Thx a lot everyone. This certainly helped me out quite a bit.
Moleps
On Thu, Jun 12, 2008 at 9:30 AM, Carlo Lazzaro <[email protected]> wrote:
> Dear Richard e Svend,
> thanks a lot for your further clarifications on this classic
> classwork-issue.
> By the way I was wondering whether it may be interesting to add to the Stata
> bookshelf a tutorial-based textbook on probability (let's say from basics to
> Bayesian probability distributions) issues related to biostatistics,
> epidemiology, social sciences and alike (with alike I mean all the research
> fields which are unknown for me) which can be tackled with Stata routines.
>
> (This is not an advertising stunt to promote myself as an author: this idea
> is indeed the offspring of my hard times in dealing with the quantitative
> world!)
>
> Kind Regards,
>
> Carlo
> -----Messaggio originale-----
> Da: [email protected]
> [mailto:[email protected]] Per conto di Svend Juul
> Inviato: gioved� 12 giugno 2008 8.51
> A: [email protected]
> Oggetto: Re: st: R: probability question
>
>
> Moleps asked, and Carlo and Richard commented:
>
> I need to calculate the probability of the following event:
> That at least one patient out of 99 suffers from kidneycancer
> (incidence 15/100000 pr year) over a time period of 15 years.
>
> ====================
>
> Carlo:
>
> searching the literature (Briggs A, Sculpher M, Claxton K. Decision
> Modelling for Health Economic Evaluation. Oxford: Oxford University
> Press,
> 2006:51) I have found out a formula for converting rate into
> probabilities:
>
> p=1-exp(-rt)
>
> where:
>
> p=probability;
> r= instantaneous rate, provided that it is constant over the period of
> interest (t)
>
> ======================
>
> Richard:
>
> That might be more accurate than what I calculated before, but luckily
> it gives pretty much the same result. For one person, the probability of
> NOT getting cancer over 15 years is
>
> p(0) = 1 - (1 - exp(-rt)) = exp(-rt) = exp(-15/100000 * 15), i.e.
>
> . di exp(-15/100000 * 15)
> .99775253
>
> The probability that all 99 people should be so lucky is
>
> . di exp(-15/100000 * 15) ^ 99
> .8003149
>
> Hence, the probability that at least one is not so lucky is
>
> . di 1 - (exp(-15/100000 * 15) ^ 99)
> .1996851
>
> Which is pretty close to my earlier calculation, which was
>
> . di 1 - ((1 - 15/100000 )^15^99)
> .19969847
>
> The difference is .00001337!
>
> I imagine there might be some other complications in these calculations.
> The rate may not be constant across time; or people may die from
> something else before they get kidney cancer. But for the problem as
> stated, it sounds like there is a 20% chance of at least one person
> getting kidney cancer.
>
> ======================================================================
>
> To get the expected proportion over 15 years, 15*15/100000 = 0.00225
> is almost right with a rare event, but as Carlo pointed out, the
> correct conversion from a rate to a proportion says:
>
> . display 1-exp(-15/100000 * 15)
> .00224747
>
> Now, the official -bitest- command does the rest of the job:
>
> . bitesti 99 1 0.00224747
>
> N Observed k Expected k Assumed p Observed p
> ------------------------------------------------------------
> 99 1 .2224995 0.00225 0.01010
>
> Pr(k >= 1) = 0.199685 (one-sided test)
> Pr(k <= 1) = 0.978786 (one-sided test)
> Pr(k >= 1) = 0.199685 (two-sided test)
>
> note: lower tail of two-sided p-value is empty
>
> Pr(k >= 1) = 0.199685
>
> Svend
>
> __________________________________________
> 1
> Svend Juul
> Institut for Folkesundhed, Afdeling for Epidemiologi
> (Institute of Public Health, Department of Epidemiology)
> Vennelyst Boulevard 6
> DK-8000 Aarhus C, Denmark
> Phone: +45 8942 6090
> Home: +45 8693 7796
> Email: [email protected]
> __________________________________________
>
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