Alexandre:
Sorry not to respond - it's been busy here.
You wrote:
> I tried to test this. In order to test this I tried to plot 3d the
> difference between variances function of two proportion.
> Here is my Maple formula:
> plot3d((10*x+10*y)/20*(1-(10*x+10*y)/20)*(1/10+1/10)-(x*(1-x)/10+y*(1-y)/10)
> ,x
> = 0 .. 1,y = 0 .. 1);
> where x=p1, y=p2, n1=n2=10
You are correct - I clearly either made a mistake some years ago, or
misremembered what I worked out then. Sorry - bad brain day.
Looking at it again, the inequality you present can be deduced easily
from the convexity of the function p(1-p). The pooled variance is
(essentially) the value of the function at ta + (1-t)b, the other is the
linear interpolation. And, clearly, the pooled variance is larger; as
the function is convex up, the linear interpolation is always less.
OK, so what *is* going on here? Checking a dozen or so sources, I
found that indeed both versions are used fairly frequently (BTW, I
myself use the pooled version, and the last few textbooks I've used do
so).
Then I did what I should have done years ago, and I tried a MINITAB
simulation. I saw that for (say) n1=n2=10, p1=p2=0.5, the unpooled
statistic tends to have a somewhat heavy-tailed distribution. This makes
sense: when the sample sizes are small the pooled variance estimator is
computed using a sample size for which the normal approximation works
better.
The advantage of the unpooled statistic is presumably higher power;
hoewever, in most cases, this is illusory. When p1 and p2 are close
together, you do not *get* much extra power. When they are far apart
and have moderate sample sizes you don't *need* extra power. And when
you have small samples it's not clear that you can *trust* the extra
power.
OK, that makes more sense. Sorry about the last attempt. If I'm still
being obtuse, somebody please straighten me!
-Robert Dawson
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