Ok. Thank you.

karena

On Thu, Sep 15, 2011 at 5:33 PM, Bert Gunter [via R] <
ml-node+s789695n3816961...@n4.nabble.com> wrote:

> You should post this to the r-sig-mixed-models  list, not here.
>
> -- Bert
>
> On Thu, Sep 15, 2011 at 1:42 PM, karena <[hidden 
> email]<http://user/SendEmail.jtp?type=node&node=3816961&i=0>>
> wrote:
>
> > Hi Dear all,
> >
> > I have some gene expression data samples from different tissue types
> > -----------------------------------------------
> > - 120 samples from blood (B)
> > - 20 samples from Liver (L)
> > - 15 samples from Kidney (K)
> > - 6 samples from heart (H)
> > -----------------------------------------------
> > All the samples are from different individuals, so there are in total 161
>
> > individuals from which the DNA was collected.
> >
> > For each sample DNA, the expression level of 200 genes were obtained, so
> > the
> > whole expression data is a 161 x 200 matrix.
> >
> > The purpose of my project is to interrogate, across the 200 genes, how
> many
> > of them have an expression level consistent across all the tissue types
> > examined, how many of them have an expression level different between any
>
> > two of the tissue types, how many of them have an expression level unique
>
> > to
> > only one tissue (i.e., gene20 is expressed at a very low level (0.1) in
> > Liver, but expressed at the same level (~0.7) across all the other tissue
>
> > types).
> > So my hypotheses are:
> > H0: Mean(B)=Mean(L)=Mean(K)=Mean(H)
> > H1: Mean(B)≠Mean(L)=Mean(K)=Mean(H) or
> Mean(L)≠Mean(B)=Mean(K)=Mean(H) or
> >      Mean(K)≠Mean(B)=Mean(L)=Mean(H) or
> Mean(H)≠Mean(B)=Mean(K)=Mean(L)
> > H2: Mean(B)≠Mean(L)≠Mean(K)≠Mean(H)
> >
> > In my analysis, the gene expression level is the dependent value (Y), the
>
> > tissue-type is the fixed effects, the inter-individual variation and the
> > batch effects are the random effects. So I was suggested to use the 'lme'
>
> > function to do the analysis.
> > We want to compare the Likelihood ratio between models based on different
>
> > hypothesis. i.e., if I want to see, for one gene, if the expression
> levels
> > are all different between any two tissue types, that is to compare H2 to
> > H1,
> > if the p value corresponding to the
> > D_H2H1=likelihoodRatio(H2)-likelihoodRatio(H1) in a chi-square
> distribution
> > is less than 0.05, then we will say H2 is accepted.
> >
> > However, my problem is, I don't know how to specify these
> parameterizations
> > in 'lme' function based H0, H1 and H2, respectively. Can anyone help me
> > with
> > that?
> >
> > Thank you very much,
> >
> > Karena
> >
> >
> >
> > --
> > View this message in context:
> >
> http://r.789695.n4.nabble.com/Questions-on-lme-function-urgent-tp3816741p3816741.html
> > Sent from the R help mailing list archive at Nabble.com.
> >
> > ______________________________________________
> > [hidden email] 
> > <http://user/SendEmail.jtp?type=node&node=3816961&i=1>mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
> --
> "Men by nature long to get on to the ultimate truths, and will often be
> impatient with elementary studies or fight shy of them. If it were possible
>
> to reach the ultimate truths without the elementary studies usually
> prefixed
> to them, these would not be preparatory studies but superfluous
> diversions."
>
> -- Maimonides (1135-1204)
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
>
>         [[alternative HTML version deleted]]
>
>
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