I think I got it... I assumed that n=100000 would be big enough to give me a sample mean which converges on the population mean, but this was a bad assumption.
Using sigma / sqrt(n) for the standard error of the sample mean, I get... ----------------------------------------------------- se.x.mean.as.perc.of.pop.mean <- (0.25 / sqrt(100000)) /0.08; se.y.mean.as.perc.of.pop.mean <- (0.25/sqrt(252)) / sqrt(100000) /(0.08/252); se.x.mean.as.perc.of.pop.mean # 0.0098821 se.y.mean.as.perc.of.pop.mean # 0.1568738 -- View this message in context: http://www.nabble.com/rnorm%28%29-converted-to-daily-tp23092444p23093105.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
