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jesseh
‎09-19-2024
  • 6 Posts
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Re: Covariate values in ESTIMATE statements when using a BY variable.

by Calcite | Level 5 in Statistical Procedures
‎12-14-2023 03:10 PM
‎12-14-2023 03:10 PM
This is perfect, thank you. I have used lsmestimate before, but it slipped my mind to use it in this way. ... View more

Covariate values in ESTIMATE statements when using a BY variable.

by Calcite | Level 5 in Statistical Procedures
‎12-12-2023 09:21 AM
‎12-12-2023 09:21 AM
I am doing an analysis using a mixed model (PROC MIXED) on longitudinal data, wherein I'm running the model for different bootstraps/imputation combinations using a BY statement. Getting the LSMEANS from each is easy, and I can estimate the model mean AT any visit for all different values of my BY variables (bootstrap/imputation combo). SAS automatically plugs in the means of the other covariates BY the variables I have given to create different LSMEANS each time.    However, one estimate of interest is the mean OVER the first six visits. For a single analysis, I have previously used an ESTIMATE statement as shown in the code below, giving equal weight over the first six visits. To produce the mean, this ESTIMATE statement needs average values of my two continuous covariates (var1 and var2) explicitly plugged in, which here are assumed to be calculated previously and stored as macro variables &mvar1 and &mvar2; it is not automatically done like in LSMEANS.   How can I most efficiently run the model over my BY variables (bootstrap sample/imputation combinations) using the mean of var1 and var2 for the specific bootstrap sample/imputation each time, the same way LSMEANS does automatically with the BY statement?   proc mixed data=mydata method = reml; by bootn imputen; class subj trt visit; model chg = trt|visit var1 var2 / ddfm=bw solution; repeated visit / type=un subject=subj; lsmeans trt|visit/ pdiff=all cl; estimate 'TRT 1 Over First 6 visits' intercept 1 var1 &mvar1 covar3 &mvar2 visit 0.166666 0.166666 0.1666666 0.166666 0.166666 0.166666 0 0 0 0 trt 1 0 0 0 trt*visit 0.1666666 0.166666 0.166666 0.166666 0.166666 0.166666 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /CL;     ... View more

Covariate values in ESTIMATE statements when using a BY variable.

by Calcite | Level 5 in SAS Programming
‎12-11-2023 05:42 PM
‎12-11-2023 05:42 PM
I am doing an analysis using a mixed model (PROC MIXED) on longitudinal data, wherein I'm running the model for different bootstraps/imputation combinations using a BY statement. Getting the LSMEANS from each is easy, and I can estimate the model mean AT any visit for all different values of my BY variables (bootstrap/imputation combo). SAS automatically plugs in the means of the other covariates BY the variables I have given to create different LSMEANS each time.    However, one estimate of interest is the mean OVER the first six visits. For a single analysis, I have previously used an ESTIMATE statement as shown in the code below, giving equal weight over the first six visits. To produce the mean, this ESTIMATE statement needs average values of my two continuous covariates (var1 and var2) explicitly plugged in, which here are assumed to be calculated previously and stored as macro variables &mvar1 and &mvar2; it is not automatically done like in LSMEANS.   How can I most efficiently run the model over my BY variables (bootstrap sample/imputation combinations) using the mean of var1 and var2 for the specific bootstrap sample/imputation each time, the same way LSMEANS does automatically with the BY statement?     proc mixed data=mydata method = reml; by bootn imputen; class subj trt visit; model chg = trt|visit var1 var2 / ddfm=bw solution; repeated visit / type=un subject=subj; lsmeans trt|visit/ pdiff=all cl; estimate 'TRT 1 Over First 6 visits' intercept 1 var1 &mvar1 covar3 &mvar2 visit 0.166666 0.166666 0.1666666 0.166666 0.166666 0.166666 0 0 0 0 trt 1 0 0 0 trt*visit 0.1666666 0.166666 0.166666 0.166666 0.166666 0.166666 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /CL;     ... View more

Re: How to test equality of three mean expressions coming from PROC MI...

by Calcite | Level 5 in Statistical Procedures
‎10-17-2023 11:29 AM
‎10-17-2023 11:29 AM
Thank you, this is perfect, I didn't know about the JOINT option. Just for any others looking at this post, here is the final code with I believe all the correct zeros and everything.      ods output lsmeans=lsms lsmestimate=lsmests; proc mixed data=eff_pop method = reml; class subj trt visit agegroup hist; model chg = trt visit agegroup trt*visit trt*agegroup visit*agegroup trt*visit*agegroup base cont hist / ddfm=kr solution outp=predict; repeated visit / type=un subject=subj; lsmeans trt*visit*agegroup; lsmestimate trt*visit*agegroup 'Difference within age group 1 at visit 4, trt 1 vs. 2' 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0, 'Difference within age group 2 at visit 4, trt 1 vs. 2' 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0, 'Difference within age group 3 at visit 4, trt 1 vs. 2' 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0/ joint; run; ods rtf close;     ... View more

How to test equality of three mean expressions coming from PROC MIXED ...

by Calcite | Level 5 in Statistical Procedures
‎10-13-2023 02:27 PM
‎10-13-2023 02:27 PM
I'm analyzing data using an MMRM with PROC MIXED. The covariates are treatment (i=1,2,3,4), visit (j=1,2,3,4,5), and age group (k=1,2,3) with three-way and all two-way interactions included, along with 3 other covariates. I'm mainly using ESTIMATE statements to provide treatment differences at visit 4, very simple. I have one hypothesis I want to test that is more complicated, and I can't figure out if I can use built-in SAS statements.    I want to test if the difference between treatment 1 and 2 is the same across the 3 age groups at week 24. So, using the (ijk) notation... H_0   =   mu_(141)-mu_(241) = mu(142)-mu_(242) = mu_(143)-mu_(243) H_a    = not all equal   It would be simple to test the equality of any two of these expressions using ESTIMATE statements, but there are three expressions here I guess necessitating an ANOVA kind of test? I'm not sure how to go about coding this test. Main modeling code is below.     ods output estimates= ests lsmeans= lsms diffs= diffs; proc mixed data=eff_pop method = reml; class subj trt visit agegroup hist; model chg = trt visit agegroup trt*visit trt*agegroup visit*agegroup trt*visit*agegroup  base  cont hist / ddfm=kr solution outp=predict; repeated visit / type=un subject=subj; lsmeans trt01pn|avisitn/ pdiff=all cl;     ... View more
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