Re: pvalues from t-distribution

To:	"'s-news@lists.biostat.wustl.edu'" <s-news@lists.biostat.wustl.edu>
Subject:	Re: pvalues from t-distribution
From:	"Lucke, Joseph F" <LUCKE@uthscsa.edu>
Date:	Wed, 10 Dec 2003 15:43:10 -0600

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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