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### Calculating Errors in Measurement

Date: 08/06/2002 at 16:02:20
From: Jeaneen Aldridge
Subject: Significant digits/calculating error in measurement

Dr. Math:

related questions.

Suppose you are told that a digital scale measures to the nearest
0.2kg and asked to find the absolute error involved in using the scale
to weigh a cat that weighs 7.4 kg.

It's not clear from the question whether 7.4 is the cat's "true"
weight or the weight given by the scale, but I'm going to assume that
it is the weight given by the scale. The way I figured it, the scale
is in increments of 0.2kg, so the cat's actual weight would fall
between 7.3 and 7.5, making the largest possible value for the
absolute error 0.1 kg. However, my friend says the actual value would
be between 7.2 and 7.6 kg, making the absolute error 0.2 kg.

The next question goes like this: a schoolroom ruler measures lengths
to the nearest millimeter. Find the absolute error in using the ruler
to measure a line of length 0.5 cm.

Here again, I'm assuming that if you can measure to the nearest
millimeter, the actual value (or the measured value; once again it's
not clear which is which) would fall somewhere between 0.45 and
0.55 cm and the absolute error would be .05 cm.

My friend agrees with me on this one, but somehow I feel that we are
both on shaky ground. I need an authoritative answer and explanation.
Hope you can help.

Also, could you explain how to calculate the relative and percent
error in cases like the two given questions, where you don't really
know the "true" value of the quantity? In both questions above, I've
assumed that since the "true" value isn't really known, you would just
determine the greatest distance that the actual value could differ
from the measured value and use that difference as the absolute error.
Every place I've looked defines relative error as the absolute error
divided by the true (or accepted) value. In this case, the true value
is unknown.  So, would you just divide the absolute error by the
measured value?

Thanks,
ja

Date: 08/07/2002 at 13:03:49
From: Doctor Peterson
Subject: Re: Significant digits/calculating error in measurement

Hi, Jeaneen.

Suppose the scale is an analog scale that looks like this:

7.0   7.2   7.4   7.6   7.8   8.0
...--+-----+-----+-----+-----+-----+--...
^

Your digital scale sort of snaps the needle to the nearest marking;
so the actual weight might be anywhere in this range:

7.0   7.2   7.4   7.6   7.8   8.0
...--+-----+-----+-----+-----+-----+--...
|<--->|
7.3   7.5

So the absolute error would be no greater than 0.1. I can't imagine
why your friend would think a cat weighing 7.2 kg could show up as
7.4.

If we took the 7.4 kg to be the cat's actual, exact weight, then the
absolute error would be 0, since the scale would show 7.4 kg. I doubt
that this is what was intended; rather, you would see 7.4 not knowing
that that was the exact weight, and would have to determine from your
observation what is the MAXIMUM possible absolute error based solely
on the precision of the scale, namely 0.1 kg.

Note that the errors we are talking about are unrelated to any
inaccuracy in calibration; if the scale were off by 5 kg, the absolute
error would be 5.1 kg, not 0.1 kg, but we have no way of knowing that.

You're right about the question concerning the ruler, and for just the
same reasons as above. Of course, a ruler doesn't really measure
anything; you read it and can often estimate more accurately than the
markings. Often it is assumed that you can recognize a value halfway
between markings, so that you could distinguish 0.45 from 0.5, and the
error would be more like 0.25. But from the wording of the problem, I
think you are intended to take it just as you did.

>Also, could you explain how to calculate the relative and percent
>error in cases like the two given the questions, where you don't
>really know the "true" value of the quantity...

Ordinarily you go by what you see, not by what you don't know, so you
would divide the absolute error by the actual MEASURED value to find
the relative error. After all, "true" values are NEVER known!

But just to increase our confidence, let's suppose that the actual
value is x, and we measured m. Then the relative error based on the
actual value would be

|m-x|/x

whereas our calculation based on the measurement would give

|m-x|/m

How far off are we? For simplicity I'll suppose that m>x, so the
difference in these two numbers is

m-x   m-x          1     1           m-x
--- - --- = (m-x)[--- - ---] = (m-x)[---] = (m-x)^2/(mx)
x     m           x     m            mx

What is the relative error in our value for the relative error? It
will be this number divided by the correct value, (m-x)/x:

(m-x)^2/(mx)
------------ = (m-x)/m
(m-x)/x

Interesting: the relative error in the relative error is the same as
the relative error itself! So for example if we found that the
relative error was 0.1, the ACTUAL relative error would be off by no
more than 0.1 of this, making it between 0.09 and 0.11: 10% plus-or-
minus 10%. That's close enough for this sort of estimate.

Does that help reassure you that we aren't losing anything by doing
our calculations based only on what we know?

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/

Date: 08/07/2002 at 20:11:28
From: Jeaneen Aldridge
Subject: Thank you

Dr. Peterson:

THANK YOU. You cannot imagine how incredibly happy I was to get your
reply to my problem. I am so relieved to have an answer that makes
sense - I have been struggling to find an answer to this problem for
several days now, and I feel as if a big weight has been lifted from
my shoulders. I teach high school math, and I am supposed to teach
significant figures and errors in measurement for the first time in a
couple of weeks. The book I'm using poses these problems with no
examples and gives several incorrect answers in the key. I've been
asking people and looking for information... you just don't know how
happy you have made me. Now I can feel confident when I present the
subject to my students. You've indirectly helped a lot of people -
thanks again and know that you have really been a great help to me and
my students. I really appreciate it.

Sincerely,

Jeaneen

Date: 08/08/2002 at 08:43:15
From: Doctor Peterson
Subject: Re: Thank you

Hi, Jeaneen.

It always helps to have several good references, so I looked for a
few good sites on this topic that will either give a good overview,
or will cover some of the details beyond what you probably have to
teach. Here are a few (found by using google.com to search for
"absolute relative error measurement"; there's a lot more):

The Vocabulary of Error Analysis - Univ. of Michigan
http://instructor.physics.lsa.umich.edu/ip-labs/tutorials/errors/vocab.html

Error in Measurements - Introduction to Chemical Sciences,
James A. Plambeck, Univ. of Alberta

Relative Errors - Numerical Methods Tutorial, MathsDirect
http://www.mathsyear2000.org/alevel/pure/purtutnumerr1.htm

Also, don't forget to search our site for the phrases "relative
error" and "significant digit"; we have several discussions that

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
Associated Topics:
Middle School Measurement

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