On Wed, 10 Dec 2003 16:06:27 -0500, Rich Ulrich <[EMAIL PROTECTED]> wrote:
> On 10 Dec 2003 07:30:40 -0800, [EMAIL PROTECTED] (Donald Burrill) wrote: > > > So far all responses have appeared to assume that the data represent > > left and right ovaries on the SAME women. But this cannot be the case, > > because the total number of left ovaries in the data is 132, and of > > right ovaries 103. So: before we can offer useful advice, we need to > > know whether the pathology being documented occurs, when it occurs, only > > in one of a woman's ovaries (for these data, anyway), or whether some of > > the 235 instances of pathology represent both ovaries for some of the > > women. If there are any of these last, those are the only cases for > > which one has paired data. [ snip, to what I posted before ... ] >If the L and R instances are disjunct, then the paired t-test > would take place with a negative correlation, and the > absolute correlation would be *larger* for a smaller total N. [ snip, rest ] Correlated? Or negatively correlated? Or assumed to be not-correlated? Ignore the scores for a moment, except Some vs None. Here are the cells, for Left versus Right, if the table is assumed to be disjunct, where 103 and 132 represent the cells with R-not-L and L-not-R. 0 + 103 132 + [the rest] This can be tested with McNemar's, and a good approximation is the difference D=(132-103) squared, divided by their sum S: And that gives a chisquared test of 3.58, which is not enough to leave anyone heavily impressed. If there were *correlation* which would mean that L&R tend to coincide, the table would have the same D, but a smaller S: 50 + 53 82 + [the rest] Here the test is 6.23 -- enough to be a bit impressive. Keep in mind: The proper test uses the proper overlap. The "independent" t-test would require a proper N for the cell marked [the rest]. I believe that if we stick in any number that is rather large, this test will come out intermediate between the two versions of McNemar that I just demonstrated. At least, that is what I get when I test the numbers as Poisson counts, using the square-root transform (so the results for each sample have SD= 0.5). Those are results while ignoring the rated severities of disease. When you put in those scores again, you can write similar models, with similar results. You cannot do an "independent t test" unless you have the extra cases for the diseaseless-cell. - Note that the question is *not* whether the 132 cases have a higher 'average score' than the 103 -- the profiles tend to match, and that is not a problem. In fact, the question might be construed as whether the 132 have a higher *total* score. -- Rich Ulrich, [EMAIL PROTECTED] http://www.pitt.edu/~wpilib/index.html "Taxes are the price we pay for civilization." . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
