Yes, its just a warning because if you
were trying to estimate both then you
have an insolvable problem!
Its intended to alert users that want
to only have one residual term (ie. most
users, most of the time) that something
has gone wrong with their model specification.
In your case, although you only have one
residual term in principle, you have two
in practice because you have no replication,
therefore your between and within-study
terms are structurally equivalent. Then
you avoid the aliasing problem by fixing
one to the known residual variance.
Cheers
Sue
> -----Original Message-----
> From: James Krieger [mailto:[log in to unmask]]
> Sent: 02 March 2004 16:00
> To: [log in to unmask]
> Subject: Re: meta-analysis in GenStat
>
>
> Sue,
>
> When I run this analysis, I get the following warning message:
>
> **** G5W0002 **** Warning (Code VC 53). Statement 1 on Line 38
>
> Command: reml [print=#,mon,eff; weights=INVSD] EFFECT
>
> More than one residual term specifed - first term found will
> be used as R
>
>
>
> Is this normal?
>
> James
>
> James Krieger, M.S.
> Science Editor, Pure Power Magazine
> http://www.purepowermag.com
> Webmaster, WSU Strength and Conditioning
> http://www.wsu.edu/~strength
>
>
>
> ----- Original Message -----
> From: "sue welham (RRes-Roth)" <[log in to unmask]>
> To: <[log in to unmask]>
> Sent: Monday, March 01, 2004 6:37 AM
> Subject: Re: meta-analysis in GenStat
>
>
> > Dear James
> >
> > This message tries to answer your question and
> > the other various comments.
> >
> > First, Stephen Senn is right, the meta-analysis
> > menu is designed for analysis of several experiments
> > together when the individual data values are available.
> > In your case, you have a single value for each study
> > with its standard error. Without any constraints, the
> > between- and within-study variation are aliased but
> > as you know the within-study variation, if you can
> > fix this, then you can estimate the between-study
> > variation.
> >
> > I think the following code should do what you want,
> > note I have assumed 19 studies, so replace 19 with the
> > relevant number:
> >
> > factor [lev=19; val=1...19] units
> >
> > calc [p=s] wt = 1/(se*se)
> >
> > vcomp [fixed=pweeks; cadjust=none] random=units+study; \
> > init=1,1; constr=fix,pos
> > reml [print=#,mon,eff; weights=wt] effect
> >
> > The units factor is created to represent the residual
> > variation, and the value of its variance component is
> > set to be 1.0. The use of the weights option in
> > REML then ensures you have the correct residual
> > variance for each study. The study component can
> > then estimate the between-study variability. As
> > pointed out previously, the CADJUST=none option
> > means that the covariate is not centered and so
> > should match the SAS output.
> >
> > With respect to your previous code:
> >
> > vcomp [fixed=pweeks; cadjust=none] random=units+study; \
> > init=1,1; constr=fix,pos
> > vstructure [terms=study; form=whole] model=unstr
> > reml [print=#,mon,eff; weights=wt] effect
> >
> > In this case, the model attempts to estimate an
> > unstructured covariance matrix across the 19 study
> > intercept values. Because there is only a single
> > data point for each study, this is not possible
> > hence the failure of the algorithm.
> >
> > I think the presence of the unstructured model
> > in the SAS code is a red herring here. I'm not
> > an expert, but I think the line
> >
> > random int /type=un subject=study
> >
> > means that separate intercepts are to be fitted
> > for each study, and that an unstructured matrix
> > is to be fitted across the effects for each subject
> > (study). Since there is only one effect (intercept)
> > per study, then the unstructured matrix becomes
> > a single variance component.
> >
> > Hope this helps,
> >
> > Sue Welham
> >
> > Rothamsted Research
> > Harpenden AL5 2JQ, UK
> >
> > Email: [log in to unmask]
> >
> >
> > -----Original Message-----
> > From: James Krieger
> > To: [log in to unmask]
> > Sent: 01/03/04 06:07
> > Subject: Struggling to adapt short piece of SAS code to GENSTAT
> >
> > Hello, everyone,
> >
> > I've been trying to adapt the following piece of SAS code
> to GENSTAT,
> > with
> > no success.
> >
> > SAS code:
> >
> > Proc mixed covtest data=meta;
> > Class studyID;
> > Weight invsd;
> > Model effect = week /s;
> > Random int /type=un sub=studyID;
> > Parms (0.1) (1) /hold=2;
> > Run;
> >
> >
> > First, I tried this in GENSTAT:
> >
> > VCOMPONENTS [FIXED=WEEK] RANDOM=STUDYID; INITIAL=0.1,1;
> > CONSTRAINTS=none,fixrelative
> > REML
> >
> [PRINT=model,components,effects,means,vcovariance,deviance,waldTests;
> > WEIGHTS=INVSD; PSE=differences;MVINCLUDE=*; METHOD=AI] EFFECT
> >
> > I get the correct parameter for the week factor. In SAS, it
> > is -0.158(0.036). In GENSTAT, it is -0.158(0.035).
> >
> > However, I do not get the correct parameter for the
> intercept. In SAS,
> > it
> > is 0.409(0.087), but in GENSTAT, I get .061(.036). I then tried to
> > impose
> > an unstructured covariance structure, just as in the SAS code.
> >
> > VCOMPONENTS [FIXED=WEEK] RANDOM=STUDYID; INITIAL=0.1,1;
> > CONSTRAINTS=none,fixrelative
> > VSTRUCTURE [TERMS=STUDYID] MODEL=unstructured
> > REML
> >
> [PRINT=model,components,effects,means,vcovariance,deviance,waldTests;
> > WEIGHTS=INVSD; PSE=differences;MVINCLUDE=*; METHOD=AI] EFFECT
> >
> > When I do this, I get numerous warning messages, including
> warnings that
> > the
> > value of the deviance at the final iteration is larger than
> at previou s
> > iterations, and that the REML algorithm has diverged and
> the parameters
> > are
> > out of bounds.
> >
> > I cannot figure out what I'm doing wrong. Is there anyone
> here that has
> > good knowledge of both PROC MIXED and REML/VCOMPONENTS that
> can tell me
> > what
> > mistake I'm making in trying to adapt this SAS code? I am new to
> > GENSTAT.
> >
> > Thank you for your time,
> >
> > James Krieger, M.S.
> > Science Editor, Pure Power Magazine
> > http://www.purepowermag.com
> > Webmaster, WSU Strength and Conditioning
> > http://www.wsu.edu/~strength
> >
>
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