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Suppose I am testing two drugs, I and II. The response profile is either I > II, I < II or I = II. I was trying to figure out if I is in fact better than II.

One way I thought of was to throw out all the ties (I == II), and then assign a 1 where I > II and a 0 where I < II. Then this boils down to a binomial distribution (which I can approximate with the normal).

But I was wondering if throwing away the ties is really appropriate... If I don't throw away ties, I can't use the binomial distribution (so would I switch to some multinomial test)?

The other thing I can think of is some wilcoxon rank sum test or mann whitney u test with ties... but I was wondering if anyone else has an idea? The concern here is that there will be a LOT of ties...

Thanks!

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  • $\begingroup$ Are the data really that coarse? Are there really only 3 states, or are there, for example, instances when I > II and then instances when I >> II? $\endgroup$ – A.M. Aug 16 '13 at 3:30
  • $\begingroup$ Hm, for now, let's assume it's really that coarse... $\endgroup$ – user1357015 Aug 16 '13 at 4:15
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It's still binomial even after throwing out the ties, and you get a more powerful test that way. But for describing the size of the difference, you should keep the ties and present the three sample proportions.

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  • $\begingroup$ It's easy to come up with proportions but I would need confidence intervals? How would I do that? Use the binomial for each proportion and then take a multiple comparison adjustment? $\endgroup$ – user1357015 Aug 16 '13 at 5:13
  • $\begingroup$ The CI for the conditional probability Pr(I > II | I <> II) is obtained the usual way, using only the untied data. The trinomial case is messier because it's a 2-dimensional region, not a simple 1-dimensional interval, and you need to know who you're talking to and how many technical corners you need to round to be understood (or, more importantly, to not be misunderstood). $\endgroup$ – Ray Koopman Aug 16 '13 at 7:03
  • $\begingroup$ So what you're saying is first do a test to see if I == II -- easy enough, a chi-square test. Now for Pr(I > II | I < II), your saying throw out all the ties and check. How are you certain that throwing away the ties is the right way to proceed... your n would go down which would effect the result. Sorry if I'm missing something obvious. $\endgroup$ – user1357015 Aug 16 '13 at 11:15
  • $\begingroup$ Your report might read something like this: "Responses to the drugs differed in x% of the cases (95% CI = (x0,x1)). When they differed, the response to I was better than the response to II in y% of the cases (95% CI = (y0,y1))." If A, B, and C are the counts for I > II, I < II, and I = II then x = 100(A+B)/(A+B+C) and y = 100A/(A+B), and the CIs follow accordingly. $\endgroup$ – Ray Koopman Aug 16 '13 at 16:59

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