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Sundar
The tscore problem is not related to glm not reporting p-values. There the problem is not knowing what finite-sample distribution is and not knowing how fast the unknown distribution converges to the normal.
In the tscore problem, the normal distribution would also be inappropriate. The actual but anonymous distribution for a tscore is given by Chen et al. It's not a t-distribution.
Joe
-----Original Message-----
From: Sundar Dorai-Raj [mailto:sundar.dorai-raj@PDF.COM]
Sent: Wednesday, December 10, 2003 3:46 PM
To: Lucke, Joseph F
Subject: Re: [S] pvalues from t-distribution
Joe,
Thanks for the link. In that case, wouldn't the normal distribution (and
thus the corresponding p-values) also be inappropriate? This is the
reason (I believe) that p-values aren't given in the summary table with
glm. However, t-scores are provided.
Personally, I'm not a big fan of p-values because of all the baggage
they carry.
Sundar
Lucke, Joseph F wrote:
> Using the t-distribution to 'standardize' scores to a 't-score" is a
> common practice in educational and psychological statistics. However,
> the t-score does not have a t-distribution. The discrepancy is large
> for large 't-scores' and small samples.
>
> Chen, G., & Adatia, A. (1997). Independence and $t$ distribution. The
> American Statistician, 51(2), 176--177.
>
> JSTOR Stable URL:
> http://links.jstor.org/sici?sici=0003-1305%28199705%2951%3A2%3C176%3AIATD%3E2.0.CO%3B2-X
>
>
> Joe
>
> -----Original Message-----
> From: Sundar Dorai-Raj [mailto:sundar.dorai-raj@PDF.COM]
> Sent: Wednesday, December 10, 2003 2:55 PM
> To: Sliwerska, Ela
> Cc: 's-news@lists.biostat.wustl.edu'
> Subject: Re: [S] pvalues from t-distribution
>
>
> Ela,
>
> Sliwerska, Ela wrote:
>
> > Hi all!
> > I am trying to convert my tscore column into pvalues. I use >
> 2*(1-pnorm(abs(tscore))) command from command window but it works for
> > particular value. I would like to transfer my column all at once.
> Can > someone help? Thanks! Ela >
>
> (Please use an informative subject line.)
>
> If "tscore" is part of a data.frame (say, "DF") then you should do the
> following:
>
> 2 * (1 - pnorm(abs(DF$tscore)))
>
> But why not use a t-distribution rather than a Z-distribution? What
> are the distributional assumptions of "tscore"? If you know the
> degrees of freedom then using the following may be more appropriate:
>
> 2 * (1 - pt(abs(DF$tscore), degrees.of.freedom))
>
> Hope this helps,
>
> Sundar
>
>
>
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