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Re: st: Friedman tests

From   n j cox <[email protected]>
To   [email protected]
Subject   Re: st: Friedman tests
Date   Wed, 23 Mar 2005 15:14:01 +0000

Various replies gave strategic advice of the form:
don't do that, do this instead.

More concretely, the one Stata implementation
of Friedman tests does not support interactions,
and indeed I find it difficult to make sense
of how such procedures, whatever their name, could
provide a framework for handling interactions.
The main idea of interactions would seem to
be that effects can be multiplicative as
well as additive, on the scale being used,
but if you throw the scale away by ranking,
I think you discard that idea too. If you
believe that ranks provide a suitable metric,
then -anova- on the ranks might be one way to
proceed; my guess is that's unlikely to be the
correct choice.

[email protected]

Ashley Harris

What I have nonparametric data for 3 groups, 1 dependent variable (ratio)
and 2 independent variables (technique and analysis type). The questions I
want to test are: does the technique matter with respect to the ratio
result?, does the analysis type matter with respect to the ratio result? is
there a difference between the groups? If so what are these differences?
However, there is the possibility of interactions, so I should test for that
as well.

I think I want to use a Friedman test, but cannot seem to figure out how to
deal with interactions, and even more simplistically, I cannot seem to
replicate the results for the sample that is in the help file (pasted
below). Has any one had these problems too and is there a solution?

                          S t u d e n t
 T E S T      1    2    3    4    5    6    7    8
 1            90   60   45   48   58   72   25   85
 2            62   81   92   76   70   75   95   72
 3            60   91   85   81   90   76   93   80

If you enter this as three variables, then you can just type in
    friedman varlist
and all is fine.

The results from the above data are:
         Friedman =   2.8889
         Kendall =   0.1376
         p-value =   0.8951

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