I found here the formula for computing the sample size $n$ of a finite population $N$
$$
n = \frac{n_\infty}{1 + \frac{n_\infty – 1}{N}}
$$
where the sample size for an infinite population $n_\infty$ is given as
$$
n_\infty = \frac{z^2 p(1-p)}{c^2}
$$
with the the z-score $z$, the confidence interval $c$ and $p$ being the proportion of the population picking a specific choice. However I found no derivation of this formula. Can someone give a mathematical derivation of this formula?
Solved – Derivation of formula for sample size of finite population
finite-populationmathematical-statisticssample-size
Related Solutions
Update: 2014-02-06: changed text to be more emphatic that fpc should not be used in a causal analysis **Update: 2014-02-04: impact of the randomized experimental design
This question has raised some fundamental issues.
You stated in your update that a researcher can control the make-up of the experimental groups. Not so. Even if one randomized an entire population, there would be imbalance, perhaps trivial, in every variable. Even with some kind of balancing algorithm, which would destroy the randomization, one can never arrange for identity of the means of the outcome variable, yet unmeasured.
You also asked Tom Lumley:
Are you saying it is legitimate to estimate the confidence interval of say, the difference between the proportion of men and women answering 'Yes' but not a p-value to determine if it is zero (i.e. to reject the null)?
I think that's what Tom meant, and I agree with its application to descriptive statistics; I'm not sure that it applies It does not apply to causal analyses, including those generated by an experiment. Your particular example is a borderline case, as you intend the results to apply to a single population at a particular time. If someone asked you to project your findings to another setting or to another time period, the confidence interval calculation probably should not include the fpc.
Some additional insight can be gained by considering the experimental design as part of the sample design. If the initial random sample is of size $n$, randomization produces two random sub-samples of size $n_1 = n/2$ and $n_2 = n/2$. (For the theory that follows, $n_1$ and $n_2$ need not be equal.) Let $\overline{y}_1$ and $\overline{y}_2$ be the means of the sub-samples; proportions are special cases. In this scenario, which conforms to the absence of a treatment effect, it can be shown (Cochran, 1977, problem 2.16, p. 48) that:
\begin{equation} Var(\overline{y}_1 -\overline{y}_2) = S^2\left(\frac{1}{n_1} +\frac{1}{n_2}\right) \end{equation}
where $S^2$ is the population variance and variation is over repetitions of the sampling and randomization. Notice: no fpc.
Update: one of the few established uses of hypothesis tests + FPCs for finite populations: lot quality assurance sampling (LQAS)
I agree with Tom's answer. Hypothesis testing rarely has a place in finite population questions, but confidence intervals certainly do. One good use of hypothesis tests per se in finite populations is lot quality assurance sampling (LQAS), which tests whether the rate of some event (e.g. vaccination) in a geographic area is too high or too low. Note that, unlike the question at hand, there is no hypothesis of zero difference. The null hypothesis is that the rate is < K, and the alternative that is it is $\geq$K. See, at Google Scholar.
Robertson, Susan E, Martha Anker, Alain J Roisin, Nejma Macklai, Kristina Engstrom, and F Marc LaForce. 1997. The Lot quality technique: a global review of applications in the assessment of health services and disease surveillance. Relation 50, no. 3/4: 199-209.
Lemeshow, Stanley, and Scott Taber. 1991. Lot quality assurance sampling: single-and double-sampling plans. World Health Stat Q 44, no. 3: 115-132.
Original Answer
Using the fpc to reduce sample size makes no sense unless intend you use it in the the hypothesis-testing statistic. But that would be an error: the fpc should not be used when testing hypotheses [added about "no difference"].
The reasoning is interesting (Cochran, 1977, p.39): It is seldom of scientific interest to ask if a null hypothesis (e.g. that two proportions are equal) is exactly true in a finite population . Except by a very rare chance, the null hypothesis will never be true, as one would discover by enumerating the entire population. Therefore hypothesis tests on samples from finite populations are done from a "super-population" viewpoint. See also Deming (1966) pp 247-261 "Distinction between enumerative and analystic studies"; Korn and Graubard (1999), p. 227.
References
Cochran, W. G. (1977). Sampling techniques (3rd ed.). New York: Wiley.
Deming, W. E. (1966). Some theory of sampling. New York: Dover Publications.
Korn, E. L., & Graubard, B. I. (1999). Analysis of health surveys (Wiley series in probability and statistics). New York: Wiley.
Confidence Interval – How to Calculate Confidence Level for a Given Sample Size and Population Size?
When constructing confidence intervals usually the size of a population is far larger than the sample size. In these cases we treat the sample as if it came from an infinite population and this simplifies the analysis a bit. For these cases the confidence interval formula is the following
Lower limit:
$$p-z\sqrt{\frac{p(1-p)}{n}}$$
For your example this is $0.5-1.96\sqrt{\frac{0.5(1-0.5)}{1406}}=0.4739$
Upper limit:
$$p+z\sqrt{\frac{p(1-p)}{n}}$$
For your example this is $0.5261$ so the 95% confidence interval for the population value of $p$ is $(0.4739,0.5261)$
Small population size
When the size of the population is small then you can make an adjustment to account for this fact. In this case the confidence interval is
Lower limit:
$$p-z\sqrt{\frac{p(1-p)}{n}\left(\frac{N-n}{N-1} \right)}$$
Upper limit:
$$p+z\sqrt{\frac{p(1-p)}{n}\left(\frac{N-n}{N-1} \right)}$$
The part under the square root is modified slightly. In your example the population is huge so it's being modified by a factor of $\frac{292456752-1406}{292456752-1}= 0.999995$. You can try calculating the modified confidence interval, it doesn't change the first four decimal places.
Small sample sizes
When you sample very few people then the methods used to derive the above formulas can be invalid. A common rule for deciding if sample size is large enough is the following:
If $np > 5$ and $n(1-p)>5$ then the sample size is large enough. Your example certainly has a large enough sample size. When the sample size is too small then you should use a different interval such as the Wilson Score interval:
$$\text{Lower limit} = \frac { 2n\hat{p} + z^2 - \left[z \sqrt{z^2 - \frac{1}{n} + 4n\hat{p}(1 - \hat{p}) + (4\hat{p} - 2)} + 1\right] } { 2(n + z^2) }$$
$$\text{Upper limit} = \frac { 2n\hat{p} + z^2 + \left[z \sqrt{z^2 - \frac{1}{n} + 4n\hat{p}(1 - \hat{p}) + (4\hat{p} - 2)} + 1\right] } { 2(n + z^2) }$$
If these formulas give a value below $0$ or above $1$ (which is an impossible value for $p$) then round them to $0$ or $1$
This one doesn't have a nice way of adjusting for a small population size. If you have both a small population size and a small sample size I'd recommend prioritizing the small population size and using the second set of confidence interval formulas I described.
Best Answer
Amazingly a pretty complete derivation was given at math.stackexchange.com at https://math.stackexchange.com/a/1357604/27609.
Another derivation can be found online at https://onlinecourses.science.psu.edu/stat414/node/264