You've got a tricky issue there.
First, I'd treat the data as continuous, and do a two way anova.
You can't do a chi-square test, because you are treating the data as
categorical, and therefore losing power.
You could do two Kruskal Wallis tests, but that won't tell you about
the interaction.
When you have an ordinal outcome variable, and want to get beyond
Mann-Whitney and Kruskal Wallis tests, the appropriate thing to do is
ordinal logistic regression. (Or ordinal probit.)
It's possible to do a 2-way anova with regression - they are the same
test, thought about in different ways. This is covered in most books
on regression (including mine :] ).
You combine these two things - do a 2-way anova as regression, in an
ordinal logistic regression sort of way. (Ordinal logistic regression
is getting a bit clever, and isn't covered in my book on regression.
My favourite book on this sort of thing is by J Scott Long, and is
called Regression with categorical and limited dependent variables. )
However, as you have empty cells, you might struggle with this (or the
program might struggle, and say you can't do it.)
It's the sort of thing that makes a nice example of how you need to
think about how you will analyse your data before you collect them.
It's surprisingly easy to design a study in which the data are hard
(as is this) or impossible to analyse,
Jeremy
On 04/04/07, David Hambrook <[log in to unmask]> wrote:
>
>
> Hello all,
> Again I feel very sheepish having to ask you what is probably a very basic
> question. I thought my stats knowledge was pretty good but perhaps not!
> Anyway, I need help deciding on the type of test I need to run in order to
> answer a particular question. My situation:
> I have a 2X3 design. I want to compare two independent groups of
> participants on an ordinal level variable with four categories (low,
> average, above average, and high - coded as 1,2,3,4 respectively).
>
> I thought that the appropriate test to use would be a chi-square test of
> independence. However, Pallant (2005) suggests that the lowest expected
> frequency in any cell should be 5 or more. After running a chi-square test,
> the SPSS output indicates that "3 cells (37.5%) have an expected count of
> less than 5". I'm not entirely sure what this means, and whether or not I
> have chosen the wrong test to start with.
>
> Can anybody offer some (easy to understand) advice?
>
> Many thanks in advance,
> Dave.
>
>
> David Hambrook
>
>
>
> MSc Student
>
> Mental Health Studies Programme,
>
> Institute of Psychiatry,
>
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>
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>
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>
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> ________________________________
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--
Jeremy Miles
Learning statistics blog: www.jeremymiles.co.uk/learningstats
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