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a large cohort study is conducted to assess the association between smoking and squamous cell carcinoma of esophagus among middle aged chinese men. during 10 years of follow up, smokers have had five times the risk of esophageal carcinoma compared to non smokers (RR=5.0, 95%CI=2.9-7.1). according to the study results, what percentage of esophageal carcinoma in smokers can be attributed to smoking?

A 25%
B 50%
C 70%
D 80%
E 90%

in UW, attributable risk percentage or etiologic fraction formula applied to this question. can anyone explain this question and the formula to apply. thanks in advance :redcheeks;
 

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a large cohort study is conducted to assess the association between smoking and squamous cell carcinoma of esophagus among middle aged chinese men. during 10 years of follow up, smokers have had five times the risk of esophageal carcinoma compared to non smokers (RR=5.0, 95%CI=2.9-7.1). according to the study results, what percentage of esophageal carcinoma in smokers can be attributed to smoking?

A 25%
B 50%
C 70%
D 80%
E 90%

in UW, attributable risk percentage or etiologic fraction formula applied to this question. can anyone explain this question and the formula to apply. thanks in advance :redcheeks;
ARP (attributable risk percentage) = RR - 1 / RR
In this case, they already gave us the RR, so: ARP = 5-1 / 5 = 4/5 = 0.8 = 80%
In plain words, this means that out of 100 patients with esophageal cancer, 80 of those have smoking as their major risk factor.
 
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