Chapter 2: Error Analysis

2.4a Propagation of error (One variable) (without calculus!)

Very often the result that you want from an experiment is not the directly measured quantity, but must be calculated from one or more measured quantities. We need to know how to get from the uncertainty in some quantity x to the corresponding uncertainty in another quantity q(x) calculated from x. The fancy name for this process is "propagation of errors."

As an example, suppose I've measured the edge length of a cube. In order to assess the random errors in the measurement process, I did it several times, and got

5.18, 5.31, 5.26, 5.16, 5.25, 5.14 cm

The mean of these measurements is 5.217 cm, with a standard error of .027 cm. If this were our final result we'd quote it as 5.22 ± .03 cm, but let's hold on to the extra significant figure for a moment, to help us see what's going on.

Now suppose that what I really want to know is the volume V = L³ of the cube. The best value I can get from my experimental result is plainly (5.217 cm)³ = 142.0 cm³; but what uncertainty should I associate with it? Well, if the true length were one standard error higher than the measured value, that is 5.217 + 0.027 = 5.244 cm, it would make the volume (5.244 cm)³ = 144.2 cm³; or if it were one standard error lower, the volume would have come out 139.8 cm³.

Whenever we put a "+ or -" on an experimental result, what we're doing is identifying a range of values within which the true value probably lies. Whatever the probability that the true volume of the cube lies between 139.8 and 144.2 cm³, in the example above, clearly it's the same as the probability that the true edge length lies somewhere between 5.190 and 5.244 cm. Since, therefore, the ±0.027 cm I'm quoting is one standard deviation (of the mean value) of x, the corresponding ±2.2 cm³ is the standard deviation of the volume. You see that what we're doing is just figuring out how much variation is caused in a derived quantity q(x) by a given variation in another quantity x, from which q is derived.

The above calculation looks something like this:
    volume V = L³ = (5.217 cm)³ = 142.0 cm³
    volume V + standard error S_v = (L+S_v)³ = ((5.217 + 0.027) cm)³ = 144.2 cm³
    |(volume V + standard error S_v) - (volume V)| = (standard error S_v) = |144.2 cm³ - 142.0 cm³| = 2.2 cm³ , where we have used the absolute value to force the error to always be positive.
    Thus, the volume is 142.0 cm³ ± 2.2 cm³, not 142.0 cm³ ± 0.027 cm.

Section 2-5