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sample mean variance

Is sample mean variance really
(N-n)/N*sigma^2/n
where N is the number of population and n is the number of samples draw from population.

Where is the sampling correction (N-n)/N from? The whole time I was using the wrong equation? I know when N gets larger, the correction term tends to 1. But when N is not a big number, the term is significant!
Original post by Jundong
Is sample mean variance really
(N-n)/N*sigma^2/n
where N is the number of population and n is the number of samples draw from population.

Where is the sampling correction (N-n)/N from? The whole time I was using the wrong equation? I know when N gets larger, the correction term tends to 1. But when N is not a big number, the term is significant!


Can't recall seeing that formula before, at least in that format - where's it from? Do you have a link, or scan of the appropriate section.
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Jundong
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Original post by ghostwalker
Can't recall seeing that formula before, at least in that format - where's it from? Do you have a link, or scan of the appropriate section.


http://www.stanford.edu/~mrosenfe/soc180B_notes_mean_variance.pdf
page 6
Original post by Jundong


Cool.

From a bit of digging around, yes, that's the full formula, using sampling without replacement. It's not so much that N is big, but rather if n << N, then the error in ignoring that factor is small. It's generally ignored if n/N < 0.05 from what I can see.

Can't find an actual proof.

A search for "finite population correction" or fpc variance, may be useful.

Edit: It's not something you need to take into account in Sixth Form.
(edited 9 years ago)

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