FPC = sqrt((

*N*-

*n*)/

*N*)

where

*N*is the population size and

*n*is the sample size. If your survey includes everyone in the population, a.k.a. a census, then

*N*=

*n*and your FPC = 0 . Thus, your standard error is zero, because you know everything about everyone, and there is no uncertainty.

In practice, the FPC can be ignored when the sampling fraction is less than 10%. However, in practice, and as a general rule, survey research organizations do not like to pass up an opportunity to make their estimates look better, and will not ignore the FPC.

For the 1990 Census, everyone responded to a certain number of questions, all of which were contained on the "Short Form." A sample of the population--the sampling rate varied across the country, but in most parts of the country the sampling rate was one-in-six--received a "Long Form" questionnaire. They answered all of questions that were contained on the Short Form, plus more detailed questions on variety of topics, such as how they commuted, what their occupation was, etc.

Including the FPC, the standard error for the estimate of the number of Whites becomes:

se(Nw) = FPC x N x sqrt(p x q/n)

= sqrt((N - n)/N) x N x sqrt(p x q/n)

= sqrt(1 - (n/N)) x N x sqrt(p x q/n)

Taking the sampling rate to be 1/6, or n/N = 1/6, this becomes:

se(Nw) = sqrt(1 - (1/6)) x N x sqrt(p x q/n)

= sqrt(5/6) x (N/sqrt(n)) x sqrt(p x q)

= sqrt(5/6) x sqrt(N/n) x sqrt(N) x sqrt(p x q)

= sqrt(5/6) x sqrt(6) x sqrt(Np) x sqrt(1 - p)

= sqrt(5 x Np x (1-p))

= sqrt(5 x Np x (1 - Np/N))

which is basically the result in the paper for the one-in-six simple-random-sample result. It may come out looking a bit more complicated than it needs to look.

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