Re: what good is "percentage error"?

[Doug Craigen <dcc@cyberspc.mb.ca>]

On Mon, 12 May 1997, John E. Gastineau wrote:

> It's time to stir the phys-l pot again. 
>
> Today I was looking through a commercial high school physics level lab
> manual, and came across the instruction to perform an experiment once
> and then determine the "percentage error" by calculating 
>
> (accepted - measured)/accepted * 100%. 

I have commonly found this in the lab book for courses that I've been
asked to teach (a common variant is "theoretical" instead of "accepted").
I've settled on telling the students that this is a misnomer, that what is
being shown is a discrepancy between your value and some other value.  The
point is then not to think of it as an error, but something to be compared
with the estimated error in your value and the "accepted" value and then
make appropriate comments (e.g. "the discepancy between our value and the
accepted value is less than the estimated error of our measurements")

As for whether to express it as a percentage, I would personally
rather leave it as is (accepted - measured).  For errors in general I
encourage students to (at least mentally) line them up vertically with the
number to get something like:
     5.193
p/m  0.004

rather than 5.193 p/m 0.004

I find that this drastically cuts down on the number of cases of things 
like 5.2 p/m 0.004 or 5.19326893404 p/m 0.004 appearing in lab reports.
Expressing the quantity as a percentage does not work well for this
purpose.

(I suppose that I should also introduce the 5.193(4) format, but I guess I
like to see them use the p/m symbol.) 

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| Doug Craigen                                                 |
|                                                              |
| If you think Physics is no laughing matter, think again .... |
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