Hi all
Sorry for the late return but I would like to post the answer to the pooled
standard deviation that I received . All the responses were similar to the
following
Try pooled Var = ((n1-1)*Var1 +(n2-1)*Var2)/(n1+n2-2)
= (310*2.548*2.548+232*2.807*2.807)/542
= 7.085978
or pooled SD = sqrt((310*2.548*2.548+232*2.807*2.807)/542)
= 2.66195
or
To get the pooled SD, you have to take a weighted average of the variances
(squares of SDs), weighted by one less than the sample sizes, and then take
the square root of the result. So here it's
Sqrt( (2.548^2*310+2.807^2*232)/(310+232)
which does come to the number you quote.
Thanks to those that responded
Regards
Kevin Elhassan
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