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The Definition of incidence rate is defined as : new event/person-time

Now there's my data：

                   Mutation       Without mutation
incidence rate    0.01345908         0.00407633

Where 0.01345908 = (N.O new event with mutation case) / (person year of all case from mutation + person year of all case from without mutation)

Where 0.00407633 = (N.O new event without mutation case) / (person year of all case from mutation + person year of all case from without mutation)

It's trivial that this two sample is not independent (probability,that's, incidence rate would influence by each case due to person year on Denominator)

So, i can't undergo two-sample proportion test. (due to dependent)

Well , i want to know how to test whether this two incidence rate reach statistic significant , how can i do in this situation ? is anyone can help me ?

Thank a lot.

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  • $\begingroup$ Why " (N.O new event with mutation case) / (person year of all case from mutation + person year of all case from without mutation)", instead of " (N.O new event with mutation case) / (person year of all case from mutation )" $\endgroup$ – user158565 Nov 23 '18 at 17:02
  • $\begingroup$ What is the null hypothesis that you want to test? $\endgroup$ – FMarazzi Nov 23 '18 at 17:03
  • $\begingroup$ Ohh sorry，it's (person year of case from with/without mutation) $\endgroup$ – 連振宇 Nov 23 '18 at 19:05
  • $\begingroup$ Ho : p1 = p2 (incidence equal) $\endgroup$ – 連振宇 Nov 23 '18 at 19:05
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You want to compare the two incident rates. Two incident rates are based on the same exposure time (person-year). Then it equals to compare the number of two kinds of events.

Assume that # of events follows Poisson distribution. The null hypothesis is N.O new event with mutation case and N.O new event without mutation case follow the same Poisson distribution.

There are different test based on approximation, and also exact test, such as poisson.test in R you can search internet to get the method and software.

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