The z-test for equality of two independent proportions is is equal to the square root(Pearson chi-square test for independence in the corresponding 2 x 2 table). The tests are equivlent and both assume a sufficiently large sample.
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of Lachenbruch, Peter
Sent: Monday, January 05, 2009 3:47 PM
To: [email protected]
Subject: RE: st: testing equality of proportions?
If you do a correlation with one variable a dichotomy, the significance test for the correlation is equivalent to a t-test. I haven't worked out the mathematics, but I assume that the chi-squared test is equivalent. If X is categorical, the math will still work, but the test may not mean much.
Tony
Peter A. Lachenbruch
Department of Public Health
Oregon State University
Corvallis, OR 97330
Phone: 541-737-3832
FAX: 541-737-4001
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Svend Juul
Sent: Monday, January 05, 2009 11:52 AM
To: [email protected]
Subject: Re: st: testing equality of proportions?
Dan Waldo wrote:
...
My subjects are classified by two nominal attributes, say X and Y. I run the tabulate procedure
svy: tab X Y , row ci stubwidth(15)
but for the life of me I cannot figure out the syntax of a command to test whether the proportion with Y==A is the same for subjects with X==Often as for those with X==Seldom. ...
====================================================
I believe you are pretty close. If Y is a dichotomous variable, you can:
svy: tabulate X Y , pearson
To lean more, look at:
help svy_tabulate
Hope this helps
Svend
________________________________________________________
Svend Juul
Institut for Folkesundhed, Afdeling for Epidemiologi (Institute of Public Health, Department of Epidemiology) Vennelyst Boulevard 6 DK-8000 Aarhus C, Denmark
Phone, work: +45 8942 6090
Phone, home: +45 8693 7796
Fax: +45 8613 1580
E-mail: [email protected]
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