Date: Fri, 20 Feb 1998 09:01:41 +0000
Reply-To: Bruce Weaver <pss091@BANGOR.AC.UK>
Sender: "SPSSX(r) Discussion" <SPSSX-L@UGA.CC.UGA.EDU>
From: Bruce Weaver <pss091@BANGOR.AC.UK>
Organization: University of Wales, Bangor.
Subject: Re: Nonparametric test for mixed model
In-Reply-To: <34ED3BCD.7BB5B341@psy.cuhk.edu.hk>
Content-Type: TEXT/PLAIN; charset=US-ASCII
On Fri, 20 Feb 1998, Helen wrote:
> Could anyone tell me what kind of test I should be use to test a mixed model?
I've
> got 1 between-subject and 1 within-subject effects. And although my
dependent
> variable is continuous, it is highly non-normal. I know how to test each
single
> effect. But don't know about their interaction effect.
>
> Any help would be much appreciate.
>
> Helen
--------------- >8 ----------------
I have a copy of an unpublished manuscript by G. Rolfe Morrison from the
Pscyhology Department at McMaster University (in Hamilton, Ontario) that
adddresses this issue. The paper is called "Rank order alternatives to
the two way analysis of variance". I'm sure he would be happy to send you
a copy. (Maybe if enough people ask him to send copies, I'll finally be
able to convince him to submit it for publication!) His e-mail address is
<morrison@mcmail.mcmaster.ca>. The snail-mail address is:
Psychology Department
McMaster University
Hamilton, Ontario
CANADA L8S 4K1
For a problem like yours, he uses the following example: Two independent
groups of subjects are given the same picture sorting task. Each subject
in Group A is shown 4 photos of "physicians" and asked to rank them from
1 to 4 in terms of trustworthiness. Subjects in Group B are shown the
same photos, but are told they are psychiatrists.
If you treat each group separately, you can use Friedman ANOVA. The test
statistic is distributed (at least approximately) as chi-square with k-1
degrees of freedom (i.e., df = 3 in this case). Because the groups are
independent, you have 2 independent chi-squares, each with df = 3. The
sum of two independent chi-squares is itself a chi-square with df = to
the sum of the df for the components. In other words, if you add
together the two Friedman test statistics, the sum is approximately
distributed as chi-square with df = 2(k-1) = 6 for this example.
It might help if I give some numbers to make the example more concrete.
For Morrison's example:
Friedman statistic for Group A = 12.0 (df = 3)
Frideman statistic for Group B = 3.9 (df = 3)
Total = 15.9 (df = 6)
The next step is to calculate the Friedman statistic on ALL THE DATA,
as if there was only one group of subjects. For Morrison example:
Friedman statistic for ALL data = 2.55 (df = 3)
This statistic provides a test of the MAIN EFFECT of the variable with
repeated measures. For Morrison's example, it is not significant.
The final step is a test of the AxB "order" interaction. The statistic
for this test is obtained by subtraction as follows:
AB statistic = Total - Main Effect
= 15.9 - 2.55 = 13.35
df = 6 - 3 = 3
This statistic is distributed (approximately) as chi-square with df = 3.
For Morrison's example, the AB order interaction is significant. This
means that the two groups are not ranking the photos the same way.
One important thing to note is that this procedure allows you to detect
"order" interactions only (i.e., where the B treatments are ordered
differently at the levels of A). It cannot detect an interaction where
the effects are going in the same direction, but the size of the effect
is larger in one case than the other.
Morrison also discusses a test for the case where you have two variables
that are manipulated between groups. His methods don't work if both
variables are manipulated within-subjects though, because the chi-square
values are not mutually independent in that case.
Hope this helps.
--
Bruce Weaver
UWB, Psychology
pss091@bangor.ac.uk
http://www.bangor.ac.uk/~pss091
