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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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