Some simplification might help.  There might be some redundancy in your RANDOM statements - these two in particular:   Random int/sub=year; Random int/sub=year group=amend*cut; Another thing I noticed and probably not as important is that row is included in the CLASS statement, but never enters into any of the other statements.  That might raise some issues with the sweep operator.   I do notice that your input dataset and your ldata set have identical names, which is probably just confusing to me, as I ordinarily don't use the ldata option.  The way I read the documentation for the ldata is that it is the coefficient matrix for the LIN(q) structure type.  Your use of LIN(1) gives a single covariance estimate without boundray constraints for the intercepts, which seems logical.  Again, I think this is going to give convergence issues as you have two different ways of getting a variance component for the intercept.  Is it possible that the second RANDOM statement should be:   random t/sub=year group=amend*cut; Finally, what happens if the variance component for st is greater than zero?  Just walking around the linear/nonlinear definition- this situation is still linear in the parameterization, just curvilinear if plotted.   SteveDenham     ... View more

Could you post the code you are using?  I believe you can get family-wise adjustments of p values for differences between lsmeans using the ADJUST= option in the LSMEANS/LSMESTIMATE/ESTIMATE/CONTRAST statement, but it will depend on how you frame the comparisons.  It may also require a GROUP= option for the random or repeated effect.   SteveDenham ... View more

I guess I don't know why you are subsetting your data, and then using PROC ANOVA to essentially just do a t test.  Your output from the first example should give a Tukey adjusted confidence interval for the difference between the levels of the variable 'leaf'.   If you want a Tukey adjust p value for the difference, shift to PROC GLM, and replace "cldiff" with "pdiff" in the MEANS statement.   SteveDenham       ... View more

It is a generalized linear model with a binomial distribution and a logit link. Or Logistic regression.   SteveDenham ... View more

If the interest here is to compare the two groups, AND you have reason to believe that the percentages are valid estimates for some population (like repeatedly measuring the given schools), you might want to consider using PROC GENMOD to analyze your data.  After converting percentages to proportions by dividing by 100, this could give you a first shot at an analysis:   proc genmod data=yourdata; class group; model activity_1_rate = group /dist=binomial type3; lsmeans group/diff ilink; run; A separate analysis would be done for activity_2_rate. @StatDave would likely also recommend following this up with the %NLmeans macro to correctly compare the means on the original scale. Pay attention to @Reeza 's comment about power - an N of 3 schools per group is only going to detect large differences in the dependent variable.   SteveDenham       ... View more

Following up on @jiltao 's comment - the code you present looks quite good for analyzing milk yield, but you may want to consider the following:   Parity is likely a fixed effect in this case, as the lactation curve is known to be different for heifers as compared to cows. Consider dichotomizing to two parity groups based on lactation number (primiparous = 1, multiparous  >1) .  Also, your code treats 'cow' as a subject, but is not included in the CLASS statement, so be sure that this variable is numeric and that the data are sorted by 'cow' and 'DIM'. Thus, you may wish to consider the following:   PROC GLIMMIX DATA = haytrial; CLASS treatment DIM block parity; MODEL milkyield=treatment*DIM*parity/DDFM=kr2; RANDOM block; RANDOM intercept/ SUBJECT=cow(block) group=parity; RANDOM DIM/RESIDUAL SUBJECT=cow(block) TYPE=ar(1) group=parity; LSMEANS parity*treatment*DIM; RUN; This enables you to fit different curves for the primiparous and multiparous animals, with different intercepts by parity and correlations in time by parity. I don't see a reason to expect the use of a multinomial distribution unless you change to a different response variable, which would be categorical.   SteveDenham           ... View more

The most likely answer is that there is either insufficient variablility or insufficient sample size to estimate a variance component for industry.  So the answer to how to get a value using a REML method is to increase the sample size, if possible.  A non-maximum likelihood method might be to use PROC VARCOMP or PROC GLM to estimate the variance using a method of moments approach.   SteveDenham     ... View more