In order to find out if the proportion of deaths in these cases is more extreme than the national proportion, I asked the question: what is the probability that the neurosurgical team at this hospital would experience 5 or more deaths in 30 procedures if they were operating at the national proportion of .05? (Then compare to the standard .05 probability).
N=30
pi=.05
P(X>=5|N=30, p=.05)=???
I then ran this in R Markdown using the sum(dbinom)() function…
sum(dbinom(5:30,30,.05))## [1] 0.01563551P(X>=5|N=30, p=.05) = 0.01563551
Since the probability of the team at this hospital experiencing 5 or more deaths is lower (.016, 1.6%) than the standard of .05, there is evidence that the proportion of deaths experienced at this hospital is more extreme than the national proportion.
N=30
pi=.05
(lambda)=N(pi)=1.5
t=1
E(Poisson)=(lambda*t)=1.5
E(Binomial)=(N*pi)=1.5
sum(dpois(5:1000,1.5*1))## [1] 0.018575941-ppois(4,1.5)## [1] 0.01857594P(Y>=5|(lambda)=1.5,t=1) = 0.01857594
1-P(Y<=4|(lambda)=1.5,t=1) = 0.01857594
Using the Poisson Model I was able to calculate the probability of the team experiencing 5 or more deaths AND 1 - the probability of the team experiencing 4 or less deaths, which both came out to 0.01857594, just about .003 above the Binomial result. Since the probability of the team at this hospital experiencing 5 or more deaths is lower (.019, 1.9%) than the standard of .05, there is evidence that the proportion of deaths experienced at this hospital is more extreme than the national proportion.