Thanks, @jiltao. That is likely the case, especially as I left out a step. After calculating the sum of squared deviations and before taking a square root, there needs to be a division by an appropriate degrees of freedom value. So we would have the deviations of the values predicted by the model from the observed values, each then squared, summed over all observations, divided by an appropriate, model-based, degrees of freedom value. That looks, at least to me, a lot like a variance due to things not in the model like unmodeled fixed or random effects. If the model was as simple as a mean, it would be the variance, wouldn't it? Taking the square root gives a standard deviation, or in the case of a general linear model, the RMSE.   Since the predicted values are empirical BLUPs, this calculated value is a measure of how closely the empirical BLUPs represent the variability in the raw data.  I really need my copy of Graybill's Theory and Application of the Linear Model to refresh my BLUP knowledge, but it is in a box in the basement somewhere...   SteveDenham     ... View more

I am not sure what you mean by t(.05,ddf).   SAS has several functions related to returning values from distributions. I suspect you may want one of these but insufficient information for sure. data example; a1 = cdf('T',0.05,2); a2 = pdf('T',0.05,2); run; Return values for the T distribution cumulative density function and probability density functions. The first parameter is the distribution name. The other parameters depend on the distribution. There are also SDF (and LOGC/P/Sdf) and Quantile functions for different pieces. 2 above is the Degrees of Freedom. ... View more

Sounds like a classic repeated measures. Seeing the data would help a bit more, but from what you describe a REPEATED statement would be the way to go. Use the TYPE= to specify the covariance structure you wish for the 4 repeated measures on each subject. If each subject received one and only one treatment, then you would not be able to estimate a SUBJECT*TREATMENT interaction. If you model TIME as a CLASS effect and have 1 observation per time point per subject, then you will not be able to model a SUBJECT*TIME interaction. It is possible to model TIME as a continuous effect and set up a random coefficients model for this data with random int time / subject=subject However, you get much more flexibility in the covariances with the REPEATED approach. ... View more

repeated age / type=cs; just estimates the residual variances and break it into two pieces. So your R matrix is a constant variance (sum of the two estimates) and zero covariance. repeated age / type=cs subject=person; estimates a block-diagonal covariance matrix R, with each block corresponds to each person. Within each block, the structure is a compound symmetry . You can add the R option in the REPEATED statement to see the R matrix in the output. If you want to model the correlations in your data, you might need to use the second specification (with the SUBJECT= option). And go with whatever this model produces for your fixed effects. ... View more

I am not really sure what is causing the difference. It could be a difference in algorithm, such as a difference in whether Period is handled as another repeated measure, with time nested in period.  To check you might try the following:   proc mixed data=Result maxfunc=5000 convf=1e-6; class subject period treatment time; model result = baseline period subject treatment time treatment*time/ddfm=kr2; repeated period time / type=UN@CS subject=subject; lsmeans treatment / pdiff=control; contrast 'linear' treatment -3 -1 1 3; lsmestimate Treatment*time 'Linear treatment at time 1' -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3 0 0 0 0 0, 'Linear treatment at time 2' 0 -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3 0 0 0 0, 'Linear treatment at time 3' 0 0 -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3 0 0 0, 'Linear treatment at time 4' 0 0 0 -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3 0 0, 'Linear treatment at time 5' 0 0 0 0 -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3 0, 'Linear treatment at time 6' 0 0 0 0 0 -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3/FTEST; run; quit; On my machine, the linear treatment at time 1 comes in with a P value of 0.0019, so I applied a multiplicity adjustment:   proc mixed data=Result maxfunc=5000 convf=1e-6; class subject period treatment time; model result = baseline period subject treatment time treatment*time/ddfm=kr2; repeated period time / type=UN@CS subject=subject; lsmeans treatment / pdiff=control; contrast 'linear' treatment -3 -1 1 3; lsmestimate Treatment*time 'Linear treatment at time 1' -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3 0 0 0 0 0, 'Linear treatment at time 2' 0 -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3 0 0 0 0, 'Linear treatment at time 3' 0 0 -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3 0 0 0, 'Linear treatment at time 4' 0 0 0 -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3 0 0, 'Linear treatment at time 5' 0 0 0 0 -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3 0, 'Linear treatment at time 6' 0 0 0 0 0 -3 0 0 0 0 0 -1 0 0 0 0 0 1 0 0 0 0 0 3/FTEST adjust=simulate(seed=1) stepdown adjdfe=row; run; quit; This gives an adjusted P value of 0.0109, which is within striking distance of the 0.0105 you are looking for, so it may have a different adjustment method   SteveDenham ... View more