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I am having difficulty in understanding the concept of Prevalence lately. Cambridge notes on Medical Statistics are not elaborate and I would really appreciate if any one of you can explain it to me.

The Question I am stuck on is related to two drugs: Ecstasy and Mephedrone.

Assume there are only 30 ecstasy related deaths in the UK per annum. If Mephedrone use is 60% as prevalent as ecstasy but half as dangerous, How many Mephedrone related deaths can you expect in the UK.

To my understanding, there is no definitive way to do this question. I haven't been taught any formulae or theory.

Can anyone explain how to tackle these kinds of questions.



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up vote 2 down vote accepted

Prevalence is the proportion of a population for which some condition is true. The population of the UK is fixed, say $N$. If the prevalence of two drugs are $p_1$ & $p_2$, the frequency of usage is the same (say $f$ times per year), and the risk of death per use is $r_1$ and $r_2$ respectively, then the expected number of deaths from each drug per year (assuming users take only one or the other, otherwise we get into a counting war over which drug kills them) are $Nf\,p_ir_i$ for $i=1,2$. Perhaps it makes sense to define a drug's prevalence not by the number of users, but by the number of uses per year, the drug has. Then we can throw away the $f$ above, since it is modeled inside the $p_i$. But in any case, your problem becomes $$Nf\,p_1r_1=30$$ $$\frac{p_2}{p_1}=\frac35$$ $$\frac{r_2}{r_1}=\frac12$$ so that $$ Nf\,p_2r_2 =Nf\,p_1r_1\cdot\frac{p_2}{p_1}\cdot\frac{r_2}{r_1} =30\cdot\frac35\cdot\frac12 =9 $$ where you can cross out all the $f$'s if they're understood as part of the $p$'s.

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Thank you bgins! – Hassan Khan Mar 21 '12 at 18:14

You have 60% as many users and half as many of them die, so you expect 30*60%*(1/2)=9 deaths per year.

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Thank you Ross! =) – Hassan Khan Mar 21 '12 at 18:14

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