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...
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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