When using an ordered factor in aov, how do I break the treatment sum of squares into all its separate degrees of freedom? In other words can I use contr.poly in aov to get the full breakdown of the treatment sum of squares? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
Breakdown of Treatment Sum of Squares
1 message · Wendy Catchpole