At 06:28 26/11/2006, you wrote:
>Below is the output for p5.random.p,p5.random.p1 and m0
>My question is
>in p5.random.p, variance for P is 5e-10.
>But in p5.random.p1,variance for P is 0.039293.
>Why they are so different?
Please do as the posting guide asks and reply to the list, not just
the individual.
a - I might not know the answer
b - the discussion might help others
You give very brief details of the underlying problem so it is hard
to know what information will help you most.
Remember, if a computer estimates a non-negative quantity as very
small perhaps it is really zero.
I think you might benefit from reading Pinheiro and Bates, again see
the posting list.
>Is that wrong to write Y~P+(1|P) if I consider P as random effect?
I suppose terminology differs but I would have said the model with
Y~1+(1|P) was a random intercept model
>Also in p5.random.p and m0, it seems that there are little
>difference in estimate for P. Why?
>
>thanks,
>
>Aimin Yan
>
> > p5.random.p <-
> lmer(Y~P+(1|P),data=p5,family=binomial,control=list(usePQL=FALSE,msV=1))
> > summary(p5.random.p)
>Generalized linear mixed model fit using Laplace
>Formula: Y ~ P + (1 | P)
>Data: p5
> Family: binomial(logit link)
> AIC BIC logLik deviance
> 1423 1452 -705.4 1411
>Random effects:
> Groups NameVariance Std.Dev.
> P (Intercept) 5e-102.2361e-05
>number of obs: 1030, groups: P, 5
>
>Estimated scale (compare to 1 ) 0.938
>
>Fixed effects:
> Estimate Std. Error z value Pr(>|z|)
>(Intercept) 0.153404 0.160599 0.9552 0.3395
>P8ABP -0.422636 0.202181 -2.0904 0.0366 *
>P8adh0.009634 0.194826 0.0495 0.9606
>P9pap0.108536 0.218875 0.4959 0.6200
>P9RNT0.376122 0.271957 1.3830 0.1667
>---
>Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
>
>Correlation of Fixed Effects:
> (Intr) P8ABP P8adh P9pap
>P8ABP -0.794
>P8adh -0.824 0.655
>P9pap -0.734 0.583 0.605
>P9RNT -0.591 0.469 0.487 0.433
> > p5.random.p1 <-
> lmer(Y~1+(1|P),data=p5,family=binomial,control=list(usePQL=FALSE,msV=1))
> > summary(p5.random.p1)
>Generalized linear mixed model fit using Laplace
>Formula: Y ~ 1 + (1 | P)
>Data: p5
> Family: binomial(logit link)
> AIC BIC logLik deviance
> 1425 1435 -710.6 1421
>Random effects:
> Groups NameVariance Std.Dev.
> P (Intercept) 0.039293 0.19823
>number of obs: 1030, groups: P, 5
>
>Estimated scale (compare to 1 ) 0.9984311
>
>Fixed effects:
> Estimate Std. Error z value Pr(>|z|)
>(Intercept) 0.1362 0.1109 1.2280.219
>
> > m0<-glm(Y~P,data=p5,family=binomial(logit))
> > summary(m0)
>
>Call:
>glm(formula = Y ~ P, family = binomial(logit), data = p5)
>
>Deviance Residuals:
> Min 1Q Median 3Q Max
>-1.4086 -1.2476 0.9626 1.1088 1.2933
>
>Coefficients:
> Estimate Std. Error z value Pr(>|z|)
>(Intercept) 0.154151 0.160604 0.960 0.3371
>P8ABP -0.422415 0.202180 -2.089 0.0367 *
>P8adh0.009355 0.194831 0.048 0.9617
>P9pap0.108214 0.218881 0.494 0.6210
>P9RNT0.374693 0.271945 1.378 0.1683
>---
>Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
>
>(Dispersion parameter for binomial family taken to be 1)
>
> Null deviance: 1425.5 on 1029 degrees of freedom
>Residual deviance: 1410.8 on 1025 degrees of freedom
>AIC: 1420.8
>
>Number of Fisher Scoring iterations: 4
>
>
>At 06:13 AM 11/24/2006, you wrote:
>>At 21:52 23/11/2006, Aimin Yan wrote:
>>>consider p as random effect with 5 levels, what is difference between these
>>>two models?
>>>
>>> > p5.random.p <- lmer(Y
>>>~p+(1|p),data=p5,family=binomial,control=list(usePQL=FALSE,msV=1))
>>> > p5.random.p1 <- lmer(Y
>>>~1+(1|p),data=p5,family=binomial,control=list(usePQL=FALSE,msV=1))
>>
>>Well, try them and see. Then if you cannot understand the output tell us
>>a) what you found
>>b) how it differed from what you expected
>>
>>>in addtion, I try these two models, it seems they are same.
>>>what is the difference between these two model. Is this a cell means model?
>>>m00 <- glm(sc ~aa-1,data = p5)
>>>m000 <- glm(sc ~1+aa-1,data = p5)
>>
>>See above
>>
>>
>>>thanks,
>>>
>>>Aimin Yan
>>
>>Michael Dewey
>>http://www.aghmed.fsnet.co.uk
>
>
>
>
>
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Michael Dewey
http://www.aghmed.fsnet.co.uk
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