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I earlier asked about Slovin's Formula (https://math.stackexchange.com/questions/1410492/what-is-the-intuition-behind-slovins-formula), and learned shortly thereafter that it was derived from this formula.

$n=\dfrac{n_0}{1+\dfrac{n_0}{N}}$,

Where $n_0=\dfrac{z^2p(1-p)}{e^2}$,

$N$ is the population size, $z$ is the standard normal variate based on the confidence coefficient, $p$ is the estimate for $P$, and $e$ is a specified margin of error.

So, breaking it down, what factors take part in determining the sample size?

(read the comments for further context)

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  • $\begingroup$ It's unclear to me what you are trying to ask. On the face of it, understanding "$n$" as referring to sample size, the formula explicitly shows it depends on $z$, $p$, $e$, and $N$. Since this is so obvious, I have to assume you are trying to get at something more meaningful--but what is it? $\endgroup$
    – whuber
    Aug 28, 2015 at 14:28
  • $\begingroup$ @whuber Yes. I sort of mean: what value does the numerator represent, and what value does the denominator represent? $\endgroup$
    – Ray James
    Aug 28, 2015 at 15:34
  • $\begingroup$ Understanding you to be seeking an interpretation of the formula--and perhaps a clear explanation of the assumptions and theory used to derive it--I am happy to upvote your question and I look forward to reading the answers. You would be able to reach more people (and perhaps get more or better answers) by editing this post to include explanations of the quantities represented by all the variables. $\endgroup$
    – whuber
    Aug 28, 2015 at 17:36
  • $\begingroup$ You may want to give a link to the previous question(s) you are referring to. $\endgroup$
    – StasK
    Aug 28, 2015 at 19:06

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The intuition is that you don't have infinity in the finite population world. You cannot have a sample more of size more than $N$. As $n_0 \to \infty$, for whatever reason (you keep making the specified margin of error smaller), $n\to N$ because there is no room to grow beyond $N$. The numerator and denominator may or may not represent anything in particular: you can multiply them both by $N$ or divide by $n_0$, which would change what is in the numerator and in the denominator... But the way the formula is written as is, $n_0$ is the required sample size from an infinite population, and the denominator is some sort of undoing the finite population correction (FPC), of SRSWOR sampling variance: $1-n/N$. Thinking again about large samples, as $n\to N$, the variance goes to zero, which denotes that you don't have any sampling error (you may still have measurement error, of course, but the finite population sampling paradigm assumes fixed and perfectly measured values, at least as the first natural science approximation to whatever complexities might be encountered in practice). The formula you gave undoes this FPC.

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  • $\begingroup$ When I show that formula to statisticians, they don't seem to be familiar with it. Is it not used often? $\endgroup$
    – Ray James
    Aug 28, 2015 at 19:58
  • $\begingroup$ Statistics is a broad discipline. There are many things that I can show to the typical faculty in stat departments, and they won't be familiar with it -- path models, instrumental variables, finite population corrections... and of course there will be tons of things these statisticians can show to me that I won't be familiar with, like slice sampling for MCMC or VC dimension. $\endgroup$
    – StasK
    Aug 31, 2015 at 2:56
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    $\begingroup$ But I was under the impression that this was one of the main formulas used for deriving a sample size. There are others? $\endgroup$
    – Ray James
    Aug 31, 2015 at 4:02
  • $\begingroup$ Also, what is the intuition behind $n_0$? $\endgroup$
    – Ray James
    Aug 31, 2015 at 15:55

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