[R] Test for trend?

[Andrew Perrin]

I think this depends on what you mean by "trend." What I would mean is
"effect of successive trials that is very unlikely to be spurious," which
is a good lay definition of statistical significance.

Given that these are multiple trials on the same subjects over time, it
seems like a mixed-effects model might be in order. Take a look at the
nlme pacakge, as well as Pinheiro and Bates' excellent treatment:

http://cm.bell-labs.com/cm/ms/departments/sia/project/nlme/MEMSS/index.html

...and John Fox's different (but also excellent) discussion:

http://www.socsci.mcmaster.ca/jfox/Books/Companion/appendix-mixed-models.pdf

Best,
Andy Perrin

----------------------------------------------------------------------
Andrew J Perrin - http://www.unc.edu/~aperrin
Assistant Professor of Sociology, U of North Carolina, Chapel Hill
clists at perrin.socsci.unc.edu * andrew_perrin (at) unc.edu


On Wed, 28 May 2003, Strecker, Stefan wrote:

> Hello R community,
>
> I would like to test for learning effects by subjects in my experiment. Each subject participates in six consecutive auction rounds of the same treatment.
> The response variable is the efficiency of an auction outcome measured by a real number. Since the efficiency increases over the six rounds, I suppose that subjects learn about the rules of the auction institution, but I would like to test for that conjecture.
>
> The prop.trend.test does not seem to be right, because the treatment does not change between the rounds, i.e. the number of trials (n) is not available. A linear regression shows a positive slope and the 99%-confidence interval shows a significant deviation from a zero slope, but I am not able to compute the exact p-value. The Cox-Stuart test for trend detects a trend but gives a p-value of 1.
>
> Isn't there a distribution-free, exact test for trend which operates on the rank-oder of the data instead of binary coded values?
>
> Please apologize for asking a rather R-unspecific question.
>
> Thanks in advance
> Stefan
> ---
> Stefan Strecker
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