Re: Determining reliability

Michel Rouzic wrote:
I need to determine the reliability of the precision of the value I
obtain for the average of a series of measurements.

Every measurement consists of measuring the exact same thing, only
there is noise in my measurements, so the only way for me to obtain a
reliable value for what i'm trying to measure is to calculate the
average of the measurements. Only once it's done, I have no idea on how
reliable the precision of this average value is, in other words, I
don't know how many signficant figures after the point I have.

So how do I find out how reliable the precision of the average value of
all my measurements is? I thought about for example perfoming 10 sets
of 10 mesurements, and "compare" (i think i mean calculate the
deviation) the average value of each, in order to get an idea of the
reliability on a set of 10 mesurements, however I'm not sure it would
indicate me reliability in a very, reliable way, well I don't know I'm
a bit confused with all that I admitt.

I just wish to find out how, for example on a single set of 100
mesurements I can determine the reliability of the average value, I
looked around but didn't find a way to do that so far.

What you want is called the "standard error of measurement" (SEM).
The sample standard deviation of n measurements of the same thing is
a legitimate estimate of the SEM. And if the measurement errors can be
assumed to normal (or close to it -- nothing is exactly normal) then
you can also get a confidence interval for the SEM: the limits of the
100C% confidence interval are s*sqrt[(n-1)/U] and s*sqrt[(n-1)/L],
where s is the sample standard deviation, and U & L are upper & lower
critical values of the chi-square distribution with n-1 degrees of
freedom, such that 50(1-C)% of the area is above U and 50(1-C)% of the
area is below L. For n = 10 and 95% confidence, the multipliers on s
are sqrt(9/19.0228) = .688 and sqrt(9/2.70039) = 1.8256.

If you have measurements of several different things, with each thing
measured several times, and if it is reasonable to assume that all the
measures have the same true SEM, then the variances can be pooled to
get a more reliable estimate of the SEM. Let ni be the number of
measures in set i, and let vi be the variance of those measures. Then
s = sqrt[(sum (ni-1)*vi)/(sum (ni-1))] is a better estimate of the
common SEM. The correponding confidence limits are s*sqrt[(n-m)/U] and
s*sqrt[(n-m)/L], where n = sum ni is the total number of measurements,
m is the number of sets of measures (i.e., the number of different
things measured), and U & L are critical values of the chi-square
distribution with n-m degrees of freedom. For n = 100, m = 10, the
multipliers on s for a 95% interval are sqrt(90/118.136) = .8728 and
sqrt(90/65.6466) = 1.1709.

.

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