I have to determine the annual probability of at least one heart attack for various groups of given data that was collected over a period of 24 years.
For men that fell in the age group 30-39, and had a relative weight score of 80-89, the 24 year incidence of at least 1 heart attack was 101 per 1,000 individuals in that category. For 100-199, it was 89 per 1,000 (The weight score is based on 100 being median weight, 80 being 20% below median weight, but that's not important here).
I need to go from 101/1000 in a 24 year span to what the probability of at least 1 heart attack would be annually. I found this answer that does what I need, except when I worked it with my data, the categories with more occurences had lower probabilites. So when I worked out 101/1000, making this equation
$1-e^{-(-ln(\frac{101}{1000}))/24}$
gives 0.0911
However, if I go for the next weight category and do
$1-e^{-(-ln(\frac{89}{1000}))/24}$
I get 0.0959
If that's the right answer, let me know, but to me it doesn't make sense for the lower annual probability to be with the group that had higher incidences over 24 years.
To find $\lambda$ you must compute the probability of $X=0$, eg the annual probability of NO heart attack.
$$ P(X=0)={\lambda^0 e^{-\lambda}\over 0!}=e^{-\lambda}, $$ considering : $$ P(X=0)=1-p=1-{101 \over 1000}=0.899, $$ in the first case.
Thus, it comes : $$ e^{-\lambda}=0.899 \rightarrow \lambda=-\ln(0.899)\approx 0.1064 $$ $$ 1-e^{-0.1064/24}\approx 0.0044, $$ and in the second case : $$ P(X=0)=1-p=1-{89 \over 1000}=0.911, $$ $$ e^{-\lambda}=0.911 \rightarrow \lambda=-\ln(0.911)\approx 0.0932 $$ $$ 1-e^{-0.0932/24}\approx 0.0038. $$