Re: [R] random effect question and glm

2006-11-27 Thread Michael Dewey
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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Re: [R] random effect question and glm

2006-11-24 Thread Michael Dewey
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

__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


[R] random effect question and glm

2006-11-23 Thread Aimin Yan
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))

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)


thanks,

Aimin Yan

__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.