[R-sig-ME] Zero cells in contrast matrix problem

Ben Bolker bbolker at gmail.com
Tue Oct 27 17:06:37 CET 2015


I think this isn't as much of a problem as you think.  lme4
automatically drops non-estimable columns in the model matrix.  The
problem is just getting it to work right with blme.  (But yes,
collapsing this into a one-way design is one way of proceeding ...)

On Tue, Oct 27, 2015 at 12:01 PM, Thompson,Paul
<Paul.Thompson at sanfordhealth.org> wrote:
> This is a problem in SAS as well. When you specify a contrast in a 2-factor design, things get screwed up with missing cells. In fact, the contrasts require a matrix multiplication, and if a cell isn't there, the matrix multiplication does not work correctly. They are not estimable, in the SAS terminology (which is standard statistical terminology as well).
>
> One approach that can be used is to convert the 2-factor design to a one-factor design, and do everything with contrasts, main effects, in teractions, etc. A little more work, but since you are specifying the contrasts (and presuming that you do so correctly), you will get estimable contrasts.
>
> Marginal effects must take the missing cells into account correctly.
>
> -----Original Message-----
> From: R-sig-mixed-models [mailto:r-sig-mixed-models-bounces at r-project.org] On Behalf Of Francesco Romano
> Sent: Monday, October 26, 2015 6:18 AM
> To: Ben Bolker; r-sig-mixed-models at r-project.org
> Subject: Re: [R-sig-ME] Zero cells in contrast matrix problem
>
> For some reason the silly bugger didn't past the full command:
>
>> revanaA<-
> bglmer(Correct~Syntax*Animacy*Prof.group.2+(1|Part.name)+(1|Item), data = revana, family = binomial, fixef.prior = normal(cov = diag(9,16))) fixed-effect model matrix is rank deficient so dropping 2 columns / coefficients Error in normal(cov = cov, common.scale = FALSE) :
>   normal prior covariance of improper length
>
> To give more info on this, it is the Animacy factor that is causing separation because two levels of it have zero counts in some cases.
>
> On Mon, Oct 26, 2015 at 12:13 PM, Ben Bolker <bbolker at gmail.com> wrote:
>
>> Well, that's a separate problem (and not necessarily a "problem").   R
>> is telling you that you have 16 separate combinations of the factors,
>> but only 14 unique combinations represented in your data set, so it
>> can only estimate 14 parameters.  Unless there is a weird interaction
>> with blme I don't know about, this should still give you reasonable
>> answers.
>>
>> On Mon, Oct 26, 2015 at 7:10 AM, Francesco Romano
>> <francescobryanromano at gmail.com> wrote:
>> > Many thanks Ben,
>> >
>> > but I tried that already:
>> >
>> >> revanaA<-
>> >> bglmer(Correct~Syntax*Animacy*Prof.group.2+(1|Part.name)+(1|Item),
>> >> data
>> =
>> >> revana, family = binomial, fixef.prior = normal(cov = diag(9,16)))
>> > fixed-effect model matrix is rank deficient so dropping 2 columns /
>> > coefficients Error in normal(cov = cov, common.scale = FALSE) :
>> >   normal prior covariance of improper length
>> >
>> > On Mon, Oct 26, 2015 at 12:06 PM, Ben Bolker <bbolker at gmail.com> wrote:
>> >>
>> >> On 15-10-26 06:56 AM, Francesco Romano wrote:
>> >> > I wonder if anyone can help with the separation problem
>> >> > originally solved by Ben Bolker (see thread).
>> >> > The model and fitting I used previously was
>> >> >
>> >> > trial<-bglmer(Correct ~ Syntax.Semantics, data = trialglm, family
>> >> > = binomial, fixef.prior = normal(cov = diag(9,4))
>> >> >
>> >> > which now has to change because the Syntax.Semantcs factor needs
>> >> > to be split into separate within-subjects variables, Syntax, a
>> >> > factor with two levels, and Animacy, a factor with four levels.
>> >> > In addition a new between-subjects factor called Group with two
>> >> > levels (native vs non-native speaker) has to be added which
>> >> > determines the following model, fit by bglmer:
>> >> >
>> >> > trial<-bglmer(Correct ~ Syntax*Animacy*Group+
>> >> > (1|Part.name)+(1|Item), data = trialglm, family = binomial,
>> >> > fixef.prior = normal(cov = diag???)
>> >> >
>> >> > What values should I use for the cov=diag portion in order to
>> >> > continue attempting convergence of a model that includes the
>> >> > random effects?
>> >>
>> >>     In general a reasonable form is normal(cov = diag(v,np)) where
>> >> v is the prior variance (generally something reasonably
>> >> large/non-informative; 9 (=std dev of 3) is probably an OK default)
>> >> and np is the number of fixed-effect parameters.  You can figure
>> >> this out
>> via
>> >>
>> >> ncol(model.matrix(~Syntax*Animacy*Group,data=trialglm)
>> >>
>> >> or multiply 2*4*2 to get 16 ...
>> >>
>> >> >
>> >> > R returns the following error because I don't know how to
>> >> > establish
>> the
>> >> > parameters when more than one
>> >> > fixed effect is involved:
>> >> >
>> >> > Error in normal(cov = cov, common.scale = FALSE) :
>> >> >   normal prior covariance of improper length
>> >> >
>> >> > Many thanks in advance for any help!
>> >> >
>> >> >
>> >> >
>> >> >
>> >> >
>> >> > On Thu, May 28, 2015 at 10:46 PM, Ben Bolker <bbolker at gmail.com>
>> wrote:
>> >> >
>> >> >>   I don't see your data -- I see a little tiny subset, but
>> >> >> that's not really enough for a reproducible example.
>> >> >>
>> >> >> This is the example given in the URL I sent:
>> >> >>
>> >> >> cmod_blme_L2 <- bglmer(predation~ttt+(1|block),data=newdat,
>> >> >>                        family=binomial,
>> >> >>                        fixef.prior = normal(cov = diag(9,4)))
>> >> >>
>> >> >> trial<-bglmer(Correct ~ Syntax.Semantics+(1|Part.name),
>> >> >>           data =trialglm,
>> >> >>          family = binomial,
>> >> >>         fixef.prior = normal(cov=diag(9,8)))
>> >> >>
>> >> >> The last line specifies an 8x8 matrix (because you have 8 fixed
>> effect
>> >> >> parameters) with a value of 9 on the diagonal, meaning the
>> >> >> priors for the fixed effects are independent and each is Normal
>> >> >> with a sd of sqrt(9)=3.
>> >> >>
>> >> >>
>> >> >> On Thu, May 28, 2015 at 3:25 PM, Francesco Romano
>> >> >> <francescobryanromano at gmail.com> wrote:
>> >> >>> Yes but this seems a bit above my head without your help. The
>> >> >>> data
>> are
>> >> >>> in
>> >> >>> the three variables at the bottom of my email but I forgot to
>> mention
>> >> >>> the
>> >> >>> random participant effect (n = 17). Thanks!
>> >> >>>
>> >> >>>
>> >> >>> Il giovedì 28 maggio 2015, Ben Bolker <bbolker at gmail.com> ha
>> scritto:
>> >> >>>>
>> >> > On 15-05-28 06:55 AM, Francesco Romano wrote:
>> >> >>>>>> Many thanks to both.
>> >> >>>>>>
>> >> >>>>>> The approaches you suggest (and others online) help one deal
>> >> >>>>>> with the separation problem but don't offer any specific
>> >> >>>>>> advice as to how getting a valid p coefficient when
>> >> >>>>>> comparing two levels of
>> the
>> >> >>>>>> model vexed by separation.
>> >> >>>>>>
>> >> >>>>>> Ben, here's the output of the bglmer which by the way would
>> >> >>>>>> be ideal since it allows me to retain the random effect so
>> >> >>>>>> that all
>> my
>> >> >>>>>> pairwise comparisons are conducted using mixed effects.
>> >> >>>>>>
>> >> >>>>>>> trial<-bglmer(Correct ~ Syntax.Semantics+(1|Part.name),
>> >> >>>>>>> data = trialglm,
>> >> >>>>>> family = binomial) Warning message: package ‘blme’ was built
>> under
>> >> >>>>>> R version 3.1.2
>> >> >>>>>>> summary(trial)
>> >> >>>>>> Cov prior  : Part.name ~ wishart(df = 3.5, scale = Inf,
>> >> >>>>>> posterior.scale = cov, common.scale = TRUE) Prior dev  :
>> >> >>>>>> 1.4371
>> >> >>>>>>
>> >> >>>>>> Generalized linear mixed model fit by maximum likelihood
>> >> >>>>>> (Laplace
>> >> >>>>>> Approximation) ['bglmerMod'] Family: binomial  ( logit ) Formula:
>> >> >>>>>> Correct ~ Syntax.Semantics + (1 | Part.name) Data: trialglm
>> >> >>>>>>
>> >> >>>>>> AIC      BIC   logLik deviance df.resid 269.9    305.5   -126.0
>> >> >>>>>> 251.9      376
>> >> >>>>>>
>> >> >>>>>> Scaled residuals: Min      1Q  Median      3Q     Max -0.9828
>> >> >>>>>> -0.4281 -0.2445 -0.0002  5.7872
>> >> >>>>>>
>> >> >>>>>> Random effects: Groups    Name        Variance Std.Dev. Part.name
>> >> >>>>>> (Intercept) 0.3836   0.6194 Number of obs: 385, groups:
>> Part.name,
>> >> >>>>>> 16
>> >> >>>>>>
>> >> >>>>>> Fixed effects: Estimate Std. Error z value Pr(>|z|) (Intercept)
>> >> >>>>>> -1.8671     0.4538  -4.114 3.89e-05 *** Syntax.Semantics A
>> >> >>>>>> 0.8121     0.5397   1.505   0.1324 Syntax.Semantics B  -16.4391
>> >> >>>>>> 1195.5031  -0.014   0.9890 Syntax.Semantics C   -1.1323
>>  0.7462
>> >> >>>>>> -1.517   0.1292 Syntax.Semantics D    0.1789     0.5853   0.306
>> >> >>>>>> 0.7598 Syntax.Semantics E   -0.8071     0.7500  -1.076   0.2819
>> >> >>>>>> Syntax.Semantics F   -1.5051     0.8575  -1.755   0.0792 .
>> >> >>>>>> Syntax.Semantics G    0.4395     0.5417   0.811   0.4171 ---
>> >> >>>>>> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’
>> >> >>>>>> 1
>> >> >>>>>>
>> >> >>>>>> Unfortunately the separation problem is still there. Should
>> >> >>>>>> I be constraining the parameter somehow? How would I do
>> >> >>>>>> that? The data is below.
>> >> >
>> >> >    Did you read the section in the URL I suggested?  Just using
>> >> > bglmer isn't enough; you also have to set a prior on the fixed effects.
>> >> >
>> >> >   Your data don't seem to be attached (note that the mailing list
>> >> > strips most non-ASCII file types).
>> >> >
>> >> >>>>>>
>> >> >>>>>> In passing I also tried brglm which solves the separation
>> >> >>>>>> problem but tells me comparison is significant which I don't
>> >> >>>>>> believe one bit (see the data below). I am pretty sure about
>> >> >>>>>> this because
>> when
>> >> >>>>>> I reveled and look at the comparisons I was able to compute
>> >> >>>>>> using glmer, these turn out to be non-significant, when
>> >> >>>>>> glmer told me they were:
>> >> >>>>>>
>> >> >>>>>>> trial<-brglm(Correct ~ Syntax.Semantics, data = trialglm,
>> >> >>>>>>> family =
>> >> >>>>>> binomial) Warning messages: 1: package ‘elrm’ was built
>> >> >>>>>> under R version 3.1.2 2: package ‘coda’ was built under R
>> >> >>>>>> version 3.1.3
>> >> >>>>>>> summary(trial)
>> >> >>>>>>
>> >> >>>>>> Call: brglm(formula = Correct ~ Syntax.Semantics, family =
>> >> >>>>>> binomial, data = trialglm)
>> >> >>>>>>
>> >> >>>>>>
>> >> >>>>>> Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept)
>> >> >>>>>> -1.6358     0.4035  -4.053 5.05e-05 *** Syntax.Semantics A
>>  0.6689
>> >> >>>>>> 0.5169   1.294   0.1957 Syntax.Semantics B  -3.0182     1.4902
>> >> >>>>>> -2.025   0.0428 * Syntax.Semantics C  -1.0135     0.6889  -1.471
>> >> >>>>>> 0.1413 Syntax.Semantics D   0.1515     0.5571   0.272   0.7857
>> >> >>>>>> Syntax.Semantics E  -0.7878     0.6937  -1.136   0.2561
>> >> >>>>>> Syntax.Semantics F  -1.2874     0.7702  -1.672   0.0946 .
>> >> >>>>>> Syntax.Semantics G   0.4358     0.5186   0.840   0.4007 ---
>> Signif.
>> >> >>>>>> codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>> >> >>>>>>
>> >> >>>>>> (Dispersion parameter for binomial family taken to be 1)
>> >> >>>>>>
>> >> >>>>>> Null deviance: 262.51  on 384  degrees of freedom Residual
>> >> >>>>>> deviance: 256.22  on 377  degrees of freedom Penalized deviance:
>> >> >>>>>> 245.5554 AIC:  272.22
>> >> >>>>>>
>> >> >>>>>>
>> >> >>>>>> MCMCglmm is too complex for me.
>> >> >>>>>>
>> >> >>>>>> Wolfgang, I tried the penalized likelihood method (logistf
>> >> >>>>>> function) but output is hard to read:
>> >> >>>>>>
>> >> >>>>>>> trial<-logistf(Correct ~ Syntax.Semantics, data = trialglm,
>> >> >>>>>>> family =
>> >> >>>>>> binomial) Warning messages: 1: package ‘logistf’ was built
>> >> >>>>>> under
>> R
>> >> >>>>>> version 3.1.2 2: package ‘mice’ was built under R version
>> >> >>>>>> 3.1.2
>> >> >>>>>>> summary(trial)
>> >> >>>>>> logistf(formula = Correct ~ Syntax.Semantics, data =
>> >> >>>>>> trialglm, family = binomial)
>> >> >>>>>>
>> >> >>>>>> Model fitted by Penalized ML Confidence intervals and
>> >> >>>>>> p-values by Profile Likelihood Profile Likelihood Profile
>> >> >>>>>> Likelihood Profile Likelihood Profile Likelihood Profile
>> >> >>>>>> Likelihood Profile
>> Likelihood
>> >> >>>>>> Profile Likelihood
>> >> >>>>>>
>> >> >>>>>> coef  se(coef) lower 0.95 upper 0.95 Chisq            p
>> (Intercept)
>> >> >>>>>> 3.2094017 0.7724482  2.9648747  3.5127830 0.000000
>> >> >>>>>> 1.000000e+00 Syntax.Semantics A  4.1767737 6.3254344
>> >> >>>>>> 0.4224696 12.0673987
>> >> >>>>>> 64.224452 1.110223e-15 Syntax.Semantics B -1.0583602
>> >> >>>>>> 0.8959376
>> >> >>>>>> -1.3963977 -0.7625216  0.000000 1.000000e+00
>> >> >>>>>> Syntax.Semantics C
>> >> >>>>>> -0.7299070 0.9308193 -1.0765598 -0.4180076  0.000000
>> >> >>>>>> 1.000000e+00 Syntax.Semantics D  0.2314740 1.1563731
>> >> >>>>>> -0.1704535  0.6479908
>> >> >>>>>> 1.156512 2.821901e-01 Syntax.Semantics E -0.6476907
>> >> >>>>>> 0.9771824
>> >> >>>>>> -1.0076740 -0.3164066  0.000000 1.000000e+00
>> >> >>>>>> Syntax.Semantics F
>> >> >>>>>> -0.8271499 0.9305931 -1.1743834 -0.5160799  0.000000
>> >> >>>>>> 1.000000e+00 Syntax.Semantics G  0.9909046 1.3787175
>> >> >>>>>> 0.5457741  1.5353981
>> >> >>>>>> 0.000000 1.000000e+00
>> >> >>>>>>
>> >> >>>>>> Likelihood ratio test=121.9841 on 7 df, p=0, n=385 Wald test
>> >> >>>>>> =
>> >> >>>>>> 5.334321 on 7 df, p = 0.6192356
>> >> >>>>>>
>> >> >>>>>> In particular, what is this model telling me? That Z (my ref
>> level)
>> >> >>>>>> and B are significantly different?
>> >> >>>>>>
>> >> >>>>>> I'm happy to try the elrm function with exact logistic
>> >> >>>>>> regression but I am not capable of programming it. Besides,
>> >> >>>>>> would it give me valid estimates for the comparison between the Z and B levels?
>> The
>> >> >>>>>> data frame should look like this:
>> >> >>>>>>
>> >> >>>>>> Outcome variable (Correct, incorrect) Predictor variable (A,
>> >> >>>>>> B,
>> C,
>> >> >>>>>> D, E, F, G, Z) Counts (E: 38,3; B: 51,0; Z: 37,7; G: 40,12; D:
>> >> >>>>>> 36,8; C:45,3; A: 34,13; F:65,22).
>> >> >>>>>>
>> >> >>>>>> Thank you! F.
>> >> >>>>>>
>> >> >>>>>> On Thu, May 28, 2015 at 2:28 AM, Ben Bolker
>> >> >>>>>> <bbolker at gmail.com>
>> >> >>>>>> wrote:
>> >> >>>>>>
>> >> >>>>>>> And for what it's worth, you can do this in conjunction
>> >> >>>>>>> with
>> lme4
>> >> >>>>>>> by using the blme package instead (a thin Bayesian wrapper
>> around
>> >> >>>>>>> lme4), or via the MCMCglmm package; see
>> >> >>>>>>>
>> http://ms.mcmaster.ca/~bolker/R/misc/foxchapter/bolker_chap.html
>> >> >>>>>>> for an example (search for "complete separation").
>> >> >>>>>>>
>> >> >>>>>>> On Wed, May 27, 2015 at 5:21 PM, Viechtbauer Wolfgang
>> >> >>>>>>> (STAT) <wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:
>> >> >>>>>>>> You may need to consider using an 'exact', Bayesian, or
>> >> >>>>>>>> penalized
>> >> >>>>>>> likelihood approach (along the lines proposed by Firth).
>> >> >>>>>>>>
>> >> >>>>>>>> Maybe a place to start:
>> >> >>>>>>>
>> >> >>>>>>>
>> >> >>>
>> >> >>>
>> http://sas-and-r.blogspot.nl/2010/11/example-815-firth-logistic-regres
>> sion.html
>> >> >>>>>>>>
>> >> >>>>>>>>
>> >> >>>>>>>
>> >> > Best,
>> >> >>>>>>>> Wolfgang
>> >> >>>>>>>>
>> >> >>>>>>>>> -----Original Message----- From: R-sig-mixed-models
>> >> >>>>>>>>> [mailto:r-sig-mixed-models-bounces at r- project.org] On
>> >> >>>>>>>>> Behalf Of Francesco Romano Sent: Wednesday, May 27, 2015 23:00 To:
>> >> >>>>>>>>> r-sig-mixed-models at r-project.org Subject: [R-sig-ME] Zero
>> >> >>>>>>>>> cells in contrast matrix problem
>> >> >>>>>>>>>
>> >> >>>>>>>>> After giving up on a glmer for my data, I remembered a
>> >> >>>>>>>>> post by Roger
>> >> >>>>>>> Levy
>> >> >>>>>>>>> suggesting to try the use non mixed effects glm when one
>> >> >>>>>>>>> of the cells in a matrix is zero.
>> >> >>>>>>>>>
>> >> >>>>>>>>> To put this into perspective:
>> >> >>>>>>>>>
>> >> >>>>>>>>>> trial<-glmer(Correct ~ Syntax.Semantics + (1 |
>> >> >>>>>>>>>> Part.name), data =
>> >> >>>>>>>>> trialglm, family = binomial)
>> >> >>>>>>>>>
>> >> >>>>>>>>> Warning messages: 1: In checkConv(attr(opt, "derivs"),
>> >> >>>>>>>>> opt$par, ctrl = control$checkConv, : Model failed to
>> >> >>>>>>>>> converge with max|grad| = 0.053657 (tol = 0.001,
>> >> >>>>>>>>> component 4) 2: In checkConv(attr(opt, "derivs"),
>> >> >>>>>>>>> opt$par, ctrl = control$checkConv, : Model is nearly
>> >> >>>>>>>>> unidentifiable: large eigenvalue ratio - Rescale variables?
>> >> >>>>>>>>>
>> >> >>>>>>>>> My data has a binary outcome, correct or incorrect, a
>> >> >>>>>>>>> fixed effect predictor factor with 8 levels, and a random
>> >> >>>>>>>>> effect for participants. I believe the problem R is
>> >> >>>>>>>>> encountering is with one level of the factor (let us call
>> >> >>>>>>>>> it level B) which has no counts (no I won' t try to post
>> >> >>>>>>>>> the table from the paper with the counts because I know
>> >> >>>>>>>>> it will get garbled up!).
>> >> >>>>>>>>>
>> >> >>>>>>>>> I attempt a glm with the same data:
>> >> >>>>>>>>>
>> >> >>>>>>>>>> trial<-glm(Correct ~ Syntax.Semantics, data = trialglm,
>> >> >>>>>>>>>> family =
>> >> >>>>>>>>> binomial)
>> >> >>>>>>>>>> anova(trial)
>> >> >>>>>>>>> Analysis of Deviance Table
>> >> >>>>>>>>>
>> >> >>>>>>>>> Model: binomial, link: logit
>> >> >>>>>>>>>
>> >> >>>>>>>>> Response: Correct
>> >> >>>>>>>>>
>> >> >>>>>>>>> Terms added sequentially (first to last)
>> >> >>>>>>>>>
>> >> >>>>>>>>>
>> >> >>>>>>>>> Df Deviance Resid. Df Resid. Dev NULL
>> >> >>>>>>>>> 384     289.63 Syntax.Semantics  7   34.651       377
>> >> >>>>>>>>> 254.97
>> >> >>>>>>>>>> summary(trial)
>> >> >>>>>>>>>
>> >> >>>>>>>>> Call: glm(formula = Correct ~ Syntax.Semantics, family =
>> >> >>>>>>>>> binomial, data = trialglm)
>> >> >>>>>>>>>
>> >> >>>>>>>>> Deviance Residuals: Min        1Q    Median        3Q
>> >> >>>>>>>>> Max -0.79480  -0.62569  -0.34474  -0.00013   2.52113
>> >> >>>>>>>>>
>> >> >>>>>>>>> Coefficients: Estimate Std. Error z value Pr(>|z|)
>> >> >>>>>>>>> (Intercept)                 -1.6917     0.4113  -4.113
>> >> >>>>>>>>> 3.91e-05 *** Syntax.Semantics A   0.7013     0.5241   1.338
>> >> >>>>>>>>> 0.1809 Syntax.Semantics B -16.8744   904.5273  -0.019
>> >> >>>>>>>>> 0.9851 Syntax.Semantics C  -1.1015     0.7231  -1.523
>> >> >>>>>>>>> 0.1277 Syntax.Semantics D   0.1602     0.5667   0.283
>> >> >>>>>>>>> 0.7774 Syntax.Semantics E  -0.8733     0.7267  -1.202
>> >> >>>>>>>>> 0.2295 Syntax.Semantics F  -1.4438     0.8312  -1.737
>> >> >>>>>>>>> 0.0824 . Syntax.Semantics G   0.4630     0.5262   0.880
>> >> >>>>>>>>> 0.3789 --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’
>> >> >>>>>>>>> 0.05 ‘.’ 0.1 ‘ ’ 1
>> >> >>>>>>>>>
>> >> >>>>>>>>> (Dispersion parameter for binomial family taken to be 1)
>> >> >>>>>>>>>
>> >> >>>>>>>>> Null deviance: 289.63  on 384  degrees of freedom
>> >> >>>>>>>>> Residual
>> >> >>>>>>>>> deviance: 254.98  on 377  degrees of freedom AIC: 270.98
>> >> >>>>>>>>>
>> >> >>>>>>>>> Number of Fisher Scoring iterations: 17
>> >> >>>>>>>>>
>> >> >>>>>>>>> The comparison I'm interested in is between level B and
>> >> >>>>>>>>> the reference level but it cannot be estimated as shown
>> >> >>>>>>>>> by the ridiculously high estimate and SE value.
>> >> >>>>>>>>>
>> >> >>>>>>>>> Any suggestions on how to get a decent beta, SE, z, and p?
>> >> >>>>>>>>> It's the only comparison missing in the table for the
>> >> >>>>>>>>> levels I need so I think it
>> >> >>>>>>> would
>> >> >>>>>>>>> be a bit unacademic of me to close this deal saying 'the
>> >> >>>>>>>>> difference
>> >> >>>>>>> could
>> >> >>>>>>>>> not be estimated due to zero count'.
>> >> >>>>>>>>>
>> >> >>>>>>>>> And by the way I have seen this comparison being
>> >> >>>>>>>>> generated using other stats.
>> >> >>>>>>>>>
>> >> >>>>>>>>> Thanks in advance,
>> >> >>>>>>>>>
>> >> >>>>>>>>> Frank
>> >> >>>>>>>>>
>> >> >>>>>>>>> [[alternative HTML version deleted]]
>> >> >>>>>>>>>
>> >> >>>>>>>>> _______________________________________________
>> >> >>>>>>>>> R-sig-mixed-models at r-project.org mailing list
>> >> >>>>>>>>> https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
>> >> >>>>>>>> _______________________________________________
>> >> >>>>>>>> R-sig-mixed-models at r-project.org mailing list
>> >> >>>>>>>> https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
>> >> >>>>>>>
>> >> >>>>>>
>> >> >
>> >> >>>
>> >> >>>
>> >> >>>
>> >> >>> --
>> >> >>> Sent from Gmail for IPhone
>> >> >>
>> >> >
>> >> >
>> >> >
>> >>
>> >
>> >
>> >
>> > --
>> > Frank Romano Ph.D.
>> >
>> > Tel. +39 3911639149
>> >
>> > LinkedIn
>> > https://it.linkedin.com/pub/francesco-bryan-romano/33/1/162
>> >
>> > Academia.edu
>> > https://sheffield.academia.edu/FrancescoRomano
>>
>
>
>
> --
> Frank Romano Ph.D.
>
> Tel. +39 3911639149
>
> *LinkedIn*
> https://it.linkedin.com/pub/francesco-bryan-romano/33/1/162
>
> *Academia.edu*
> https://sheffield.academia.edu/FrancescoRomano
>
>         [[alternative HTML version deleted]]
>
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