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Hello Allstat,
I had many replies to this query - a selection is listed below. For those
interested, the proposed study which motivated the question would involve
observing no. of road accidents in London on an "exposed" tube strike day as
compared with a "paired & unexposed" day (ie one week before or after or
both). Data is available for 4 tube strike days (and correspondingly for
paired days). The hypothesis, in case you haven't guessed, is that more
accidents occur on days of tube strikes.
I'm not requesting a reply here, but thought it would be of academic
interest to some of those who replied. Of course, if you feel that the
original approach was fundamentally flawed, then please feel free to point
out!
Many thanks to everyone who replied,
Neil Walker
ORIGINAL QUERY:
Hello Allstat,
An apparently simple question. Does anyone know of a parallel to the
paired t-test where the data is in the form of counts and I assume Poissonly
distributed? I've found no direct reference from my limited library or from
googling. Any comments would be much appreciated.
Thanks in advance,
Neil Walker
Westminster PCT
------------------------
That is a generalized linear model with a Poisson distribution and log link
function. This could be done in SAS for instance with proc genmod.
Paul R. Swank, Ph.D.
Professor, Developmental Pediatrics
Medical School
UT Health Science Center at Houston
------------------------
Sounds to me like you are interested in a test involving the difference of
two Poisson random variables. At least, that would be the closest parallel
to the paired t-test with normal variates: you form the difference within
each pair, and then work with the distribution of those differences.
Johnson, Kotz, and Kemp (1993) Univariate Discrete Distributions, talk about
such a distribution (difference of 2 Poisson r.v.'s) on pp 191 - 192. I
haven't gone through the pf in detail, but it looks reasonably managable.
They do say "the approach to normality is rapid", for what that's worth.
(What does "rapid" mean?) So, you MIGHT do OK using the standard paired
t-test.
Alternatively, you could use a generalized linear mixed model (GLMM).
Cheers,
Scott
===================================
Scott D Chasalow
Monsanto Company
Email: [log in to unmask]
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At 12:45 26/09/02 -0500, you wrote:
>That is a generalized linear model with a Poisson distribution and log link
>function.
No it isn't. A standard GLM doesn't account for the within-pair
correlation. A GLM approach would have to be implemented via GEEs to
account for this.
Regards,
Richard
=====================================================================
Richard E. Chandler
^^^^^^^^^^^^^^^^^^^
Room 145, Dept of Statistical Science, University College London,
1-19 Torrington Place, London WC1E 6BT, UK
Tel: +44 (0)20 7679 1880 Fax: +44 (0)20 7383 4703
Internet: http://www.ucl.ac.uk/Stats (department)
http://www.homepages.ucl.ac.uk/~ucakarc (personal)
email: [log in to unmask]
------------------------
Niel,
Conditional Poisson Regression, which conditions on the total number of
events for each stratum, might be something worth considering. For paired
data, each
pair would be a stratum. It is not strictly analagous to the paired t-test
in
that the paired t-test is equivalent to a random stratum (pair) effect,
whereas
Conditional Poisson Regression would be interpretted as a fixed stratum
effect.
But its better than nothing as a way of taking pair (stratum) effects into
account in the model.
Andrew Dunning
Andrew J. Dunning
Senior Statistician
Wyeth Research
(845) 602 4313
------------------------
Yes. Given two poisson counts x1, x2 with expected values mu1 and mu2,
x1 is Binomial n=x1+x2 and p = mu1/(mu1+mu2).
mu1 = mu2 only if p = 1/2.
Use a standard binomial test for x1 from, with n= x1+x2, p = 1/2.
------------------------
The basic idea is that for each pair (x,y), if the null hypothesis being
tested is that the mean difference is zero (ratio of means equals 1), then
x be treated as binomial with "n" equal to x+y, and null "p" equal to 1/2.
So the full analysis would be that for a series of binomial responses.
If the null ratio of means is some constant other than 1, say k, then the
binomial null "p" would be mean(x) divided by mean(x)+mean(y).
As a previous e-mail said, this can be done "blindly" as a loglinear model
in SAS or S-Plus or R.
DVH, UCSB
------------------------
I think this is standard - just condition on the total counts (n_i) within
each pair and you have conditionally independent Binomial(n_i,0.5) r.v.'s.
Alternatively use a simple chi-squared test - I'll leave you to think of the
details!
Jim.
------------------------
The two-sample Poisson test is done via the binomial.
If X is Poisson (mean mu_1) and, independently, Y is Poisson (mean mu_2),
then X+Y is also Poisson. Given X+Y = n, say, X is binomial with
index n and parameter (probability) mu_1 / (mu_1 + mu_2). So a
test of mu_1 = mu_2 is equivalent to testing a binomial for p = 1/2.
So if x=5 and y=12, you see if 5 is far into the tail of B(17,0.5).
This extends in a natural way to random samples rather than single
observations X and Y, since the totals will still be Poisson. Also
you can test hypotheses such as mu_1 = 2 mu_2 by testing p=1/3.
Let me know if the above is too cryptic, and I can try to
expand.
Eryl Bassett
------------------------------+------------------------------------------
E. Eryl Bassett, | Web: http://www.ukc.ac.uk/ims/statistics/
Institute of Mathematics | e-mail: [log in to unmask]
and Statistics, | phone: +44 (0) 1227 764000 ext 3798
Univ. of Kent at Canterbury, | +44 (0) 1227 823798 direct
Canterbury, Kent, UK, CT2 7NF | fax: +44 (0) 1227 827932
------------------------
hi Neil,
have a look in Instat [freeware] at
http://www.rdg.ac.uk/ssc/instat/instat.html
open any worksheet from the library and under simple models u find 1 & 2
sample Poisson.
the 1-sample has exact CI, formulae from Armitage & Berry 1992.
The 2-sample has a normal approx CI only, exact CI for a difference between
I have never seen.
Alessandro Leidi
Statistician, Statistical Services Centre
------------------------
McNemar's test could be what you are looking for.
Covered in most general stats texts.
------------------------
Hi Neil
You could try the useful result (See Cox, Analysis of Binary Data)
that if n1 and n2 have Poisson distributions n1/(n1+n2) is Binomial.
So tp test is E(n1)=E(n2) you simply test if n1/(n1+n2) could come
from a Binomial distribution with expectation 0.5.
Mike
Professor Mike Campbell
Institute of Primary Care and General Practice
University of Sheffield
Community Sciences Centre
Northern General Hospital
Sheffield S5 7AU
Tel 0114 271 5919
Fax 0114 242 2136
e-mail [log in to unmask]
http://www.shef.ac.uk/michaelcampbell/
http://www.shef.ac.uk/medical-statistics/
------------------------
For each pair, condition on the sum, so you deal with
binomial with appropriate parameters. The binomial
data can be analysed using logistic regression.
The test of equal mean reduces to testing whether
the constant term is zero. (See Exercise 6.13 in Pawitan,
In All Likelihood.)
------------------------
Dear Neil
You could use the non-parametric equivalent of the paired t-test: the
Wilcoxon
signed ranks test. Note that this tests the null hypothesis of equal
distributions between the two groups. Hence p<0.05 may result if the mean
counts
are equal but there is a marked difference in the tails of the two groups of
observed data.
The suggested Poisson regression would allow more complexity in the analysis
(you could adjust for confounders for example) but you would need to find
some
way of incorporating the paired structure of the data into the analysis. I
don't
know SAS, but the easiest (and not necessarily the best) way to do this in
S-Plus and Stata is to use the "cluster(id)" option.
Finally, is using the paired t-test out of the question? This test is based
on
the differences between the sets of paired measurements, which you will find
(due to central limit theorem) are more normally distributed than the
individual
measurements. Of course, *more* normally distributed may still be nowhere
near
normally distributed.
With best wishes, Chris
__________________________
Chris Metcalfe
Research Associate
MRC Biostatistics Unit
Institute of Public Health
University Forvie Site
Robinson Way
Cambridge
CB2 2SR
e.mail : [log in to unmask]
telephone : 01223 330 394
fax : 01223 330 388
------------------------
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