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- Newsgroups: bit.listserv.stat-l
- Path: sparky!uunet!elroy.jpl.nasa.gov!sdd.hp.com!wupost!spool.mu.edu!torn!nott!cunews!tgee
- From: tgee@alfred.carleton.ca (Travis Gee)
- Subject: Re: X^2 interaction
- Message-ID: <tgee.726413695@cunews>
- Sender: news@cunews.carleton.ca (News Administrator)
- Organization: Carleton University
- References: <STAT-L%93010516044921@VM1.MCGILL.CA>
- Date: Thu, 7 Jan 1993 13:34:55 GMT
- Lines: 58
-
- In <STAT-L%93010516044921@VM1.MCGILL.CA> EJOHNSO3@UA1VM.BITNET writes:
-
- >I'd appreciate your insight in interpreting the results of a 2X3 contingency
- >table.
-
- >I've run the standard chi-square test with significant results (p<.01), but I
- >see this as analogous to an omnibus ANOVA--it tells me something is happening
- >among these six cells, but it doesn't tell me what; therefore, my question is
- >whether there is anything like a test for main effects or interaction for a
- >chi-square?
-
- I have an unpublished paper lying about where I argue that such a test
- can be done on the following rationale:
-
- 1) chi-square is a distribution of z-squares, summed:
- 2) therefore (o-e)^2/e is an approximation to (x-mu)^2/s^2
- ____________
- 3) thus, \/ (o-e)^2/e is an approximation to z and so
- 4) the "important" cells where the effect is occurring will have a big
- "z" score (from 3) ) which is interpretable as a "z", and the
- "important" rows/columns will have sum(z^2) that exceed some critical
- value, with df (# of cells in row/col. ) - 1.
- 5) This leads us to an ANOVA-like situation where an a-priori test for
- the effect of a row or column would find sum(z^2) and test as a X^2
- with r-1 or c-1 df, using perhaps a Bonferroni-like adjustment to the
- alpha rate. Series of tests using both rows and columns would have to
- be adjusted further, because of the non-independence of the tests.
-
-
-
- >My gut feeling says that I can divide the critical chi-square
- >by my number of cells and then compare each cell chi-square score to this new
- >ciritcal value. Is this acceptable? I've not been able to find in the literat-
- >ure any comments about interpreting contingency tables, and so any comment you
- >could make will be a great assistance.
-
- I don't see why the average-per cell would be a suitable critical
- value, because it is tied to the final test statistic, not the value
- expected under the null hypothesis. Thus, if ONE cell has a
- monstrously large value, and a second cell shows a marginal effect,
- the second cell would not be considered significant because the effect
- was larger elsewhere. Hence the effects are not independent by
- definition.
- The hypothesis-testing model works from a null prior distribution
- - in the case of X^2, the distribution of Z^2. I believe that my
- formulation is correct, however, I would greatly appreciate any
- feedback from fellow netters who have an opinion. Many thanks in
- advance!
-
- ((((((((((((((((((((((((((((((((((((((((((((((((((((((((((
- Travis Gee () tgee@ccs.carleton.ca ()
- () tgee@acadvm1.uottawa.ca () ()()()()
- () (Carleton address () ()
- () preferred) ()()()()()()()()()
- Recent government figures indicate that 43% of all statistics
- are utterly worthless.
-