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- Path: sparky!uunet!gatech!darwin.sura.net!paladin.american.edu!auvm!UCLAMVS.BITNET!IARSPAL
- Message-ID: <STAT-L%93010613025728@VM1.MCGILL.CA>
- Newsgroups: bit.listserv.stat-l
- Date: Wed, 6 Jan 1993 10:02:00 PST
- Sender: STATISTICAL CONSULTING <STAT-L@MCGILL1.BITNET>
- From: "Peter A. Lachenbruch" <IARSPAL@UCLAMVS.BITNET>
- Subject: contingency tables
- Lines: 19
-
- The recent query about the chi-square for contingency tables may be addressed
- in an alternative way. Using a log-linear model, the chi-square (slightly diff-
- erent from the Pearson chi-square) is a test for interaction. The Pearson chi-
- square is also a test of interaction: the expected values are computed under th
- e assumption of independence. (The test can also be developed as one of homo-
- geneity - the proportions in the rows or columns are the same in all groups -
- the tests are the same).
- The natural question next arises in partitioning the chi-square. I think
- this makes sense for the homogeneity test, but not for the independence test.
- There are some fairly old articles (Cochran in the Annals and in Biometrics in
- the early 1950s and Gabriel in JASA in 1967 I think) which give ideas on this
- problem. Gabriel's idea is especially appealing to me, although it has never
- become very widespread. Gabriel showed that if you used the likelihood ratio
- chi-squared, you could collapse, select rows, etc. and compare the resulting
- statistic to the chi-square critical value for the full table (d.f.=(r-1)(c-1))
- and the significance level would be retained. The problem is that the test
- isn't very powerful. Perhaps something like the modified LSD test would work:
- Test subtables only if the overall chi-square test is significant.
- Tony Lachenbruch
-