Propagation of Uncertainty
Author: J. M. McCormick
Last Update: August 27, 2010
Every measurement that we make in the laboratory has some
degree of uncertainty associated with it simply because no measuring device is
perfect. If a desired quantity can be found directly from a single measurement,
then the uncertainty in the quantity is completely determined by the precision
of the measurement. It is not so simple, however, when a quantity must be
calculated from two or more measurements, each with their own uncertainty. In
this case the precision of the final result depends on the uncertainties in each
of the measurements that went into calculating it. In other words, uncertainty
is always present and a measurement’s uncertainty is always carried through all
calculations that use it.
One might think that all we need to do is perform the
calculation at the extreme of each variable’s confidence interval, and the
result reflecting the uncertainty in the calculated quantity. Although this
works in some instances, it usually fails, because we need to account for the
distribution of possible values in all of the measured variables and how that
affects the distribution of values in the calculated quantity. Although this
seems like a daunting task, the problem is solvable, and it has been solved, but
the proof will not be given here. The result is a general equation for the
propagation of uncertainty that is given as Eqn. 1.2 In Eqn. 1 f
is a function in several variables, xi, each with their own
From Eqn. 1, it is possible to calculate the uncertainty
in the function, Δf, if we know the uncertainties in each variable and
the functional form of f (so we can calculate the partial derivatives
with respect to each variable). It is easier to understand how this all works by
doing several examples.
Example 1: f = x + y (the
result is the same for f = x – y).
Let the uncertainty in x and y be Δx
and Δy, respectively. Taking the partial
derivatives with respect to each variable gives:
The uncertainty in f is then
Example 2: f = x•y (also works for f
Again let the uncertainty in x and y again
be Δx and Δy,
respectively. Taking the partial derivatives with respect to each variable
The uncertainty in f is then
This result is more commonly written by dividing both
sides by f = x•y to give
Although the idea of error propagation may seem
intimidating, you have already been using it since your first chemistry class
when you applied the rules for significant figures in calculations. These rules
are simplified versions of Eqn. 2 and Eqn. 3, assuming that Δx
and Δy are both 1 in the last decimal place
quoted. The formal mathematical proof of this is well beyond this short
introduction, but two examples may convince you.
If we add 15.11 and 0.021, the answer is 15.13 according
to the rules of significant figures. This assumed that Δx
= 0.01 (x = 15.11) and Δy = 0.001 (y
= 0.021), substituting these values into Eqn. 2, we get
basic statistics, we know that the uncertainty begins in the first non-zero
decimal place, which in this case this means that the last significant figure in
the sum is the 1/100ths place. According to the rules for propagation
of error the result of our calculation is 15.13 ± 0.01, exactly what the
significant figure rules gave us.
If we had multiplied the numbers together, instead of
adding them, our result would have been 0.32 according to the rules of
significant figures. Again assuming Δx = 0.01 and
Δy = 0.001, and using Eqn. 3, we can determine
Δf as follows.
Once again we see that the uncertainty begins in the
second decimal place, which gives the same result as the significant figures
The significant figure rules are important to know and use
in all chemistry calculations, but they are limited in that they assume an
uncertainty in the measured quantities. So while the significant figure rules
are always to be used in any calculation, when precision matters a propagation
of error analysis must also be performed to obtain an accurate prediction of the
uncertainty arising from the precision of the measured quantities.
CHEM 120, you
have measured the dimensions of a copper block (assumed to be a regular
rectangular box) and calculated the box's volume from the dimensions. In that
exercise you were given an
that allowed you to calculate the minimum uncertainty that could be expected in
the box's volume based solely on the uncertainties in the measured dimensions,
now derive that equation using the procedure given above.
Let x, y and z be the box's length,
width and height, respectively, and the uncertainties be Δx,
Δy, Δz. Since
V = x·y·z, we can use Eqn. 1 to determine the
uncertainty in the volume (ΔV), which results
in Eqn. 4. We know that
and can then make these substitutions in Eqn. 4 to give Eqn. 5.
Dividing both sides by V gives Eqn. 6 and
simplifying gives Eqn. 7 (which you probably could have guessed from the form of
Eqn. 1 and Eqn. 3). Multiplying both sides by V then gives the equation used in
the CHEM 120 Determination of Density exercise.
Note that there are several implications of Eqn. 7.
First, if one side has a large uncertainty relative to the length of that side
(such as when one side is very short), then this side will dominate the
uncertainty. Second, when the volume is large and the uncertainty in measuring
a dimension is small compared to the uncertainty in the measurement, then the
uncertainty in the volume will be small. The experimental implication of this
is that, if you want the smallest uncertainty in a box's volume, make sure it is
a big box, with no unusually short side and use the most precise measurement
You have measured the volume and mass of a set of regular
wooden blocks and have fit a graph of their volume as a function of their mass
to a straight line using the regression package in Excel. What is the predicted
uncertainty in the density of the wood (Δd) given the uncertainty in the
slope, s, of the best fit line is Δs and the uncertainty in the
intercept is Δb? Note that you have also seen this
before in the CHEM 120 Determination of Density exercise, but now you can
The relationship between volume and mass is
This is a linear equation (y = s•x + b) where
Note that b does not affect
the value of d and so Δb has no effect on Δd.
The relationship between Δs and Δd can be calculated by simply
substituting d in place of f and s in place of x in Eqn. 3
We could have also have used Eqn. 1. First we need to
find the first derivative of the density with respect to the slope, which is
Substituting this into Eqn. 1 gives
which rearranges to
. Recognizing the relationship between s and d, this simplifies
This problem is the simplest example of how one determines
the uncertainty in a quantity extracted from a best-fit line. In general you
will have the uncertainty in the slope and intercept and the relationship
between each of these to the desired quantities. It is then a simple process to
apply Eqn. 1, where f is either the slope or intercept.
Propagation of Uncertainty through a Calibration Curve
A situation that is often encountered in chemistry is the
use of a calibration curve to determine a value of some quantity from another,
measured quantity. For example, in
120 you created and used a calibration curve to determine the percent by
mass of aluminum in alum. In that exercise, we did not propagate the uncertainty
associated with the absorbance measurement through the calibration curve to the
percent by mass. However, in most quantitative measurements, it is necessary to
propagate the uncertainty in a measured value through a calibration curve to the
final value being sought. The general procedure is quite straight-forward, and
is covered in detail in
CHEM 222. Therefore, only a very basic review of the fundamental equations
and how to implement them in Excel will be presented here. You are referred to
any analytical chemistry textbook for more details.3
For a linear least squares analysis we need to define
several parameters. We will assume that the equation of a straight line takes
the form y = mx + b (where m is the slope and b
the intercept) and that the x values are known precisely. Let there be
N individual data points (so there are N ordered pairs xi,
yi) in the calibration curve. Further, let ymeas
be the average response of our unknown sample based on M replicate
measurements, and let Smeas be the standard deviation of the
result from the calibration curve. Note that Smeas is the
standard deviation associated with the x value (xmeas)
corresponding to ymeas, and should not be confused with Sr,
the standard deviation about the regression. We can then draw up the following
table to summarize the equations that we need to calculate the parameters that
we are most interested in (xmeas and Smeas).
Location on Regression
Under the ANOVA heading it is the entry in
the row labeled "Total" in the "SS" column.
Coefficient listed under “X Variable 1”.
SLOPE(known y's, known x's)
Coefficient listed under “Intercept”.
INTERCEPT(known y's, known x's)
"Standard Error" under the Regression
STEYX(known y’s, known x’s)
Table 1. Relationships between standard equations
encountered in a linear least squares analysis and the Excel regression package
output and Excel commands. Note that arg in the Excel command refers to
a range of cells over which the command is to be calculated (e. g., E5:E10).
Although one could enter formulas in various cells to
calculate all of the intermediate parameters needed to determine Smeas,
it is not necessary. One only needs to have a cell in which to enter the number
of replicate measurements on the unknown (M) and then it is possible to
calculate Smeas using only the STEYX, SLOPE, INTERCEPT, COUNT,
DEVSQ and SQRT Excel functions. For example, in the spreadsheet shown in Fig.
1, cell D16 contains the formula
which calculates Smeas directly from the
potential as a function of temperature data. Adding a cell that will contain
ymeas (cell D17 in Fig. 1), allows calculation of xmeas
value (cell D18) and its uncertainty at 95% confidence (cell D19). Click
here to review how this is done using Smeas and Student’s
t. Note that instead of using N in the calculation of the
uncertainty from Smeas, one must use N-2 because two
degrees of freedom have been used to find the slope and the intercept.
Figure 1. An example of an Excel spreadsheet that
may be used to calculate an x value (temperature, in this case) from a
measured y value (potential) along with the uncertainty in the measured
x value at 95% confidence.
Propagation of Uncertainty of Two Lines to their
Sometimes it is necessary to determine the uncertainty in
the intersection of two lines. This problem is not trivial and the reader is
referred to the literature for more details.4
- 1. Click here to obtain this file in PDF format (link not yet active).
- 2. Andraos, J. J. Chem. Educ. 1996, 73, 150-154.
here to view this article on the Journal of Chemical Education web page
(Truman addresses and J. Chem. Educ. subscribers only).
- 3. Skoog, D. A.; West, D. M. and Holler, F. J. Fundamentals of
Analytical Chemistry, 5th Ed.; Saunders College Publishing:
New York, 1988; p. 39-42.
- 4. Carter Jr., K. N.; Scott; D. M.; Salmon, J. K. and Zarcone, G. S.
Anal. Chem. 1991, 63, 1270-1270. Click
here to view
this article in PDF format on the Analytical Chemistry web page (Truman
addresses and Analytical Chemistry subscribers only).