| Date: | Mon, 27 Nov 2000 10:51:01 -0000 |
| Reply-To: | dorothy@spss.ie |
| Sender: | "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU> |
| From: | Dorothy Kelly <dorothy@SPSS.IE> |
| Subject: | FW: Z test Algorithm or how to compare 2 frequencies |
| Content-Type: | text/plain; charset="iso-8859-1" |
**Hello all.
**
**I'm forwarding a request from a colleague below. Intially, I want to
**ensure that the formula he is suggesting is correct for what he
**wants to do - he wants to compare the male and female
**distributions for a particular cluster and see if there is a
**significant difference. He thinks that the formula below is the
**key to the question (it seems to be the formula behind normal
**curve approximation to binomial values if my memory serves me
**well, but it's been a long time since I thought in terms of
**formulae!). If he is approaching things correctly, how can this
**be done using SPSS/syntax??
**
**Thanks so much for your help
**
**Dorothy Kelly
****I want to be able to compare 2 frequencies coming from two sub
**sample and
****conclude (or not) on a significant difference among them.
****This is usually the kind of request I would have after having run
****a cluster
****analyses and cross tabling the clusters with an other
**categorical variable
****(let's say gender for the sake of arguing...).
**** So the formula use for this is:
****
****z= P1-P2/(sqrt(p*(1-p)*A)
****
****With
****
****A=(1/n1)+(1/n2)
****p=((n1*p1)+(n2*p2))/(n1+n2)
****
****The tested hypothesis are:
****H0: p1=p2 At a population level
****H1: p1<>p2
****
****Hence at a 95% level of confidence
****
****if z > 1.96 , I can conclude that P1 & P2 are significantly
**different from
****each other .
****
****The question is: How do I perform this test with SPSS
****The other question is ho do I automate this task:
****
****Such as: Cluster Variable by all the othe variables I want to
**perform the
****test on...
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