {1}{100}\right)^2 }\\ 0.13 &= \sqrt{{0.001}^2 +{\sigma_E}^2 + {\sigma_{AB}}^2}\\ \end{align}$$


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