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From: Josh on 24 Nov 2009 21:07
Hi, can someone help with proving that
the sum of the squared differences is equal to the sum of the squared deviation from the mean times the size of the population? that is

SumSum(i<j=1 to m)(z_i-z_j)^2= m*Sum(z_i-average(z))^2.
where average(z) is the average of z over the m units?

thanks,
Josh
From: Ray Vickson on 25 Nov 2009 11:21
On Nov 25, 4:07 am, Josh <rosen12j...(a)googlemail.com> wrote:
> Hi, can someone help with proving that
> the sum of the squared differences is equal to the sum of the squared deviation from the mean times the size of the population? that is  
>
> SumSum(i<j=1 to m)(z_i-z_j)^2= m*Sum(z_i-average(z))^2.
> where average(z) is the average of z over the m units?

Nice homework problem. What have you done so far? Have you tried a few
small m values, like m = 3 or m = 4?

R.G. Vickson

>
> thanks,
> Josh

From: Josh on 6 Dec 2009 23:48
thanks. Actually not a homework problem at all but a small line in a much longer proof of a survey sampling estimator.

would this be a correct proof:

First change the left hand side to
(1/2)* SumSum(i,j= 1 to m)(z_i-z_j)^2
then add and subtract the average of z over the m units, that is

(1/2)* SumSum(i,j= 1 to m)[z_i-average(z)-(z_j-average(z)]^2

I think the rest follows quite neatly.
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