Yes. If you calculate the residuals (fo - fe)/sqrt(fe) [that is, the square root of the contribution of the cell to the chi square statistic) then on the null hypothesis that fo = fe, these are normally distributed with mean zero and standard deviation given by sqrt((1-rowtotal/grandtotal)*(1-coltotal/grandtotal)). You can then do a simple Z test.
This of course assumes the expected cell frequencies are large enough...... This set of tests corresponds to testing the multiple dummy variables involved. I find it very useful - it makes the nature of the relationship between the variables much more precise. Regards, Alan Gerald Kaminski wrote: > > Hello, All: > > A general question concerning Chi-squared independence results. > > In AOV, when there is significance, we can do pair-wise comparisons, > build > confidence intervals, etc. Is there a comparable technique for > Chi-square > independence tests other than "eye-balling" the contribution to the > Chi-square > stat from each cell and looking for large changes? > > Thanks, > > Gerry > > . > . > ================================================================= > Instructions for joining and leaving this list, remarks about the > problem of INAPPROPRIATE MESSAGES, and archives are available at: > . http://jse.stat.ncsu.edu/ . > ================================================================= -- Alan McLean ([EMAIL PROTECTED]) Department of Econometrics and Business Statistics Monash University, Caulfield Campus, Melbourne Tel: +61 03 9903 2102 Fax: +61 03 9903 2007 . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
