For the moment ignore the fact that tests within workers are not indepedent
(if I'm high for lead I'm more likely to be high for mercury) and that tests
across workers are not independent (if I'm exposed to zinc, the guy next to me
probably is as well). There are 78 workers and 8 metals for 624 total tests.
The crudest method you could use to get an 'overall' test of exposure would
be to treat the tests as all independent with probability of success 0.05
and count a person's test as a 'success' if they fell in the top 5% of the
reference 'standard population' for a given metal. Under these assumptions
the total number of workers in the top 5% over all tests would be distributed
bin(624, 0.05), so you could get a single p-value from that.
For example, 20 workers total in the top 5% over all tests would be p = 0.98, but
>= 40 positive worker-tests would be p < 0.05; >= 60 would be p < 0.0000007, and
>= 100 would be p < 5 x10^-25.
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John Dawson
Postdoctoral Scholar
University of Alabama at Birmingham
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