"Rich Ulrich" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> On Mon, 2 Dec 2002 11:04:24 +0100, "Daniel Hoppe" <[EMAIL PROTECTED]> wrote:
>
> > I've a question regarding a goodness-of-fit to a Poisson distribution. I
> > have the following observations from a buying process which I expect to
be
> > Poisson.distributed:
> >
> > 0 1 2
> > 182 15 3
> >
> > The mean and my estimator for lambda in the poisson distribution
therefore
> > is 0.105 with the following expected frequencies:
> > 0 1 >1
> > 180.0649 18.90681 1.028281
> >
> > Now I would like to run a chi-squared goodness-of-fit test. But for this
> > test the expected frequencies should be >= 5, so I would need to join
> > classes "1" and ">1". If I have two classes and one estimated parameter,
the
>
> One fuller statement of the 'rule' says that you want no more
> than 25% of the Expectations less than 5, and none less than 0.
>
> What is relevant here is that we are talking about rules-of-thumb.
> The penalty for violating a rule of thumb is that your test is
> not distributed as ideally as you would like, so that (for instance)
> it might tell you p = .01 when it should be p= .04 .
>
> The function and reason for this particular rule of thumb
> is to keep you from dividing a big deviation by a tiny
> denominator and getting a huge chisquared contribution
> for that cell.
>
> For your data, you don't have a X2 that suggests non-fit.
>
> > degrees of freedom should be 2 - 1 - 1 = 0 for the test which leads me
to
> > the assumption that this is not a good idea and that I'm missing
something.
> ...
>
> Does this computed test have 1 d.f. then? Seems right....
>
> --
> Rich Ulrich, [EMAIL PROTECTED]
> http://www.pitt.edu/~wpilib/index.html
There is a standard test for a Poisson distribution , Fisher's Dispersion
test ,which no longer seems to be widely known.
My reference to it (Cochran 1954 , Biometrics !0,417) is from Volume 1 of
my old copy of Kendall and Stuart' s "Advanced Theory of Statistics, volume
1 .
Fisher's Index of Dispersion
=Sum of Squared deviations / Mean
for a Poisson random variable this has a ChiSquare distribution on n-1 df.
For your data, the index of dispersion is 24.975/0.105 = 236.14 .
This can be referred to ChiSquare tables on 199 df .
I do not have such tables but approximating with a Normal with mean 199
and variance 398 leads to a Z-score of 1.861 . This is likely to be the most
powerful test of the hypothesis as there is no loss of information from
grouping.
There is some evidence of departure from the Poisson model ,but the
question of whether the discrepancy is of practical importance is a
different question to whether it is statistically significant.
I hope this helps
Jim Snow
.
.
=================================================================
Instructions for joining and leaving this list, remarks about the
problem of INAPPROPRIATE MESSAGES, and archives are available at:
. http://jse.stat.ncsu.edu/ .
=================================================================
