Conference rusure::math

>	the relevant statistical question(s)

What is a/the relevant probability distribution?

What is the relative probability of 2 incidences vs 6 incidences?

>	the statistically valid answer(s)

The Poisson distribution, which is defined over the integers r = 0,1,2... and 
takes the form 
	 	  r     -la
      P(X=r) = 	la /r! e

where la (for lambda) is an appropriate parameter. 

For a peak at r=2, la (according to my tables) is about 2.5, and 
P(X=2) ~ .25 while P(X=6) ~ .027, so the relative probability is ~9:1. 
The result is extremely sensitive to changes in la, which my tables only
show at .5 intervals, so use a large grain of salt.

    Re:     <<< Note 1481.1 by CIVAGE::LYNN "Lynn Yarbrough @WNP DTN 427-5663" >>>
                            -< Poisson looks right >-

>	the relevant statistical question(s)

What is a/the relevant probability distribution?
> The Poisson distribution, 
    
    Good choice.  Some folks might start with a binomial distribution,
    figuring there are 5000 person-years sampled here, and 6 showed brain
    cancer, but once you try the computation, you would need to approximate
    the binomial with a Poisson anyway.
    
>For a peak at r=2, la (according to my tables) is about 2.5, and 
    
    Good interpolation.  My tables (la in steps of 0.1) show that although
    2 is a max between la=2 and la=3, The peak is roughly symmetric around
    2 when la=2.45 approx.  By the usual convention "2 is expected" implies
    that the mean is 2, not that the mode is 2, so la=2.0 might be a better
    answer.  Nevertheless, I will choose la=2.5, for a reason which I will
    keep secret.

    

>What is the relative probability of 2 incidences vs 6 incidences?

>P(X=2) ~ .25 while P(X=6) ~ .027, so the relative probability is ~9:1. 
>The result is extremely sensitive to changes in la, which my tables only
>show at .5 intervals, so use a large grain of salt.
    
    This is the right answer, but I think it is the wrong question?
    
    Here is an answer to an alternative question.  The probability of 6 or
    more cases of brain cancer is 0.0419, and the probability of fewer than
    6 cases is 0.9581.
    
    
    Why would most statisticians prefer this latter question/answer pair?
    
    What other question does it imply?
    
    What does the number 0.0419 suggest about my secret reason for choosing
    to work with la=2.5?
    
    What is implied by the word 'relevant' in .0?

>    This is the right answer, but I think it is the wrong question?

Hmm. I thought it was pretty good. :-)
    
>    What does the number 0.0419 suggest about my secret reason for choosing
>    to work with la=2.5?

No clue.
    
>    What is implied by the word 'relevant' in .0?

Perhaps what you are driving at is the fact that population size is NOT 
relevant. You get the same numbers for population of 50 as for 5000.

    1) How did this come to light?
    
    Since it was based on a "news report", we know the criteria that
    selected it for reporting, ie, "man bites dog is news".
    
    Relevant question is ..
    
    How often will some (any) life-threatening disease be found in a
    space-time cluster more frequently than expected by simple
    epidemiological ratios?
    
    Answer, ..
    
    beats me!
    
    Dick

.4 is closing in on the answer from the opposite direction, but I will 
continue to follow the direction Lynn started us on.

Here is the relevant question, as many statisticians would define it.


Define two hypotheses:

	H0 (the null hypothesis) - the process generating brain cancer deaths 
	in this building is no different from that in the country as a whole

	H1 (the alternate hypothesis) - the process generating brain cancer 
	deaths in this building is different from that in the country as a whole

One question is, what is the probability that 6 or more cases would be seen,
assuming the null hypothesis?

The answer is 0.0419.  This is less than 5%, which is a common threshold for
problems of this kind.  That is why I chose to stick with la=2.5, because it
gave a convenient number for illustration.

Another question is, what number of cases corresponds to a given threshold?

Using the tables again (or integrating the Poisson function), we can show

	cases	threshold

	6	5%
	8	0.5%
	9	0.1%


What is the right threshold to use?  This is an embarrassing question for
statisticians who use the method above.

.4>    How often will some (any) life-threatening disease be found in a
>    space-time cluster more frequently than expected by simple
>   epidemiological ratios?

As stated, this question has an easy rough answer: about half the time.  Half
of all samples should be above average, half should be below.  There is a 
complication that we are dealing with a skewed distribution, but we can 
ignore that.

The question should be more properly stated: how often will we see a sample
that exceeds the threshold we have set?  The answer is equal to the threshold.
If our threshold is 5% then in a sample of random diseases and random 
populations, we should see about one signal in twenty, assuming the null 
hypothesis.

In other words, the data in .0 could well be due to pure chance.

We have not exhausted the statistical possibilities of .0, but perhaps we have
exhausted ourselves.

If not, consider the following:

	What is the right threshold?
	Why do we care?
	How many disease-population pairs are there in the US?
	Where did that la=2.5 come from, and is it relevant to this problem?
	How else could we approach the problem?

>	Where did that la=2.5 come from, and is it relevant to this problem?

This comes directly from the statement of the problem and the decision to 
model with a Poisson distribution: of all such distributions, the one with
its peak at 2 incidences has la=2.5. Different lambda, different
expectation. Given the model, it is not merely relevant, it's determined.

RE: .6 (Lynn)

    Setting the mode ("peak") to 2 is not exactly wrong, but it unorthodox.

    You have interpretted the phrase "the expected number ... is just two
    cases" as meaning "the most likely number of cases is two".  Generally
    the phrase would be interpretted to mean that the average or mean
    number of cases is 2.  In fact, in statistics the term "expectation" is
    basically equivalent to "mean" (there is some complication in usage --
    times when a statistician would use one and not the other, but they are
    mathematically equivalent).

    The lambda of a Poisson distribution with a mean of two is two.  In
    general the mean of a Poisson distribution is equal to lambda.

    For those who are lost --

    The binomial distribution answers the question: if I look at N things
    (in this case, N = "about" 500 workers) selected from a much larger
    population (in this case, something like all similar workers in the
    country) some proportion, p, of whom have some characteristic (the
    characteristic in this case being, will get a brain tumor in a given
    ten year period, and p is apparently 2/500 = 1/250 = .004), how likely
    is it that I will find that at least some particular number, r, of the N
    things looked at will have the characteristic (in this case r is 6).

    The Poisson distribution answers a somewhat different question.  If
    something (e.g., some member of a group of workers gets a brain tumor)
    is equally likely to happen at anytime, and it happening at some
    particular time does not effect the liklihood of it happening at any
    other time (technically violated in this case since you have removed
    someone, at least for a time, from the population and thus decreased
    the probability that it will happen again immediately -- however,
    the change in the population size is small and the person is likely
    to be replaced fairly soon, so it is not unreasonable to "pretend" it
    applies), and you know how many times it will happen on the average
    (lambda, either 2 or 2.5 here) in a given period of time, how likely
    is it that it will occur at least some particular number of times (6)
    in the same period.

    It seems that both distributions are applicable so which should be
    used?  It doesn't really matter since both will give essentially the
    same answer.  If you take the binomial distribution and let N increase
    to infinity and let p shrink at the same time so that p*N (= the
    average or expected number of things you will find with the given
    characteristic) stays the same, then the binomial distribution becomes
    the Poisson distribution with lambda = p*N.  Since the binomial
    distribution gets hard to calculate directly for large N and small
    p, in such cases the Poisson distribution is used to approximate the
    binomial.

					Topher

REL .5 (Wally)

>What is the right threshold to use?  This is an embarrassing question for
>statisticians who use the method above.

    Despite constant claims by some militant Bayesians, traditional
    statisticians do not find this question the least bit embarrasing.  The
    question is no more embarrasing for a traditional statistician than the
    question of the choice of priors for a Bayesian.  Even most Bayesians
    (like myself) will say that given the assumptions of traditional
    statistics, the function of the threshold (signifcance criteria) is
    well dealt with.

    That doesn't mean that this isn't an embarrasing question for some --
    its just that the questions are not directed at the main body of
    tradional statisticians.  Specifically, the question is embarrasing for
    the scientist or engineer (or statistician) who does not think about
    their choice, but just blindly selects "the standard".  It is also
    embarrasing for those few militant traditionalists who attempt to argue
    the "moral" superiority of traditional over Bayesian methods because
    they are completely "objective" (most traditionalists, even most
    militant traditionalists, however, do not make such silly claims -- the
    argument is over where subjectivity should enter statistics, not
    whether or not it could/should be expunged entirely).

					Topher

    I think I read Wally's question in .5 a bit differently than Topher.
    First, here's my cut at the "right question":
    
    >	What are the chances of *ME* being affected by all this?
    
    Anyhow, I wonder if we should bother with a 5%, or even 1% figure here.
    If something is "unusual" at the 5% level of signifigance that hardly
    qualifies it for a slot on national news.  Even at the 1% level there
    must be thousands of false positives that make the news.  These false
    positives can lead to huge expenditures of money to correct nonexistent
    problems and pass highly restrictive, expensive legislation that erodes
    our national competitiveness.
    
    Which is not to say we should ignore all such situations, but rather
    recognize the certainty that some "unusually high" sickness/mortality
    rates will crop up for no reason other than pure chance.  Not easy when
    our innumerate society relies on witch hunts and magic bullets to address
    today's problems.
    
    If you can't select a good level of signifigance then maybe you should 
    be asking a different question.  See above :-)
    
      John

RE: .9 (John)

    Wally says that you should be embarrassed, John, since statisticians
    who respond to that question with essentially the same answer as you do
    are, according to him, embarrassed.  Traditional statisticians say
    that the correct threshhold to use depends on what is going to be done
    with the results, with the relative costs of a type I vs a type II
    error, and even on the prior plausibility (N.B.: *not* probability)
    of the null vs the alternate hypothesis.  Modern usage encourages the
    treating "p-values" as the results of a study rather than a yes/no
    result gotten by seeing if the p-value exceeds some particular
    threshhold.  This allows each user of the results (e.g., reader of a
    scientific paper) to choose their own thresshold -- or different
    thressholds under differing conditions.

    The real problem in this case is not the appropriateness of any
    particular value of the threshhold.  The question is the meaningfulness
    of the statistical procedure used in the evaluation.  *If* the
    situation were one for which this technique was appropriate then a .01
    or even a .05 thresshold would not be too far out of line.  The problem
    is that this is extremely unlikely to be he case here.

    This type of analysis would be appropriate if:

	1: This were an isolated incident rather than part of a series of
	   investigations.

	2: The choice of the study site was made independent of the known
	   number of cases at that site.

    What is really going on is that the health officials are doing a fairly
    reasonable procedure for screening "cases" to be looked at further, for
    more substantial evidence of a problem.  They look at those cases in
    detail which would have been considered as showing a significant
    deviation if they had been selected in isolation and "randomly" (at
    least, w.r.t the number of cases).  This is a reasonable screening
    procedure -- somewhat arbitrary, but less arbitrary than almost any
    alternative.

    Its the media which have been premature about taking these results
    as meaningful -- and the problem really isn't the "significance level"
    as such.

				    Topher

    I agree with Topher mostly, but I would point out that so much of the
    "tail probability" calculation is really rough and ready approximation
    that you can't take the numbers too seriously.  Empirical distributions
    differ most from normality in the tails and the tail areas are what the
    tests of significance and confidence levels are based on.  I think we
    are generally pushing the statistical technology to its limits here. 
    We need models that take some other effects into account or demonstrate
    the robustness of the methods to unknown parameters.
    
    Bayesian methods are the most satisfying esthetically, but we have to
    quantify some opinions we normally don't pay much attention to.  As in
    everything else, there is no free lunch.  You must think harder and
    calculate longer to use the natural Bayesian methods.
    
    Yours for a free lunch,
    
    Dick