Answer:
A) 0.83 for lung cancer and 0.18 for coronary thrombosis.
Explanation:
For the population etiologic fraction due to smoking, we use the formula:
[tex] = \frac{P_s(RR_s-1)}{(1+P_s(RR_s-1))}[/tex]
For lung cancer:
Given:
[tex] P_s = 55percent = 0.55 [/tex]
We first find RRs which is the relative risk of dying from cancer for a smoker compared to a non smoker.
[tex] RR_s = \frac{num of smokers death}{num of non-smokers death}[/tex]
[tex] = \frac{71}{7}[/tex]
= 10.1429
Therefore population etiologic fraction due to cancer wil be:
[tex] \frac{0.55(10.1429-1)}{1+0.55(10.1429-1)}[/tex]
[tex] = \frac{5.029}{6.029}[/tex]
= 0.8341
Population etiologic fraction of cancer due to smoking is 0.834
•For coronary thrombosis:
Let's use the formula:
[tex] \frac{P_s}{RR_s-1}{1-P_s(RR_s-1}[/tex]
Let's first find the RRs (relative risk of dying from cornary thrombosis for a smoker compared to a non smoker)
[tex] = \frac{599}{422} [/tex]
[tex] RR_s = 1.4194 [/tex]
The population etiologic fraction of cornary thrombosis due to smoking, will be:
[tex] = \frac{0.55(1.4194-1)}{1+0.55(1.4194-1)}[/tex]
[tex] = \frac{0.231}{1.231} [/tex]
= 0.187
The population etiologic fraction of cornary thrombosis due to smoking is 0.187
Therefore the answer is option (A) 0.83 for lung cancer and 0.18 for coronary thrombosis