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Needink statistics help.	by Esteis	2005-08-04 15:10:19
4 (n/t)	by Imp	2005-08-04 15:10:42
How, not what.	by Esteis	2005-08-04 15:13:50
Wouldn't it be...	by williamashbless	2005-08-04 15:18:42
...sqrt(sum((xk-xbar)^2)/(n+1)) like any other standard deviation? xbar would be the average value of all the 0's and 1's, which ends up being 51 / 1626. xk is each of the individual 0's and 1's. The standard deviation is the square root of the variance, which is the sum of the squares of the deviations of each data point from the mean, divided by the number of elements (well, plus 1 when its a discrete finite sample, if I recall, I"m a little rusty).
[ Reply ]
I tried that, but I got an insanely large value..	by Esteis	2005-08-04 15:21:10
Riiiiiight, n-1. Good catch. (n/t)	by williamashbless	2005-08-04 15:23:27
It depends	by basher20	2005-08-04 16:28:23

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