[EMAIL PROTECTED] (Greg Heath) wrote in message
news:<[EMAIL PROTECTED]>...
> Rich Ulrich <[EMAIL PROTECTED]> wrote in message
>news:<[EMAIL PROTECTED]>...
> > On Wed, 2 Oct 2002 15:21:14 +0100, "Jacek Gomoluch"
> > <[EMAIL PROTECTED]> wrote:
> >
> > > I have a question concerning the calculation of the confidence interval of
> > > the mean of a measured variable.
> > >
> > > I am running several (m>30) simulation runs with same parameters but
> > > different random seeds. In each of these simulation runs, n samples (usually
> > > several thousand) of a measured variable are taken. In my simulations, the
> > > samples within each simulation run can not be assumed to be independent.
> > > However, according to the literature I read so far, the sample means of the
> > > different simulation runs are normally distributed (due to the central limit
> > > theorem). In the literature, it is assumed that the number of samples (n) is
> > > the same for each simulation run.
> > >
> > > With this assumption the confidence interval of the mean is calculated by:
> > >
> > > overall_sample_mean +/- z_value* ( standard_dev_of_the_sample means)
> > >
> > > However, in my experiments the number of samples (n) is not the same for
> > > each simulation run, but slightly different (e.g. 3105, 2934, 3050,...).
> > >
> > > My question: does this matter? Or can I still use the above formula?
> >
> > I hope you are computing the <SD_of_means> by looking at
> > those means as a set of numbers -- and *not* by looking at
> > the formula for Standard Error.
> >
> > So, you can take the SD of any set of numbers, and apply
> > the formula for their CI.
>
> In this case each number is a mean which has it's own CI. It seems to
> me that the grand CI should consist of two parts: That due to the
> spread of the individual measurements about the group mean and that
> due to the spread of the group means about the grand mean.
>
> If not, what am I missing?
Oh.
The grand CI is *not* associated with E{ (xi-mu)^2 } (the spread of
*individual* measurements, xi, about the population mean, mu.)
It is associated with
E{ (m0-mu)^2 } = s^2/N0
where
mu = population mean
s^2 = population variance
N0 = total number of measurements
m0 = grand sample mean
= (1/N0)SUM(j=1,Ng){ Nj mj }
Ng = number of groups
Nj = number of measurements in the jth group
mj = sample mean of the jth group.
The population variance is estimated from
s0^2 = [1/(N0-Ng)]SUM(j=1,Ng){ (Nj-1) sj^2 }
where
sj^2 = sample variance of the jth group.
Hope this helps.
Greg
.
.
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