First, your PROC MIXED code does not seem to be appropriate. It looks to me that you wanted to fit a random coefficients model? If so, TIME should not be a CLASS variable, and you should have the SUBJECT= option in the RANDOM statement. Then with this modified RANDOM statement, your REPEATED statement with TYPE=UN structure is likely over-fitting your data and might need to be removed.   proc mixed data=trial; class idno ; model GLOBALCOG = time WMFA WMH ICV brainvol AGE RACE GENDER/s chisq OUTPM=WORK.PM RESIDUAL; random intercept time / subject=idno; run;   Then you might use approaches illustrated in the following usage note to obtain subject-specific intercept and slope estimates.   37109 - Obtaining subject-specific parameter estimates and tests for a random coefficients model or HLM with PROC MIXED or PROC GLIMMIX   Hope this helps, Jill ... View more

Here is the related information from the MIXED documentation.   EFFECT=effect specifies an effect according to which observations are grouped. Observations sharing the same level of the effect are removed from the analysis as a group. The effect must contain only class variables, but need not be contained in the model. Removing observations can change the rank of the ( X' V-1 X)- matrix. This is particularly likely to happen when multiple observations are eliminated from the analysis. If the rank of the estimated variance-covariance matrix changes or its singularity pattern is altered, you would receive that message in the Log, and no influence diagnostics are computed. I suspect that is the case here. Thanks, Jill ... View more

Those are different ways of modeling correlations in your data, and all look reasonable to me. You might examine the Fit Statistics table. The model with the smaller AICC or BIC indicates a better fit to your data. Thanks, Jill ... View more

non informative prior (Jeffreys prior) does not guarantee the Bayesian estimates match the classical approach. If you can send in the data, maybe we can look into it further. Thanks, Jill ... View more

I agree with @PaigeMiller -- the SLICE statement is much easier to use. For example -- proc mixed data=rawdata; class mice group week; model col1 =  group week week*group ; repeated / subject=mice type=cs; slice week*group / sliceby=group sliceby=week; run; For your CONTAST statement, see if the following statement works for you -- contrast "Group effect at time 2"      group -1 0 0  1 week*group 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0,     group 0 -1 0  1 week*group 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 1 0 0 0,     group 0  0 -1 1 week*group 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 1 0 0 0;   Hope this helps, Jill ... View more

What the ESTIMATE statement does is to compare the conditions averaged over the three time points. From the output you sent, it seems to me that the average is pretty close between the control and the two treatments, so the results make sense to me. If you want to compare the conditions at a specific time point, you might get the result you expected. For example, estimate 'two tx vs control at time 1' condition -1 0.5 0.5 condition*time -1 0 0 0.5 0 0 0.5 0 0; --Jill ... View more

Thanks for providing additional information. Please see if the following ESTIMAT statements work for you -- estimate '2 tx vs control'  condition -1 0.5 0.5; estimate 'tx1 vs tx2' condition 0 1 -1; Please give it a try and let us know how it goes. Thanks, Jill ... View more

If you send in the Class Level Information table from the PROC MIXED output, that would be helpful for us to provide the ESTIMATE syntax. Also, do you want two treatments vs. control averaged overall all time points? or two treatments vs. control at time1 compared with two treatments vs. control at time2? For the two treatments comparison, do you want it to be averaged over all time points? or some specific time points? Or some something else? Please be more specific. Thanks, Jill ... View more

The plus signs are the means for each group. SAS Help Center: Summary Statistics Represented by Box Plots Thanks, Jill ... View more

Sorry I got the law of logs wrong 😞  But you got the idea: there could be two different ways of computing a statistic under certain assumptions. The results are not exactly the same due to random errors in the generated data. How CV's are used in the computation of the power and sample size is provided in the documentation -- https://go.documentation.sas.com/doc/en/statcdc/14.3/statug/statug_power_details68.htm   ... View more

You can have two different formulas for the same thing. Just like if you have a series of positive values, x1, x2, ... xn. You can use the formula x1+x2+...+xn to compute the sum. You can also use the formula exp(log(x1)*log(x2)*....*log(xn)) to compute the sum. Mathematically they are equivalent.  Same thing with the CV formulas you referred to under the log-normal assumption. I wrote the following program to show you how empirically you can get almost the same value between using these two formulas for CV --   data one; do sample=1 to 2000; call streaminit(2366); do i=1 to 2000; logx=rand('normal'); x=exp(logx); output; end; end; run; proc means data=one noprint; by sample; var x logx; output out=out(drop=_freq_ _type_) mean(x)=meanx std(x)=stdx var(logx)=var_logx; run; data out; set out; cv_x2 = sqrt(exp(var_logx)-1); cv_x = stdx/meanx; run; proc means data=out mean; var cv_x cv_x2 ; run;   Hope this helps, Jill   ... View more

They are both valid. ... View more

You are correct, so is the SAS documentation- sigma1* is the standard deviation "of the bivariate normal distribution of the log-transformed data". So it is the same as your SDLog.   ... View more