Accidents before, by year:
  5  11   6  13  15  11  14   9  13  12   7  11  16  11  15  11   6 *19*  6   8
         22  30  34  39  40  34  36  34  32  30  34  38 *42* 37  32  36  31  33   3 year total

Accidents after, by year:
11  15  14  20  24  21  14  15  15  12  14   8  18  12  18  18  20  13  12  13
         40  49  58  65  59  50  44  42  41  34  40  38  48  48  56  51  45  38   3 year total

Suppose that some work is done following the year in
which 19 accidents occured. What is the probability that
the following year will see fewer than 19 accidents?

(Code: http://ex-parrot.com/~chris/tmp/poisson)

  n P(n, before)       P(n, after)
             cumul.             cumul.
-- -------- --------  -------- --------
15 0.034718 0.951260  0.102436 0.568090
16 0.021699 0.972958  0.096034 0.664123
17 0.012764 0.985722  0.084736 0.748859
18 0.007091 0.992813  0.070613 0.819472 <----
19 0.003732 0.996546  0.055747 0.875219

Answer: 0.82.

This is pretty disastrous. But don't worry, apparently in
reality three-year totals are used.

What about in the three year case? Let's say changes are
implemented in the case where 42 accidents occur in a
three-year interval. What is the probability that fewer
will take place in the next three years?

  n P3(n, before)      P3(n, after)
             cumul.             cumul.
-- -------- --------  -------- --------
38 0.024169 0.935156  0.036333 0.166459
39 0.018591 0.953747  0.041923 0.208382
40 0.013943 0.967690  0.047163 0.255545
41 0.010203 0.977893  0.051765 0.307310 <----
42 0.007288 0.985180  0.055462 0.362772

Answer: 0.31.

This is a bit better -- by comparison. But note that if you
were to apply this process to a bunch of sites with an
accident rate of 10/year, then you'd still fail to spot your
mistakes on approximately one in three occasiouns.

--
Chris Lightfoot, chris at ex dash parrot dot com; http://ex-parrot.com/~chris/
I don't care who the audience picks/
I'd rather be killed with a big sharp stick....
   (`How'd The Date End?', The Mr. T Experience)