From: Spencer Graves <spencer.graves_at_pdf.com>

Date: Sat 16 Jul 2005 - 23:33:18 EST

> fit12 <- BTm(rbind(cond1, cond2)~..)

> Dev12 <- (fit1$deviance+fit2$deviance + -fit12$deviance)

> pchisq(Dev12, 2, lower=FALSE)

[1] 0.8660497

Date: Sat 16 Jul 2005 - 23:33:18 EST

Have you considered "BTm" in library(BradleyTerry)? Consider the following example:

> cond1 <- data.frame(winner=rep(LETTERS[1:3], e=2),

+ loser=c("B","C","A","C","A","B"), + Freq=1:6) > cond2 <- data.frame(winner=rep(LETTERS[1:3], e=2), + loser=c("B","C","A","C","A","B"), + Freq=6:1) > fit1 <- BTm(cond1~..) > fit2 <- BTm(cond2~..)

> fit12 <- BTm(rbind(cond1, cond2)~..)

> Dev12 <- (fit1$deviance+fit2$deviance + -fit12$deviance)

> pchisq(Dev12, 2, lower=FALSE)

[1] 0.8660497

This says the difference between the two data sets, cond1 and cond2, are not statistically significant.

Do you present each subject with onely one pair? If yes, then this model is appropriate. If no, then the multiple judgments by the same subject are not statistically independent, as assumed by this model. However, if you don't get statistical significance via this kind of computation, it's unlikely that a better model would give you statistical significance. If you get a p value of, say, 0.04, then the difference is probably NOT statistically significant.

The p value you get here would be an upper bound. You could get a lower bound by using only one of the three pairs presented to each subject selected at random. If that p value were statistically significant, then I think it is safe to say that your two sets of conditions are significantly different. For any value in between, it would depend on how independent the three choices by the same subject. You might, for example, delete one of the three pairs at random and use the result of that comparison.

There are doubtless better techniques, but I'm not familiar with them. Perhaps someone else will reply to my reply.

spencer graves

Rafael Laboissiere wrote:

*> Hi,
**>
*

> I wish to analyze with R the results of a perception experiment in which

*> subjects had to recognize each stimulus among three choices (this was a
**> forced-choice design). The experiment runs under two different
**> conditions and the data is like the following:
**>
**> N1 : count of trials in condition 1
**> p11, p12, p13: proportions of choices 1, 2, and 3 in condition 1
**>
**> N2 : count of trials in condition 2
**> p21, p22, p23: proportions of choices 1, 2, and 3 in condition 2
**>
**> How can I test whether the triple (p11,p12,p13) is different from the
**> triple (p21,p22,p23)? Clearly, prop.test does not help me here, because
**> it relates to two-choices tests.
**>
**> I apologize if the answer is trivial, but I am relatively new to R and
**> could not find any pointers in the FAQ or in the mailing list archives.
**>
**> Thanks in advance for any help,
**>
*

-- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA spencer.graves@pdf.com www.pdf.com <http://www.pdf.com> Tel: 408-938-4420 Fax: 408-280-7915 ______________________________________________ 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 Sat Jul 16 23:37:08 2005

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