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From |
"Carlo Lazzaro" <carlo.lazzaro@tin.it> |

To |
<statalist@hsphsun2.harvard.edu> |

Subject |
R: st: R: probability question |

Date |
Thu, 12 Jun 2008 09:30:52 +0200 |

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: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] Per conto di Svend Juul Inviato: giovedì 12 giugno 2008 8.51 A: statalist@hsphsun2.harvard.edu 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: sj@soci.au.dk __________________________________________ * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**Follow-Ups**:**Re: R: st: R: probability question***From:*"moleps islon" <moleps2@gmail.com>

**References**:**Re: st: R: probability question***From:*Svend Juul <SJ@SOCI.AU.DK>

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