Thank you for your answer. I feel better now
On 11 May 2001 22:29:37 -0700, [EMAIL PROTECTED] (Donald Burrill)
wrote:
snip
>> For me first approach looks more like t test.
> I presume you mean, in the sense that a pooled variance estimate
>is used. Is this a problem, for some reason?
The only approach to deal with z test for means that I have seen so
far was using s^2 = s1^2/n1 + s2^2/n2 formula. t test is always
using pooled variance. Both z test and percentages comparison test are
using normal distribution. Thus, intuitively I was considering them as
basically the same with only difference in variance calculations. My
problem is that using weighted p for one and not using pooled s^2 for
another seemed inconsistent with that idea.
Now you are saying that pooled variance may be used in z test. That
answers my question but as usual rises a few others. When would you
use pooled variance in z test instead of sum an vice verca? What are
we really testing: just two means or whether those two samples come
from the same population? Could you give me any reference with a book
dealing with pooled variance in z test?
>
>> On the other hand the chi2 is derived from Z^2 as assumed by first
>> approach.
> Sorry; the relevance of this comment eludes me.
I meant that it may be shown
chi2=sum(z^2)=sum((xi-ni*p)^2/ni*p*q)=sumsum((xij-nij*p)^2/nij*p)
where p is weighted percentage.
The trick would not work if we used the sum of variances.
But as you have said that approach was irrelevant to our hypothesis.
This rather confirms my feeling.
>> Finally, I would like to know whether the second formula is ever used
>> and if so does it have any name.
>
>"Ever" is a wider universe of discourse than I would dare pretend to.
>Perhaps colleagues on the list may know of applications.
>I would be surprised if it had been named, though.
>
Thank you for your time,
Alexandre Kaoukhov
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