Let’s assume that a hospital’s neurosurgical team performed N procedures for in-brain bleeding last year. x of these procedures resulted in death within 30 days. If the national proportion for death in these cases is pi, then is there evidence to suggest that your hospital’s proportion of deaths is more extreme than the national proportion?

Pick your own values of N, x, and pi. The x is necessarily less than or equal to N, and pi is a fixed probability of success. The probability should be greater than or equal to x. Then model both as a Binomial and a Poisson, and provide your R code solutions. Do you get similar answers or not under the two different distributional assumptions, and can you guess why ?

The number of procedures (N) of the last year will be 50. With 5 fatalities (x) in the last 30 days. The national average is .15 (pi)

# Define Variables
N <- 50   # size
x <- 5    # num deaths in 30 days
range <- 5:50
pi <- .15 # prob of success

# BINOMIAL
binomial <- round(sum(dbinom(
              x = range,
              size = N,
              prob = pi
            )),4)

# POISSON:   # 1 - P(X<=2 | n=10, pi=.2)   # lower tail TRUE option (by default)
## NOTE ARE CALCULATING:  P(X>=3) = 1 - P(X<=2)
l = N * pi
l
## [1] 7.5
poisson <- round(sum(dpois(
            x = range, 
            lambda= l
           ) ) ,4)
cat("The Binomial probability is",binomial)
## The Binomial probability is 0.8879
cat("\nThe Poisson probability is",poisson)
## 
## The Poisson probability is 0.8679
diff <- binomial - poisson
cat("\nThe difference between the two is",diff)
## 
## The difference between the two is 0.02

The answers between the two distributions are similar. This is due to the poisson distribution being a limiting case of the binomial. When Lambda is close to np, the probabilities calculated tend to be similar.