Before I realized PROC PLM was a thing for getting simple slopes, I was taught how to get them from the proc mixed output by combining the estimates available there.  I'm now noticing that one estimate is consistently different.    Let's say we have data A B and M, where A is categorical with 2 levels, B is categorical with 2 levels, and M is continuous.    I want to get the slope for M at:  A1B1 A2B1 A1B2 A2B2 (reference)   The default reference for categorical is the last value so A2B2 is the reference. If you calculate by hand from the solution for fixed effects, the values are the same for A2B2, A2B1, and A1B2 but wildly different for A1B1 and I can't for the life of me figure out why. For example: A2B2 is -.2300 (same in PLM) A2B1 is -.2300 + 1.4324 = 1.2033 (1.2024 in PLM) A1B2 is -.2300 + .4836 = .2555 (.2535 in PLM) using the same method, A1B1 would be -.2300 + -1.3125 = -1.542 using PLM, A1B1 is .3735   Here's my syntax for the estimates: estimate 'medinc x Target1 x stim1' medinc 1 medinc*target 0 1 medinc*stim 0 1 medinc*target*stim 0 0 0 1, 'medinc x Target1 x stim2' medinc 1 medinc*target 0 1 medinc*stim 1 0 medinc*target*stim 0 0 1 0, 'medinc x Target2 x stim1' medinc 1 medinc*target 1 0 medinc*stim 0 1 medinc*target*stim 0 1 0 0, 'medinc x Target2 x stim2' medinc 1 medinc*target 1 0 medinc*stim 1 0 medinc*target*stim 1 0 0 0 /e;   Does anyone understand the difference? What is the value in the solution for A1B1 if not the comparison to the joint reference? ... View more

I've checked through the archives and found some similar questions but none quite what I am confused about, I will try to explain it as clearly as possible.    I am running a model with a continuous variable X, a Categorical variable Y with 3 levels, and a Categorical variable Z with 3 levels in proc mixed with the straightforward model  A = X|Y|Z. I understand how SAS dummy codes the categorical variables and how to estimate the simple slopes within conditions/ use the e option to see which groups get which weights in the estimate, but I am unclear about how these numbers map onto a regression equation in the form of regression weights.    This would be something like A = b0+  b1*X + b2*Y1 + b3*Y2 ... b*Z3 + b8*Y1*Z1 + b9*Y1*Z2 ... b17*Y3*Z8 + e   So b1 would be the slope of X averaged across all 9 unique dummy code combinations and then slopes for each unique conditions of Y (b2-b4) and Z (b5-b7) as simple slopes would be...the value you get from the Estimate command? estimate "name" X 1 X*Y 1 0 0; estimate "name2" X 1 X*Y 0 1 0; estimate "name3" X 1 X*Y 0 0 1; or does the value generated by the estimate command already include b1? So the actual regression weight would be something like "estimate 'name' X 0 X*Y 1 0 0" even though this doesn't actually work in SAS?   Similarly for the 3-way interaction, would b9 (for Y1*Z2) be the combination of the estimate at the reference (Y3 and Z3 by default in SAS) + the estimate for Y1 and Z2? Or...?   I guess another way to put it would be: For the made up output below, you would have these estimates for X*Y but these are all at the reference group for Z (Z3 by default). If you use the estimate command like above, is the value estimated the same as those below, except instead of being at the reference group, it is averaged across all levels of Z? If yes, that simplifies things a little, but given that the base estimate for X is included in the estimate command (X 1), I wasn't sure if this was the case.  X            4 X*Y2          -2 X*Y3          2 X*Y1          0   ... View more