dyk33
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1000 car accidents happened over the last two years in a town with population of 1,000,000. 600 were alcohol related. 400 died because of the accidents of which 50% were alcohol related. what is the one year incidence of EtOH related death?
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| sk
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200?
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| uscan
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100?
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| asmi
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:roll:
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| dyk33
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well, i thought at first 100 per population. however, don't you have to assume that there is no movement in the population within the last two years to correctly calculate the incidence rate? and can you simply divide the number in half to get one year incidence rate? Thus, without these qualifiers, i'm guessing it cannot be calculated. any other thoughts?
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| phanul75
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Hi, I would say 200, like u said incidence as new cases per year per population. Is it important if u use term incidence or incidence rate. I thought 'rate' means per 100 000.
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| dyk33
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I meant "incidence per population" per year not incidence rate.
having said that, and sounds a little dumb, would you divide simply the two year incidence in half to get one year incidence?
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| uscan
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To Dyk33
Do you have the correct answer?
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| asmi
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i think it should be 200/1,000,000 ..
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| dyk33
Forum Newbie
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sorry, i don't have the answer. i though someone here in this forum could answer this question. asmi, why do you think it is 200, instead of 100? and could someone explain why "unable to determine" is NOT an answer?
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| asmi
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cause specific mortality rate = deaths from alcohol/total population
:arrow: from Q , 50% of 400 deaths = 200 .were due to alcohol .
:arrow: 200/1,000,000
:arrow: those 600 is alcohol related accidents not deaths.
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| asmi
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mortalilty rate refers to incidence rate only
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| phanul75
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Sorry, I got confused :oops: . I wrote 200 which is wrong.
400deaths/2 years = 200 per 1 year, now 50% are ETOH related = 100
:wink:
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| dyk33
Forum Newbie
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i agree with your answer. i don't have the answer but i think it is 100. what you have to ASSUME as it is not stated in the question is that the same number of population is susceptible to EtOH in both years and that there was no change in the population within the last two years.
also, when you divide the 200 by 2 to get the ONE year incidence, what you are actually getting is average annual incidence, (but the question does not specify this either.)
i got a very similar question from NBME self assessment test, (which is why i don't have the correct answer. sorry). One of the choices was actually "not able to calculate". I was confused because of the lack of stated assumptions.
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| asmi
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from my knowledge...
i think in no way you can change or divide the population ...
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| dyk33
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to calculate incidence, you have to know the susceptible population. for example, let's say 100 people have AIDS in a town of 100,000. and 10 new cases of AIDS every year. the incidence of AIDS in a year is
10/ (100,000 - 100) = 10/ 99,900
what i was not clear on my question was whether you can make tha assumption that the same number of people are susceptible each year for EtOH related death and that there is no change in number of population.
So, was the question asking for this concept (hence the answer is not able to calculate) or just a matter of division? i think i'm reading too much into this question....hope i don't get this question on the real thing.
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| Quinn
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the answer to the first question should be 200 which is the number new cases of alcohol related death over the past year.
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