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Why should the rules for propagating significant digits not be applied to averages?
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01/16/00
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These two facts are too often forgotten when the rules for adding, subtracting, multiplying, and dividing measurements
are applied to propagate error through a calculation:
- A measurement or a result calculated from measurements is written so that only the final digit is uncertain.
- The actual scatter in the data reflects precision, not the number of digits in each individual piece of data.
Ignoring these last two statements may cause some trouble with calculations that involve repeated measurements.
Consider the following very controversial problem from the first version of this site's significant figures quiz:
A person steps on and off a bathroom scale 5 times, balancing themselves slightly differently every time. The five weights on the
scale are 152.1 lbs, 154.7 lbs, 151.0 lbs, 153.5 lbs, and 152.3 lbs. What is the average weight, correctly rounded?
If you dutifully follow the guidelines, your reasoning goes something like this:
- Adding up the weights gives a total weight of 763.6 lbs, which has 4 significant figures.
- To get the average, divide by 5. The 5 is an exact number, so you don't lose any significant figures by doing this.
The exact answer is 152.72 lbs
- The answer rounded to 4 digits is 152.7 lbs.
That implies that you know the weight to the nearest tenth of a pound.
But do you really know the weight this precisely? Is the tenths place really the first uncertain digit?
This is one of those cases where you have to keep the definition of significant figures in mind.
You have a series of replicated measurements, so you can see that the precision implied by the scale isn't the same as the precision you're actually getting, because the person is balancing themselves slightly differently every time. Even though the measuring device has a scale that allows you to read to the nearest 10th of a pound, the measurements are changing in pounds place. Pounds are the first uncertain digit, so the reported weight should be 153 lbs.
Think this is just hair-splitting? Consider another example.
Three replicate combustion analyses of a purified drug sample show that the sample contains
60.080, 60.160, and 60.000 percent carbon. The average by following the rules for adding significant figures
is 60.080, which implies that the uncertainty in the average is in the thousandths place.
The uncertainty is actually in the hundredths or even the tenths place (+/- 0.08 around
the average). If the number 60.080% C alone appears on a court report, it would tend to
support the conclusion that the sample was cocaine hydrochloride (60.08% C) rather than aspirin
(60.00% C) because the apparent error is much smaller than the difference between carbon percentages for these two substances. However,
if the actual uncertainty was used to round the answer, it's obvious that you
can't tell whether the sample was cocaine or aspirin from the data given. (Better still: provide an error estimate along with the average, and collect enough data to do a decent statistical hypothesis test!)
If you'd like to learn more about when the rules of thumb apply, and see other cases where the rules of thumb lead to incorrectly rounded results, see the following references. (J. Chem. Ed. = The Journal of Chemical Education).
- D. A. Skoog, D. M. West, F. J. Holler, Fundamentals of Analytical Chemistry, 6th Ed., Saunders, 1992, pp 28-29.
- B. L. Earl, "A comment on significant figures and propagation of uncertainty", J. Chem. Ed., 65, 186, 1988.
- D. M. Graham, "A refinement of the extreme value rule for significant figures", J. Chem. Ed., 65, 660, 1988.
- L. M. Schwartz, "Rules for propagation of significant figures", J. Chem. Ed., 64, 471,
1987.
- S. Stieg, "Rules for propagation of significant figures", J. Chem. Ed., 64 471, 1987.
- L. M. Schwartz, "Rules for propagation of significant figures", J. Chem. Ed., 62, 693, 1985.
Author: Fred Senese senese@antoine.frostburg.edu |