Re: Comparing fractions (or proportions)

Jack Tomsky wrote


*** What happens if pX < dO or if pY > 1-dO? Then
their repective contributions to variance estimate
become negative ***


My response


How is your High School Algebra?

I wrote (Jan 20, 2007 6.37 PM)

***
They think they correct this by using
__V(pX-pY) =
__= pX*(1-pX)/n1 + (pX-d0)*(1-pX+d0)/n2 ______(2)
or similarly considering pY instead pX,
__.= (pY+d0)*(1-pY-d0)/n1 + pY*(1-pY)/n2______(2 a)



BELIEVE THIS Jack

__pX < dO makes the 2nd term negative BUT NOT
NECESSARILY the variance V(pX-pY).
By the other hand
WE MUST TAKE INTO ACCOUNT THAT NECESSARILY
______ 0 < p(X)< 1 ______0 < p(Y) < 1
Hence
_______-1 < dO < 1

This answers your no completed question:

*** Without writing out all the details, I'm not sure
what happens when the sample pX < dO or pY > 1-dO.
It's possible that phat(X) becomes dO or one,
respectively.
Jack ***


__________licas (Luis



Under Ho, all outcomes have a positive probability. Put X=0 (or equivalently pX=0) into their formula. Then their variance estimate becomes -dO*(1+dO)/n2, which is negative.

When you take the square-root of V in the formula for z, it becomes imaginary. The standard deviation must be real and non-negative; thus, this estimate is inadmissible.

Since both samples contribute to the estimate of p(X) under Ho, you need to use the likelihood function to obtain an admissible estimate of p(X) and consequently of V.

Jack
.