On Dec 5, 12:35 pm, BobA5835
wrote:
Our young programmer told me this as a further explanation
[....]
The largest positive number in Excel is 9.99999999999999 * 10^307

That is the largest positive constant that you can enter. The largest
postive value as a result of computation is a little more than
1.7976931348623157E+308, which is displayed as 1.79769313486232E+308.

Yes, the equal sized Groups do produce a numerical Value equal to the Value
of the Geomean in an 'n' of 273 data points. Hence my second question to a
mathmatician: Are these actually the same Value?

First, considering the constraints of your data (up to a value of 15,
if I understand your first posting correctly), I think you would have
to settle for groups of 262, not 273. 15^262 is about 1.367E+308, but
15^263 overflows. You might know that your average data point less
than 15; you said you computed the geomean of 273 data successfully.
But I think it would be risky to depend on that. It's a judgment
call.

As for the mathematical question, it has already been answer: yes,
the geomean of k geomeans of n data each is equal to the geomean of
k*n data [1]. But remember that n must be the same for all "sub-
geomeans". If you choose 262 for n, the total number of data must be
a multiple of 262(!!).

joeu2004: Thank You for a second workable formula ditto on showing it to my
programmer.

Harlan's formula, using AVERAGE, is equivalent and arguably cleaner.
You can use exp/ln or 10/log. Frankly, that's the form that I always
use. I don't know why I chose not to use AVERAGE in my formula.
Brain fart!

"You don't need to be a mathematician." Well, actually, you do in this
case. I'm asking for an expert opinion. We are in a regulated industry.
I'm a biologist, not a mathematican.

I would not qualify as an "expert mathematical opinion" in a court of
law.

HTH.


-----

Endnotes:

[1] Proof. For simplicity, G(...) means GEOMEAN(...).

G( G(y[1,1],...,y[1,n]), ... ,G(y[k,1],...,y[k,n]) )

=( G(y[1,1],...,y[1,n]) *...* G(y[k,1],...,y[k,n]) ) ^ (1/k)

=( (y[1,1]*...*y[1,n])^(1/n) * ... * (y[k,1]*...*y[k,n])^(1/n) ) ^ (1/
k)

= ( y[1,1]*...*y[1,n] *...* y[k,1]*...*y[k,n] ) ^ (1/n) ^ (1/k)

= ( y[1,1]*...*y[1,n] *...* y[k,1]*...*y[k,n] ) ^ (1/n*k)

=G(y[1,1],...y[1,n], ... ,y[k,1],...,y[k,n])