From: Rafael Laboissiere <laboissiere_at_cbs.mpg.de>

Date: Mon 18 Jul 2005 - 04:12:08 EST

# under condition 2

c2 <- t (matrix (c (c2AA, c2AB, c2AC,

chisq.test (matrix (c(p1, q1, p2, q2), nc = 2))

Date: Mon 18 Jul 2005 - 04:12:08 EST

- Jonathan Baron <baron@psych.upenn.edu> [2005-07-16 11:49]:

> I suspect that there are more direct ways to do this test, but it

*> is unclear to me just what the issue is. For example, if there
**> are many subjects and very few stimuli for each, you might want
**> to get some sort of measure of ability for each subject (many
**> possibilities here, then test the measure across subjects with a
**> t test. The measure must be chosen so that you can specify a
**> null hypothesis. It must be directional.
**>
**> If you have a few subjects and many trials per subject, then you
**> could do a significance test for each subject.
**> You want a directional test, because you have a specific
**> hypothesis, namely, that the correct answer will occur more often
**> than predicted from the marginal frequencies in the 3x3 table.
**> (I assume it is a 3x3 table with stimuli as rows and responses ad
**> columns, and you want to show that the diagonal cells are higher
**> than predicted.) One possibility is kappa, which is in the vcd
**> package, and also in psy and concord, in somewhat different
**> forms.
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# under condition 1

c1 <- t (matrix (c (c1AA, c1AB, c1AC,

c1BA, c1BB, c1BC, c1CA, c1CB, c1CC), nc = 3))

# under condition 2

c2 <- t (matrix (c (c2AA, c2AB, c2AC,

c2BA, c2BB, c2BC, c2CA, c2CB, c2CC), nc = 3))

where "cijk" is the number of times the subject gave answer k when presented with a stimulus of class j, under condition i.

The issue is to test whether subjects perform better (in the sense of a higher recognition score) in condition 1 compared with condition 2. My first idea was to test the global recognition rate, which could be computed as:

# under condition 1

r1 <- sum (diag (c1)) / sum (c1)

# under condition 2

r2 <- sum (diag (c2)) / sum (c2)

The null hypothesis is that r1 is not different from r2. I guess that I could test it with the chisq.test function, like this:

p1 <- sum (diag (c1)) q1 <- sum (c1) - p1 p2 <- sum (diag (c2)) q2 <- sum (c2) - p2

chisq.test (matrix (c(p1, q1, p2, q2), nc = 2))

What do you think?

I also thought about testing the triples like [c1AA, c1AB, c1AC] against [c2AA, c2AB, c2AC], hence my original question.

-- Rafael ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.htmlReceived on Mon Jul 18 04:20:02 2005

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