Xiao Li <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]In other words, when n << N, the effect of N is negligible on the total SD. So N _is_ included in the equation, but it has little effect. If you work out the hypergeometric form of the equation, and then run N up very large, for constant n, you will see that the SD approaches its value (determined without N, above) asymptotically . Try it with what Joe said.
Let N be the population size. Let p denote the proportion of this population with a certain characteristic (ie. has diabetes, divorced, homosexual, etc.) Let's say we are trying to estimate p using a sample of size n. So we collect our sample and calculate p^ to be the proportion of elements in our sample with our characteristic of interest. Then we calculate the standard of deviation of p^ (also known as the standard error) because it gives us an idea of how close our sample proportion is to the real proportion.
Now, my statistics textbook says the standard of deviation is given by SD(p^) = root[(p^(1-p^))/n] when sampling with replacement or when the sample size is significantly smaller than the population size.
Jay
My question is why isn't there a N variable in our equation for
SD(p^)? According to this formula, if we are trying to figure out the
proportion of people with diabetes in Berkeley, California with sample
size n, we would get the same standard of error as if we are trying to
figure out the proportion of people with diabetes in the world with
sample size n. (In the former, the N is a lot smaller than in the
later).
I think this is a case of the hypergeometric distribution, instead of
binomial. The hypergeometric takes into account cases of
non-infinite N.
------ Joe ------- [EMAIL PROTECTED]
-- Jay Warner Principal Scientist Warner Consulting, Inc. 4444 North Green Bay Road Racine, WI 53404-1216 USA
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