saisat <[EMAIL PROTECTED]> wrote in sci.stat.edu:
>I have a "peer" group of distributions with the following data
>
>Distribution1 : Mean-M1, Standard Deviation-SD1, Total number of data
>points - N1
Work backwards. Mean = (sum of x)/n; variance = stdev^2 =
[(sum of x^2) - (sum of x)^2/n] / (n-1).
>From these formulas and the mean and standard deviation and sample
size, you can find the sum of the x's and the sum of the x^2's.
>...
>Distributionn : Mean-Mn, Standard Deviation-SDn, Total number of data
>points - Nn
>What is the most accurate way to determine the overall mean and the
>standard deviation of the above disributions.
Do the same thing for the other groups. Then add up all the sums of
x's, sums of x^s', and n's, and compute the overall mean and stdev
using the formulas.
This is the basic technique, but there are shortcuts. Look in your
textbook under "combining means and variances" or a similar topic.
>Also, I need to remove any distributions that are not similar to the
>rest of the distributions in the "peer" group.
Huh?
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com/
"You find yourself amusing, Blackadder."
"I try not to fly in the face of public opinion."
.
.
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