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For examples, if I measure length with a meter stick with the smallest unit of 1mm, can I use the uncertainty of 0.5mm in this formula: $$\Delta C = \sqrt{\left(\frac{\partial C}{\partial x_1}\Delta x_1\right)^2 + \left(\frac{\partial C}{\partial x_2}\Delta x_2\right)^2 + \hspace{0.3cm}...}$$ The reason I ask is that according to my understanding the uncertainties used in the formula carry with them a probabilistic nature. Assume the measurements are distributed normally, then $\Delta x_1,\Delta x_2,...$ can be thought of as 1 standard deviation from the mean, which gives the range that the true value can fall into 68% of the time. In contrast, the 0.5mm estimated is the maximum error that we can have(true value fall into this range 100% of the time). So how can I fit this uncertainty of 0.5mm into this formula? Assume I use 0.5mm directly into the formula, what will the result ($\Delta C$) mean? I doubt it can be considered to be either the maximum error possible or 1 standard deviation from the mean

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The modern framework for the evaluation of measurement uncertainty is given by a series of guides prepared by the Joint Committees for Guides in Metrology (JCGM) that can be found on the BIPM website.

The idea is that a measurand is modelled by a random variable (even in the case of a single measurement), and a measured quantity value $x$ is thus considered as a realization of a random variable $X$. The result of a measurement should then represent the information available about the random variable associated to a measurand, either as a probability density function or, more succinctly, as a representative value (e.g., the mean or the median) and a measure of uncertainty (spread) like the standard deviation. Scientific judgement and statistics (classic or Bayesian) offer techniques to assign probabilities to events and properties to random variables.

When the spread of a random variable $X$ is expressed as a standard deviation, it is called standard uncertainty and denoted by $u(x)$.

In the case of an indirect measurement modelled by a measurement function of the type $Y=f(X_1,\ldots,X_n)$ the law of propagation of uncertainty for uncorrelated input quantities $X_1,\ldots,X_n$ is given by

$$u(y) = \sqrt{\sum_{k=1}^n\left(\frac{\partial f}{\partial x_k}\right)^2u^2(x_k)}.$$

There are only two assumptions underlying the above equations: i) that the input quantities $X_1,\ldots,X_n$ are uncorrelated (there is also a more general formula for correlated quantities); ii) that the spread of the input quantities is sufficiently small that the function $f$ can be approximated as a first-order Taylor series (there are methods for the propagation of uncertainty when the linearity assumption fails). You don't need to assume a Gaussian distribution. The above equation should be used whether the uncertainty of the input quantities has been evaluated from statistics on repeated measurements or from scientific judgement on a single measurement.

In this framework, you can assign a probability distribution to a ruler measurement by using scientific judgement. In the language of metrology this is called a Type B evaluation of the uncertainty.

Now, let's suppose that we have a ruler with marks that are 1 mm apart. If you tell me that you measured a length of, say, 100 mm, I can think: given the marks, if the length were greater than 100.5 mm, he would have said 101 mm; conversely, if the length were less than 99.5 mm, he would have said 99 mm. Without further information, I can model the length $l$ as a random variable $L$ with uniform distribution between 99.5 mm and 100.5 mm, with half witdh $\delta l=0.5\,\mathrm{mm}$. The standard uncertainty would then be (see §4.3.7 of the GUM)

$$u(l) = \frac{\delta l}{\sqrt{3}} \approx \frac{0.5\,\mathrm{mm}}{\sqrt{3}}\approx 0.3\,\mathrm{mm}$$

However, I can also think in a more refined way: If the length is 100.5 mm, there is a 50% chance that you say 100 mm and 50% chance that you say 101 mm, and when increasing the length between 100.5 mmm and 101 mm there is less and less probability that you say 100 mm (the same when going from 100 mm to 99). So, a better assumption for the probability distribution of $L$ would be that of a triangle distribution with half width $\delta l=1\,\mathrm{mm}$. With this assumption, the uncertainty evaluation yields a slightly greater result. In fact, from the linked GUM document, figure 2(b), we have

$$u(l) = \frac{\delta l}{\sqrt{6}} \approx \frac{1\,\mathrm{mm}}{\sqrt{3}}\approx 0.4\,\mathrm{mm}$$

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Seems a reasonable question. You are right, the error addition formula you quote assumes independent uncertainties and the interpretation of the result as a 68% confidence interval requires that the $\Delta x_i$ in the formula are the standard deviations of a normal distribution.

This works for any factor of the Gaussian sigma. If your error is really a 2 or 3-sigma error then your final combined error would also be a 2- or 3-sigma error - but only if the uncertainties were actually normally distributed. In this case, likely they are not.

So, here is what you could do. (i) Assume that 0.5mm error is actually a 2-sigma uncertainty (95% confidence) and divide this by 2 to estimate a 1-sigma error and then use the standard formula.

(ii) Adopt an empirical approach. If you have several things to measure, then measure them each $\geq 5$ times and take the standard deviation of your result. Use this as the 1-sigma uncertainty in the error formula above.

Even the above will be crude approximations. Unfortunately estimating uncertainties is often not an entirely rigorous process. Just explain and attempt to justify what you are doing.

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  • $\begingroup$ No, that equation does not assume a normal distribution. $\endgroup$ – Massimo Ortolano Oct 27 '16 at 8:53
  • $\begingroup$ @MassimoOrtolano Agreed - only the interpretation of the terms and result does. $\endgroup$ – Rob Jeffries Oct 27 '16 at 10:17

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