In a study which analyses the effect of Lithium on suicide rates, the results were the following:

  • Placebo group: 3 suicides in 83 patients
  • Lithium group: 0 suicides in 84 patients

My first approach would be to apply the Chi Square test, which does not result in a significant difference (chi=3.0184).

The authors however, suggest the following, finding a significant difference between the groups (p=0.049):

As a post hoc analysis, differences between intervention groups with regard to completed suicides were examined based on determining the probability of zero events in the lithium group on the expectation of 3 ⁄ 83 events in the placebo group on grounds of a Poisson distribution

Is this a valid approach, what am I missing?

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    $\begingroup$ Consider 'Fisher exact text'. While 3 suicides are tragic and regrettable, the count is not large enough to show statistical significance. // Chi-squared statistic does not have dist'n $\mathsf{Chisq}(1)$ because of low expected counts in some cells. And P-value 0.049 is just barely significant, even if it were accurate. // Admittedly, not exactly the same thing, but you couldn't declare a coin biased, based on three Heads out of three tosses. $\endgroup$ – BruceET Jul 19 '18 at 2:13

Yes - the Poisson distr. measures "arrivals", meaning the count of a variable occurring over a given interval. The recurrence of suicide is the event in this study. The "hazard ratio" refers to the rate of arrival, or the lambda parameter of the Poisson distribution.

Rather than just measuring group deviance from mean, as a chi-sq distr. would, modelling the data as Poisson directly measures the cross-group suicide recurrence rate itself.

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    $\begingroup$ Could you please explain how this justifies the implicit assertion that the difference is "significant" with p = 0.049? Given--as the abstract says--the 167 patients were randomized into those two groups, and such randomization will place all 3 suicides into the same group with a probability of $41/167\approx 25\%,$ it's difficult to see how this result could possibly be considered significant. $\endgroup$ – whuber Jun 29 '18 at 0:13
  • $\begingroup$ Let me correct myself: I believe from the abstract the researchers refer to two statistical analyses: (1) a survival analysis, which would use a contingency table and measure the suicide rate as group variable, and was found to be insignificant, and (2) a test of incidence rate, which would measure the suicide rate in time units, and thus follow a Poisson (rate of arrival) distribution, and which was found to be significant. I can't see the full article, so I can't explain any more beyond the abstract. $\endgroup$ – Big Dogs Clothing Company Jun 29 '18 at 1:52
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    $\begingroup$ What seems invalid to me, is to set a theoretical fixed death rate of 3/83 based on the Placebo Group, thereby halving the variance. The death rate in the Placebo group could be randomly too high and the one in the Lithium Group randomly too low. $\endgroup$ – Constantin Volkmann Jun 29 '18 at 7:05
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    $\begingroup$ I agree to @ConstantinVolkmann . Doing that seems as invalid as -for example- trying to compare two groups using a one sample t-test by using the mean of one group as the null hypothesis for the other group - instead of performing a two sample t-test. $\endgroup$ – Pere Jun 29 '18 at 10:07
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    $\begingroup$ I'm not all that familiar with incidence rate testing, but it appears to be a two-sample test on Poisson parameters suited for a placebo/treatment design (statsdirect.com/help/rates/compare_crude_incidence_rates.htm). I don't understand the bit about "halving variance" - could you please elaborate? $\endgroup$ – Big Dogs Clothing Company Jun 29 '18 at 17:17

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