I won't have time to respond in any detailed way for several days. If you are inclined to wait, you can devote some effort to the concept of narrow inference that I mentioned in a previous response. I think it will allow you to make the donor comparisons you are interested in; if you disagree once you've had a chance to think about it in more detail, let me know.   Other quick thoughts:   Because TRT has only two levels, you do not need to use an ESTIMATE statement. Instead, you can use   lsmeans trt / diff;   At least for the problems I work with, it is not uncommon for variance estimates to be set to zero by the procedure, which produces the "Estimated G matrix is not positive definite" message. See   http://support.sas.com/resources/papers/proceedings12/332-2012.pdf   Usually, I interpret a set-to-zero estimate as being a variance that is very small; many references exist on this topic. In my opinion, there is not one "right" way to deal with this modeling result. SAS offers the NOBOUND option; some people recommend dropping the term from the model. Personally, I do not drop the term from the model if it represents a design element in the experiment. I often use NOBOUND, but not always.   The ESTIMATE statements in your code do not work because DONOR is a random effects factor and cannot be included in ESTIMATE statements in the way that you have specified. Do some reading about "narrow inference".   Regarding    repeated donor / subject=donor(trt tpt) type=cs; First, repeated donor / ... assumes that donor is a fixed effect factor, which I do not think is the case. Second, donor(trt tpt) assumes that donor is nested within both trt and tpt, which definitely is not true. These are mixed model issues that are addressed in the text SAS System for Mixed Models, so you may find clarification there once you have access to that.   ... View more

I'm not convinced that I have a correct understanding of your experimental design. Your message suggests to me that you might be drawing subsamples from each bottle at each timepoint. Let me know whether this is right:   For each of 8 donors, you make 6 bottles of fecal sample. 3 bottles are control (trt=2) and 3 bottles are treatment (trt=1). At 0h (tpt=1) you extract 2 aliquots (or something of that nature) from each bottle and measure various metabolites. At 16h (tpt=2) you extract another 2 aliquots from each bottle and measure various metabolites.    Once we are straight on the design, we can move forward to details about the statistical model.   Meanwhile, I will say that if donor is your replicating factor, then donor must be a random effects factor and donor cannot be in the MODEL or LSMEANS statements. However, it is possible to get predictions that are specific for individual donors using narrow inference; see Ch 6 in SAS System for Mixed Models and  https://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_intromod_a0000000337.htm   I agree, Steve Denham is a great resource, very generous with his time and expertise. ... View more

If you are using a distribution other than the normal, you likely are observing the differences between the "marginal mean" and the "binomial p" as illustrated in Figure 2 and discussed in https://dl.sciencesocieties.org/publications/aj/abstracts/107/2/811 The concept applies for distributions other than binomial. ... View more

First off, looking at your posting history, I see that you are working with some challenging models. I highly recommend two resources:  (1) https://www.sas.com/store/books/categories/usage-and-reference/sas-for-mixed-models-second-edition/prodBK_59882_en.html (the first volume of the new edition of which will soon be available, I saw a preprint last month, so excited!), and (2) https://www.amazon.com/Generalized-Linear-Mixed-Models-Applications/dp/1439815127   and   cultivating a statistician as a close colleague--ideally someone you can physically sit with at a table to discuss your research. These models are difficult and persnickety. It is treacherous (and naive) to assume that you can know enough to do a good job with statistics when your main interest is elsewhere (as it should be if you are not a statistician). This forum is not a substitute for real, problem-specific statistical advice.   You are missing a fundamental understanding of the distinction between fixed and random effects, which studing the references above may remedy; plus this paper http://onlinelibrary.wiley.com/doi/10.2307/1941729/abstract has a great discussion of fixed/random issues.   I am assuming below that observations at time 0h and 16h were obtained on the same bottle. If this is not a correct assumption, then everything below would need adjustment.   Assuming that I understand your design correctly, this is where I would start; it is not necessarily where I would end. If your response variables follow the normal distribution, then you could accomplish a similar model using MIXED, but GLIMMIX is my preferred procedure because it is more flexible.     proc glimmix data=have; class trt tpt donor bottle; model y = trt tpt trt*tpt; random donor donor*trt bottle(donor trt) tpt*donor*trt; run; Note that I have added a factor for bottle identification that does not currently exist in your dataset.      Donor is your replicate. As such, it must be a random effects factor and should not be in the MODEL or LSMEANS statement. That should address the NON-EST problem, I think.   BOTTLE is a subsample. If you had balanced data, then I would recommend computing a mean for the variable over the 3 bottles, and then using the mean as the response in a simpler model which drops BOTTLE. You could still try this approach; in practice, unless the unbalance is extreme, I've found it makes little difference.   With only two repeated measurements on the same bottle, the choice of covariance structures is reduced to whether variances are the same at 0h and 16h, or not (i.e., homogenous variances). Compound symmetry CS is equivalent to AR(1); the choice is CS versus CSH.   My experience with GLIMMIX (or MIXED) is that it tends to give me the denomDF that I think are appropriate, but I always sketch out what I expect in advance. Sometimes I specify with the ddf= option, but not often. You want to be very comfortable (and knowledgeable) about the statistical model before you override the default. (But you should not trust the default, by default.)   Assuming that the response (conditional on the predictors) follows a normal distribution, you should get the same result using MIXED as GLIMMIX. The challenge lies in first defining the correct model. ... View more

In your original post, you said that   " at pre intervention, each patient was exposed to greater and greater levels of an energy drink and heart beat was measured in response. Same was done post intervention."   which led me to believe that heart rate was a consequence of the quantity or concentration (say, of caffeine) in the energy drink, that you had multiple observations of heart rate on each subject at different levels of some energy drink metric, and that you were interested in the regression of heart rate on that energy drink metric.    Do you have multiple observations on each subject at different levels of some energy drink metric, or do you have only one observation (e.g., heart rate associated with the total quantity of energy drink consumed)? Is "trigger" the same as energy drink, or is it something else, like "treatment"?   We need a better description of the study protocol.   With respect to wanting to compare regressions but not necessarily wanting to compare slopes: Are you implying that you only want to compare intercepts or some other statistic that represents overall level? What if, for example, the regressions are linear and the slopes are not equal? What would you compare then? I'm perplexed.   ... View more

The answer to your question is that by default, this model assumes a normal distribution, and the Pearson Chi-Square/DF is the residual variance; it is not a measure of overdispersion. For a normal distribution, there is no such thing as overdispersion. (You probably are thinking about a Poisson distribution.)   Is the attached dataset the entire collection, or do you have more observations?   If these 18 are all you have, then your model is attempting to estimate way too many parameters, regardless of which distribution is used. The data don't conform very well to your description. The code you give will not run with the dataset you attach. YEAR, MAGE, and FEMALEAGE are totally confounded (tell me which YEAR, and I can tell you which MAGE and FEMALEAGE without error). With one exception, CAMP and FEMALEID are confounded. There are 3 levels of female genotypes, but two levels have only n=1. MALEID=1130273 has data for both GROUPs; all other birds have data for only one GROUP. Age is not a random effect. It is wise to put random effects factors in the CLASS statement.   In summary, you likely will need to completely rethink the analysis of these data.     ... View more

An analysis of covariance model might work, if what you want is to estimate the means at 15, 30, 45, 60, 90, and 120 minutes for a common value of baseline. (The baseline time=0 value would be the covariate.)   Because (presumably) of the repeated measures on a subject at 0, ..., 120 min, the model would be more complicated than your standard ANCOVA (which would have data for time=0 and time= a second value). In particular, the relationship between data at time 15 and time 0 might be stronger than the relationship between data at time 120 and time 0 because noise intrudes as time passes--in other words, the slope of the regression of the response on baseline (time=0) might decrease with later times.   This is speculative and untested, but I would consider proc glimmix data=have; class time subjectID; model response = time baseline time*baseline; random time / subject=subjectID type=<some covariance structure, maybe ar(1)> residual; lsmeans time / at mean; /* or some other value of baseline */ run; Issues to consider are -- the nature of the relationship between response and baseline at each time (e.g., linear) -- an appropriate covariance structure for the repeated measures within a subject -- normality and homogeneity of variance (assuming normal distribution) -- what a sensible "common" value for baseline might be, in context   Alternatively, you could compute a variable that represents deviance from baseline, either absolute (i.e., subtract the baseline value) or relative (e.g., divide by the baseline value). Or maybe a random coefficients model.  ... View more

I think you actually are wanting an analysis of covariance (ANCOVA)-like model, in which you fit a regression for each level of a categorical predictor. To do that, one of the predictor variables will be a categorical (aka CLASS) variable and another will be a continuous variable.   In your scenario, I would guess that rank should serve as the continuous variable; if so, rank should not be in the CLASS statement. And I would guess that subgroup should serve as the categorical variable; if so, subgroup should be in the CLASS statement.   One parameterization for an ANCOVA-like model is class subgroup; model outcome = rank subgroup rank*subgroup; This model fits a linear regression of outcome on rank for each level of subgroup. The interaction provides a test of H0: slopes equal. If you understand how the categorical variable is coded, you can algebraically compute the intercepts and slopes for the two regressions.   An equivalent parameterization is class subgroup; model outcome = subgroup rank*subgroup / noint; The parameter estimates for subgroup are the intercepts, and the parameter estimates for rank*subgroup are slopes; no algebra required. The p-value for each slope estimate tests H0: slope=0.   Generally, we use the GENMOD procedure when the response variable follows a distribution other than the normal; for normal distribution we can use GLM (among others). But GENMOD does normal; if you don't specify a distribution option on the MODEL statement, then a normal distribution with an identity link is used by default.   An important note: the linear regression is on the link scale! If you are using normal distn/identity link, then the measurement units for parameter estimates (including slope) are the same as the original measurement units for outcome. If you are using GENMOD because outcome follows a distribution other than normal (and your code accidentally omitted that additional and necessary information), then the parameter estimates will be on the link scale (e.g., logit, log). A regression that is linear on the link scale will not be linear on the inverse link (original) scale except for the identity link. This feature has implications for interpretation.   There's a lot going on here, and in my opinion, ANCOVA is trickier than it looks at first glance, not to mention the potential complications of generalized linear models. Also, not to mention your desire to "control for other covariates". You'll want to devote time to gaining familiarity with the methodology as well as with SAS software. ... View more

The syntax repeated day / subject=id(trt); corresponds to an experimental design in which each subject (id) is randomly assigned to one level of trt, and multiple observations are made on each id at each level of day. As such, ids are nested within levels of trt: each id belongs to only one level of trt.    id(trt) identifies a random effects factor that has one level for each unique combination of id and trt. It does not mean that trt is random; it's just a syntax shortcut.   Alternatively, you could assign each subject a unique id value and use repeated day / subject=unique_id; which is the point that @Ksharp is making.   Even with unique values for subjects, I still use the nested syntax because it does a better job of giving me the denominator df that I think are appropriate.   I note that you have time in the MODEL statement. I bet this is a repeated measures factor as well, probably multiple times within each day. If so, your model is wrong, along with your understanding of the MIXED syntax. But that's another topic.   I highly recommend  https://www.sas.com/store/books/categories/usage-and-reference/sas-for-mixed-models-second-edition/prodBK_59882_en.html It will tell you most of what you need to know.   ... View more

This paper is a good resource for the technical details https://support.sas.com/resources/papers/proceedings11/351-2011.pdf   The challenge will be deciding what you want to compare to what, which is context-specific and thus falls to you, the researcher.   If your intent is to compare each of 36 treatment means to each of two control means, then you could consider mean comparisons controlled for Type I error using the Dunnett method, possibly with additional adjustment due to having two controls rather than one.   Good luck! ... View more

With only 10 observations in your posted dataset, split 2 and 8 between outcomes, I don't know that you'll be able to extract much from an analysis, but....   If you plot the response against each predictor, it is clear that complete separation is due to X1. A solution can be obtained using Firth's penalized likelihood. See   https://pdfs.semanticscholar.org/4f17/1322108dff719da6aa0d354d5f73c9c474de.pdf and http://support.sas.com/kb/22/599.html     data have; input a$ y x1 x2; datalines; 1. 1 1.489611900786800 -0.486983894512530 2. 1 0.887638190472230 -0.899961461187430 3. 1 -0.328400349680380 0.320480850960210 4. 0 -1.283346136073470 0.314729922388780 5. 1 -0.014384666024895 -1.040793737862780 6. 1 1.005337941612940 0.385444205622100 7. 1 0.403112850999760 0.797554772638080 8. 1 1.432508077938930 -0.553701810045310 9. 1 0.137341139238340 0.177313212434980 10. 0 -1.341507064615280 0.042985039337917 ; run; proc sgplot data=have; scatter x=x1 y=y; run; proc sgplot data=have; scatter x=x2 y=y; run; proc logistic data = have desc; model y = x1 x2 / firth; run;   ... View more