Precision of a measuring device, by definition, is the repeatability you
find using the device (like a tolerance on a milling machine for example).
It is different from the accuracy of the device (how close the measurement
is to the true value). To obtain the precision of something like a measuring
device, I don' t know of any other method than collecting an adequate sample
size on one standard test piece and calculate standard deviation, as you
suggest.
However, to me, when we talk about precision of an experiment, it is
referring to the power (often ignored), defined as 1-beta, where:
Beta = P(type II error)
Thus, power can be interpreted as P(reject null hypothesis|hypothesis is
false)
Beta, however, is also dependent on sample size and standard deviation.
The point: if your experiment is to determine difference within the samples,
it sounds like a randomized complete block design will work well, using
differing samples as blocks. If there are differences in samples, a post-hoc
test can be used to determine which one(s) are significantly different
statistically.
Thus, the precision would be the magnitude of difference you can reject the
null hypothesis with a specified power. This can be done via tables or with
many statistical software packages.
Any other thoughts on the topic out there?
Dave
David Hebert
Research Engineer, Motion Analysis Laboratory
Shriners Hospital For Children
Springfield, MA 01104-2396
413-735-1265
dhebert@shrinenet.org
-----Original Message-----
From: Barker, Dan (RGH) [mailto
an.Barker@RGH.SA.GOV.AU]Sent: Monday, March 27, 2000 6:04 AM
To: BIOMCH-L@NIC.SURFNET.NL
Subject: Precision of an experiment
Dear Biomech-l subscribers
To calculate the precision of a measuring device or experiment it is usual
to repeat the test a number of times ON ONE SPECIMEN, calculate the mean and
SD of the variable of interest i.e. strain, and then calculate the
coefficient of variation. CoV=standard dev./mean.
Is there however another method to assess precision when (for example 4)
DIFFERENT SAMPLES are tested twice each i.e. 8 measures (4 paired
measurements)? Therefore instead of multiple repeated tests of one specimen,
I have multiple specimens with only one repeated measure of each.
Thanks in advance
Dan Barker
Dan.Barker@rgh.sa.gov.au
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