Rich Ulrich <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...

[ snip, previous]

> Okay.  I think you are asking about describing a small fraction,
> and putting a Confidence Limit around it.  That part of the 
> problem, the numeric part, is not too hard.

That's it

> For a small proportion, we can consider the "counts"  to be
> numbers that are distributed as Poisson:  And in that case,
> the square root of the count is very  close to being "normal"
> with standard error  of  1/2.

>From my small knowledge, it looks like an application of the Central
Limit theorem. The distribution of the sample's means is closing to a
"normal" distribution, whatever the distribution of the population is.

> Next step: the  usual 95%  CI  is built   by taking 
> the mean,  +/-   twice the SE -- which would
> thus be  +/-  1.0  added to the square root of the count.

Does SE means Square Error ?

> Example:  If the Poisson count  (something under 20% of 
> what was sampled)  is 25, then the 95%  CI  on counts 
> is (16, 36)    since that is the range implied by 5.0 +/- 1.0  .
> 
> You write the CI  most readily on 'counts'  but it translates
> directly to fractions.  It works as  25 out of 500, or out 
> of 5000, or whatever.

Thanks for this easy method. I can apply it whenever without
complicating my mind with big numbers!

On my side, I have continued my research with this Central Limit
Theorem. I was wondering if your reasoning had the same basis. Here is
what I have found :

The Central Limit theorem gives us this relation :
The standard deviation 's' of the sample (n items) can be linked to
the standard deviation 'S' of the population (N items) by the
following rule :
s = S/square root(n) * square root((N-n)/N-1))
The term (square root((N-n)/N-1))) is enough close to 1 to be dropped
if n/N<1/7 AND n>30.
In this case, the result would depend on the count (n)! I am then
using the "reduced normal law" to set the CI...


With my deepest thanks,

Louis Tillier.
.
.
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