> The function and reason for this particular rule of thumb > is to keep you from dividing a big deviation by a tiny > denominator and getting a huge chisquared contribution > for that cell. >
Understood. > > degrees of freedom should be 2 - 1 - 1 = 0 for the test which leads me to > > the assumption that this is not a good idea and that I'm missing something. > > Does this computed test have 1 d.f. then? Seems right.... > OK, if I use three classes I have 3-1-1=1 d.f.. But if I had only two classes, could I run the test anyway? On the one hand my textbooks says that classes must be >= 2, on the other it says that d.f. are #classes - number of estimated parameters - 1 which is 0 for the two-classes case. This is what I don't really understand right now... Thanks for your reply, Daniel . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
