A rough approximation to the c.i. of a proportion (prevalence) is
p \pm 2*sqrt(p*(1-p)/n)
So if you want an absolute precision of 3% (assuming that this means a confidence interval of the form p \pm 3% you are facing to solve:
sqrt(p*(1-p)/n) = 0.015
for p=0.12.
But the most illustrative would be to do this calculation for a range of prevalences:
n <- 100:1000
p <- seq(6,18,3)/100
prc <- function(n,p) 2*sqrt(p*(1-p)/n)
aa <- outer( n, p, FUN=prc )
matplot( n, aa, type="l", lty=1, col="black", lwd=2,
xlim=c(100,1100), xlan="N", ylab="Precision (2*sd)" )
text(1050,aa[length(n),],p*100,adj=1)
abline(h=0.03,col="red")
______________________________________________
Bendix Carstensen
Senior Statistician
Epidemiology
Steno Diabetes Center A/S
Niels Steensens Vej 2-4
DK-2820 Gentofte
Denmark
+45 44 43 87 38 (direct)
+45 30 75 87 38 (mobile)
b...@steno.dk http://BendixCarstensen.com
www.steno.dk
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