Thank you very much Ted! I have been looking at your simplification for
more then an hour, but I don't see how you did it.
Could you perhaps, if it is not to much work, explain me how you reduced
H? It would help me to understand what I am realy doing.
Looking at the result, it seems indeed that H does add more information
than sd already did. Intuitively I thought the square of the sum of all
possible differences would not be related to the standard deviation.
Looking at your result it seems it is related by a factor sqrt(2*(n-1))
so there is no special point in calculating H and I know I cannot trust
my intuition anymore.
Thanks again!
Kind regards,
Stef
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