I was asked by a fellow teacher at my school about significant figures and I gave her my answer to her problem and she was not satisfied, and through
our conversation I am not altogether happy with my answer either. I am somewhat interested in what you all thought about this particular problem. (its
actually a kinematics physics problem)
335 = v + (9.81)(30)
solve for v, (units removed for brevity)
My answer to the problem is that the 30 has only 1 significant figure so if I were to round before continuing, the product would be 300, but carrying
all digits through calculation you end up subtracting 335 from 294.3 to get 40.7 which is rounded to 40 due to the one significant figure present in
the original values.
Her argument is (I hope I capture it correctly) that the significance of the product of 9.81 and 30 lies with the hundredths place (300) and
that significance carries through to the end (again not rounding to the end) and her answer would be to round 40.7 down to zero because is is closer
to a real value than to round up to 100.
please tell me what you think.watson.fawkes - 11-11-2011 at 15:52
I was asked by a fellow teacher at my school about significant figures and I gave her my answer to her problem and she was not satisfied, and through
our conversation I am not altogether happy with my answer either.
The rules for significant digits don't
work well when subtracting values that are close together. Ever since I learned this is high school, I was dissatisfied that the rules just didn't
capture what was really going on. There was a chemical calculation I remember that I carried out with full +/- error bounds the whole way. By the end
of it the error was as large as the answer.
So then I got older and figured out that these rules are simplifications for random variables. To say that a value is "30 to 1 significant figure" is
saying that you have a random variable whose center is 30 and whose standard deviation is about 10. Similarly 335 is a random variable centered at 335
with deviation about 1. Unless you have some reason to believe otherwise, you should always assume that the variable is a Gaussian. (I never did work
out what the proper constant factor between the actual standard deviation and the width of the Gaussian.) Go read up on the Central Limit Theorem
(CLT) for why. Indeed the CLT is the reason that the significant digits regime works at all.
Cutting to the chase, the standard deviation for a random variable multiplied by a constant the product of the constant and the deviation, so the
product (9.81)(30) has deviation 98.1. So your teacher's right about that.
On the other hand, the center of the random variable is indeed at 40 or so, which means something like 40 +/- 100, a random variable which just
doesn't fit into the significant digits regime well.
Note: I've left out pretty much all mathematical sophistication in this treatment.Stifle - 11-11-2011 at 16:04
We were discussion the variability of the worst piece of information in the problem, the 30 could possibly be >25 or >35. Does this problem need
to be addressed according to CLT or random variables if the information happens to be a measurement? watson.fawkes - 11-11-2011 at 17:53
Does this problem need to be addressed according to CLT or random variables if the information happens to be a measurement?
Yes. Now that's a very good question actually, because it's not at all obvious that a "measurement" should actually be random. It's
a measurement, after all, something that's supposed to be fixed. What's really happening with every measurement, classical or quantum, is that you are
gauging a physical quantity by examining its interaction with a measurement device. The device has a certain mathematical model that allows use to
deduce things by using it. Yet the model is not perfectly realized. Not only are there tolerance deviations, temperature coefficients, but also purely
non-linear dynamics such as small motions under friction (in short, they're fractal). The reason that a Gaussian random variable is usually justified
is that there are lots of these factors that all add up together, and thus approach a Gaussian by the CLT.DerAlte - 11-11-2011 at 22:44
"(its actually a kinematics physics problem)
335 = v + (9.81)(30)
solve for v, (units removed for brevity)"
The units are obvious to any student of physics: g=9.81 m/s^2. Hence the problem was:
An object falling freely under the earth's gravity attains a velocity of 335 m/s after 30 seconds. What was its initial velocity?
Answer, 40.7 m/s. No clue, assumedly, was given to the sigificance of any numbers in this problem so none should be assumed. The assumption would
naturally be that 30 and 335 were not integers but taken as exact real numbers unless other wise stated. Had the 9.81 figure been stated as
9.81+/-0.01 three figure accuracy would have been correct as 40.7 +/- 0.3
Put it into a computer and you might have a problem depending what the program's number routine does: never mix integers and floating point unless the
program lets you. But if you need a computer for this one, as I suspect many modern students might, you deserve the answer it gives!
Now if 9.81 had been in the denominator, things are different. By examination 981 has factors 9x109 and 109 is prime and hence produces a recurring
decimal. So the student would be forced to round off the answer - or spend the rest of his life trying to calculate it. Or, if clever enough, present
the answer as a fraction of 981.
Regards
Der Alte
Magpie - 11-11-2011 at 23:15
Der Alte, I like your approach.
I was taught in engineering school to "never give away digits." So, rounding before doing any calculation would just seem wasteful and strange.
[Edited on 12-11-2011 by Magpie]Stifle - 12-11-2011 at 11:54
My only issue with DerAlte's explanation is that should the assumption of the data be that accuracy of measurement is implied in the written form
given in the problem, what convention should be followed? I was always taught to assume that the writer of the problem gave the information according
to significant figures unless bounds were specified.DerAlte - 12-11-2011 at 21:11
I get your drift, Stifle. But in the problem do not the two figures 30 and 350 suggest +/-5? Yet the figure for g suggests +/- 0,01? Which of these is
your poor student to take? 30+/-5, 350+/-5 or 9.81+/-0.01? There are twenty seven possible answers, all differing. The intelligent student would take
the numbers to be exact, as far as they go. The extremes are (if I calculated correctly) v=+95 or v=-13.7 - are you going to allow all such answers?
Common sense has to prevail.
Regards,
Der Alte unionised - 13-11-2011 at 02:03
"My answer to the problem is that the 30 has only 1 significant figure "
As far as I can see, there is no indication of whether that's true or not.
The imprecision on the numbers isn't given in the question and so you cannot realistically assign an imprecision to the answers.
The reason you and the teacher don't agree is that it's a poorly specified question and is therefore ambiguous.
In any event, while number of sig. fig. is a useful rule of thumb, it's not used so much for expressing error margins in professional science. White Yeti - 13-11-2011 at 13:59
I absolutely hate all the fuss that is made over significant figures. I had a teacher once that asked us to round intermediate answers to a correct
amount of significant figures in a multistep problem. This really struck a nerve because the purpose of solving a problem is to find an answer, not
demonstrate that you can use 3 or 4 significant figures throughout a problem -and be off by a small amount. From then on, I wrote down the amount of
significant figures required, and did the calculations with the exact numbers on my calculator, but don't tell anyone
It makes sense to report your final answer with a reasonable number of significant digits, but making a fuss out of them is just plain stupid.Stifle - 14-11-2011 at 12:25
Thank you so far for your comments and answers so far friends.
To address DerAlte's last comment, "The intelligent student would take the numbers to be exact, as far as they go".
My purpose to teaching significant figures is to get students thinking about the value of the measurements that they are taking in the classroom in
preparation for realistic treatment of errors later on. Treating the last significant digit as a best guess is a very strong thing to be able to get
into student's heads, especially when you start to bring real research and real lab work into the class later on.
The issue becomes, how to get students to treat this level of accuracy (significant figures and rules of calculating) with a level of respect that
they can use to build upon....
To address White Yeti's comment, I feel the same way about it, though given properly written questions, students should be able to follow through and
write an answer that is as good as the information given in the problem assuming that the writer of the problem wrote the numbers to be as accurate as
significant figure rules would imply. unionised - 14-11-2011 at 13:05
I recently had to explain to a colleague that reporting the result of a calculation as (something like) 12385.07 was absurd, not least because it was
the estimate of a town's population and I'm not sure what 0.07 of a person is meant to look like.
I'm sure that, had I asked them to, they could have reported it to 3 sig fig. The problem is that they simply hadn't thought about it.fledarmus - 14-11-2011 at 13:20
There are two major problems with the way people are taught to count significant figures. One has caught you here. What is the precision of the number
30? Should it be thought of as 3.0x10 (2 significant figures) or 3x10 (1 significant figure)? Without knowing how the measurement was obtained, there
is no way of determining how precise the measurement is: +/-0.5 or +/-5
The other problem is with integers. If you count 127 people entering a store, what is your level of precision? It is infinite. It isn't +/- 0.5 people
- there aren't any half people for you to have counted. Of course you could have made a mistake in your counting, but that is an error of accuracy,
not precision - your measurement, even if it is wrong, is still precise to an infinite degree.
I'm having a number of headaches with an electronic lab notebook I am working with now, that supposedly helps me out by doing all my calculations for
me. Unfortunately, it hasn't figured out that 1 equivalent of my limiting reagent is infinite precision, and even if I enter 2.000 equivalents for the
reagent I am adding, it will reduce that to 1 digit. It also figures the 300 mL of solvent that I added to be 1 significant digit, and won't give my
molarity past 1 significant digit. I can either lie and tell it I used 300.0 mL, or give up and use a calculator. I use a calculator a lot.
