Thank you both (@PaigeMiller and @ SteveDenham) for the replies. To rephrase my question in model terms: I am interested in looking at the differential effect of exposure on white vs. non-white patients. In the summary equation below, E=exposure, R=Race, and P=period. y = b(0) + b1*P + b2*E + b3R + b4*E*P + b5*P*R + b6*E*R + b7*P*E*R. The effect on whites should be (b2 + b4), for non-whites should be (b2 + b4 + b6 + b7). Therefore, the differential effect should be (b6 + b7). In my model, this would correspond to the coefficients for exposure*race + period*exposure*race. My SAS model code is below, setup as a linear regression so that model coefficients should provide a direct estimate of the proportion of patients with Medicaid insurance in this particular model. proc genmod data=have; class year (ref='2011') White (ref='1') exposure (ref='0') period (ref='0') confounder1 hospst/; model medicaid= White exposure period exposure*period white*period white*exposure white*exposure*period confounder1; repeated subject= hospst/type=ind; lsmestimate white*exposure*period 'estimate' 1 -1 -1 1 -1 1 1 -1; run; I may be mistaken, but it seemed that if I could specify the correct contrast that incorporated both model terms, then I could obtain the p-value for both statements being true (i.e. b6 + b7), relative to 0. For example: contrast 'period:expansion + period:expansion:race = 0' expansion*period*race 1 -1 -1 1 -1 1 1 -1, expansion*period 1 -1 -1 1; I obtained the lsmestimate statement above for the triple interaction by (1, -1)*(1,-1)*(1,-1), and I have found that this lsmestimate statement gives the correct p-value for the interaction specified, as reflected in the main model output. I would have thought that the exposure white*exposure could be specified with (1 -1 -1 1). However, when I test that as an independent statement, it does not give the correct p-value as compared to the model coefficient.
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