[DrVanNostrin]

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Intuitively this doesn't make sense to me. Logic would seem to dictate that if we had 1326 ping pong balls and 47 of them were red (but we didn't know they were red). It seems like we would have to take many more samples from the group to get an accurate representation of those red balls, especially if in some light the red ball appeared to be white...


Now suppose that 663 of the balls are red. If we were to take samples from that group it would converge toward actual much faster. Atleast that is how I picture it working in my mind.

It seems like the bigger portion of the population the test group you are trying to get information about is, the less trials it would take to get an accurate representation of that group.

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The number you're looking at is the coefficient of variation; that is, the ratio of SD to mean. The coefficient of variation relates to the % uncertainty.

But we're dealing with the absolute uncertainty (which happens to have units of %), not the % uncertainty.

Instead of looking at how many balls are red you can look at how many are not red. Knowing P(not red) is the same as knowing P(red). Having a confidence interval for P(not red) is the same a having a confidnece interval for P(red). The size of your confidence interval for P(not red) must equal the size of your confidence interval for P(red).

Var(X) = E(X^2) -[E(X)]^2

If the probability X = 1 is p and the probability X = 0 is 1-p (as it is in this case)...

E(X^2) = p

E(X) = p; [E(X)]^2 = p^2

var(X) = p - p^2 = p*(1-p)

sd(X) = sqrt(var(X)) = sqrt(p*(1-p))

If you define your random variable to be the average, not the total, sd is proportional to 1/sqrt(n)