In article <asfb66$9b1$[EMAIL PROTECTED]>, Daniel Hoppe
<[EMAIL PROTECTED]> writes
>Dear all,
>
>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
>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.
>
>Could someone kindly give me a hint, how I could test for a
>poisson-distribution in this case?
>
I note that if you consider the data conditional on the total number of
events observed, what you have under the Poisson hypothesis is
distributed as if you had scattered 21 balls (events) at random amongst
200 boxes (observations).

While you might be losing information by conditioning, this allows you
to Monte-Carlo or even calculate exact probabilities quite easily. I
looked at the total number of times two balls lay in the same box by
writing a throwaway Java Monte-Carlo program and got a right tail
probability > 0.08. If you have kept the relative order of your
observations you should be able to test for trend as well - look at
SUM_i i * (# of events in observation i).

In 'The Art of Computer Programming' Volume 2, Section 3.3.2 under
'Collision Test' Knuth describes how to calculate the exact
probabilities for a 'collision statistic' which you can think of as
based either on the number of times a ball entered a box which was
already occupied, or on the number of occupied boxes at the end. 

Another approach would be to present 200 (very sparse) counts to some of
the software packages that do exact tests and ask them to test the
hypothesis that all the boxes were equally likely to receive balls. In
fact, I believe that the right tail probability of the number of pairs
of balls in the same box is > 0.0842
-- 
A. G. McDowell
.
.
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