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(I know it is not quite appropriate to quote it as "part 2", but since the question has been dormant for quite a while, I hope by doing this will rise peoples' attention again, you may have a look of part I here.)

I have come across an article online talking about the case similar to mine, that most of the time the case count is zero, make sometimes when the case number increases to one, it already shoots above the control level and consider the case as "out of control".

Since c-chart will be easier for my bosses to read and interpret, I wonder if the method is sound, or did anyone have some more official reference on this method? (I have googled quite a while but I can find nothing)

The article can be found here.

An to further my question, I want to ask one more thing: for the assumption of c-chart that the case count needs to follow a Poisson distribution, is it applicable to all lambda (i.e. mean of case count)?

Thanks again.

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  • $\begingroup$ Better than naming it Part 2 (or n), make a link to the previous question inside this. $\endgroup$
    – user88
    Commented Aug 27, 2010 at 9:10
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    $\begingroup$ The answer I supplied to your previous question suggests a version of the "c-chart" that is more rigorously justified, quite simple, and accordingly is easy to interpret. (It also raises the logically earlier question of whether control charting is an appropriate technique for you.) You did not respond to this, but now you re-raise the question, so I am wondering why: why do you think this solution will not work for you? $\endgroup$
    – whuber ♦
    Commented Aug 27, 2010 at 15:10

1 Answer 1

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C-charts basically works like this: you collect all your observations (this is an approximation of getting only observations from "normal" periods) of a case count, then fit Poisson distribution to it (so here you apply all Poisson assumptions, mainly that each case occurance is independent to the others), and finally test all case counts with H0 that they are just from this fitted distribution. So, if you have majority of zeros, it will just drive $\lambda$ of fit near zero; for $\lambda<0.01$ even one case will be something strange (on $p$-value=1%).

EDIT: In the article you've linked the whole fitting is done just by taking mean as $\lambda$, while testing just by $3\sigma$ criterion based on the fact that Poisson's variance is also $\lambda$. Still it is sufficient for this case.

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    $\begingroup$ P-values are not appropriate measures for control charts, because of the serial multiple comparisons involved and the possibility (likelihood, in this case) of serial correlations. Control chart properties are better assessed in terms of expected lengths of out-of-control (OOC) runs and in-control runs under the null and alternative hypotheses, respectively. On this basis, letting even one case trigger an OOC is almost never a good procedure. $\endgroup$
    – whuber ♦
    Commented Aug 27, 2010 at 14:12
  • $\begingroup$ I was talking about the p-value of one case, so I think it is correct. Of course I agree with a general claim that this can (and should) be done better, still from the voice of the article lockhart cited I can conclude he seeks for statistics "light". $\endgroup$
    – user88
    Commented Aug 27, 2010 at 15:04
  • $\begingroup$ I agree that your p-value calculation is correct, but the issue I am raising concerns whether that's the right way, or even a good way, to think about and design control charts. For those familiar with adjusting for large numbers of multiple comparisons, a single p-value can be an excellent guide, but for others this way of thinking might be misleading. $\endgroup$
    – whuber ♦
    Commented Aug 27, 2010 at 15:14

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