Hello Peter,
Thank you very much for the fresh approach!
I will go with it to the researcher.
Thanks again,
Tal
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On Thu, May 6, 2010 at 4:47 PM, peter dalgaard <pda...@gmail.com> wrote:
>
> On May 6, 2010, at 1:42 PM, Tal Galili wrote:
>
> > Hi Joris,
> > Thank you for taking the time to answer.
> >
> > This data is of a test done for 39 subjects (from 2 groups) over 12
> weeks.
> > And the questions I would like to answer are:
> > 1) Did the test results changed over time?
> > 2) Did the group effected the test results?
> > 3) Did the effect of time differ for each group?
> >
> > I understand that the general limitation of using repeated measures anova
> > here is (obviously) that even if one get's a significant "effect" of
> time,
> > the analysis doesn't give any clue as to how time influences the test
> (the
> > same goes for the interaction term).
> > But a more appropriate tool would probably be some sort of GAM lm, which
> is
> > based on models I don't have much understanding of (yet).
> >
> > I am using this test since the researcher for whom I am doing the
> analysis
> > asked me to use it (since this is what was done in the previous work on
> > similar data, done by someone else).
> >
> > Due to the current stage of my ignorance, and the researchers tendency
> > towards this analysis - I am not sure how to proceed.
> >
>
> You may be able to get through with anova.mlm (little-a anova) and
> sphericity assumptions. However, I wouldn't trust the results.
>
> These data are nowhere near normally distributed, and with the size of the
> data set and the pattern of many series of straight 4s, I don't think anyone
> has a chance of figuring out how this affects the p-values.
>
> I'd rather do something like this (with the original "dat", before
> jittering):
>
> First look at the average patterns per group:
>
> > aggregate(dat[-1],dat[1],mean)
> DC week6 week7 week8 week9 week10 week11
> 1 control 4 4 4.000000 3.900000 3.900000 3.900000
> 2 head (20g) 4 4 3.894737 3.789474 3.736842 3.736842
> week12 week13 week14 week15 week16 week17
> 1 3.900000 3.900000 3.900000 3.850000 3.850000 3.750000
> 2 3.736842 3.684211 3.526316 3.421053 3.368421 3.315789
> > matplot(t(aggregate(dat[-1],dat[1],mean)[-1]))
>
> which looks promising and roughly linear. However, the slopes might differ
> between subjects and this would be the appropriate variation to gauge the
> mean slope differences against. So let's compute the individual slopes:
>
> > slope <- apply(dat[-1],1,function(x)coef(lm(x~I(1:12)))[2])
>
> We can compare these between the groups with a t test:
>
> > t.test(slope~dat$DC)
>
> Welch Two Sample t-test
>
> data: slope by dat$DC
> t = 1.6138, df = 27.189, p-value = 0.1181
> alternative hypothesis: true difference in means is not equal to 0
> 95 percent confidence interval:
> -0.01217805 0.10203819
> sample estimates:
> mean in group control mean in group head (20g)
> -0.01800699 -0.06293706
>
> However, looking more carefully at the data, we realize that many slopes
> are exactly zero, so a nonparametric test might be in order. It doesn't
> change anything, though:
>
> > wilcox.test(slope~dat$DC)
>
> Wilcoxon rank sum test with continuity correction
>
> data: slope by dat$DC
> W = 232, p-value = 0.09845
> alternative hypothesis: true location shift is not equal to 0
>
> Warning message:
> In wilcox.test.default(x = c(-2.36672330631823e-16, -2.36672330631823e-16,
> :
> cannot compute exact p-value with ties
> --
> Peter Dalgaard
> Center for Statistics, Copenhagen Business School
> Solbjerg Plads 3, 2000 Frederiksberg, Denmark
> Phone: (+45)38153501
> Email: pd....@cbs.dk Priv: pda...@gmail.com
>
>
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