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The methods of the last page, in which we derived a formula for the sample size necessary for estimating a population proportion \(p\) work just fine when the population in question is very large. When we have smaller, finite populations, however, such as the students in a high school or the residents of a small town, the formula we derived previously requires a slight modification. Let's start, as usual, by taking a look at an example.
A researcher is studying the population of a small town in India of \(N=2000\) people. She's interested in estimating \(p\) for several yes/no questions on a survey.
How many people \(n\) does she have to randomly sample (without replacement) to ensure that her estimates \(\hat
\) are within \(\epsilon=0.04\) of the true proportions \(p\)?
We can't even begin to address the answer to this question until we derive a confidence interval for a proportion for a small, finite population!
An approximate (\((1-\alpha)100\%\) confidence interval for a proportion \(p\) of a small population is:
We'll use the example above, where possible, to make the proof more concrete. Suppose we take a random sample, \(X_1, X_2, \ldots, X_n\), without replacement, of size \(n\) from a population of size \(N\). In the case of the example, \(N=2000\). Suppose also, unknown to us, that for a particular survey question there are \(N_1\) respondents who would respond "yes" to the question, and therefore \(N-N_1\) respondents who would respond "no." That is, our small finite population looks like this:
If that's the case, the true proportion (but unknown to us) of yes respondents is:
while the true proportion (but unknown to us) of no respondents is:
Now, let \(X\) denote the number of respondents in the sample who say yes, so that:
if \(X_i=1\) if respondent \(i\) answers yes, and \(X_i=0\) if respondent \(i\) answers no. Then, the proportion in the sample who say yes is:
Then, \(X=\sum\limits_^n X_i\) is a hypergeometric random variable with mean:
It follows that \(\hat
=X/n\) has mean \(E(\hat
)=p\) and variance:
Then, the Central Limit Theorem tells us that:
follows an approximate standard normal distribution. Now, it's just a matter of doing the typical confidence interval derivation, in which we start with a probability statement, manipulate the quantity inside the parentheses, and substitute sample estimates where necessary. We've done that a number of times now, so skipping all of the details here, we get that an approximate \((1-\alpha)100\%\) confidence interval for \(p\) of a small population is:
By the way, it is worthwhile noting that if the sample \(n\) is much smaller than the population size \(N\), that is, if \(n
and the confidence interval for \(p\) of a small population becomes quite similar to the confidence interval for \(p\) of a large population:
A researcher is studying the population of a small town in India of \(N=2000\) people. She's interested in estimating \(p\) for several yes/no questions on a survey.
How many people \(n\) does she have to randomly sample (without replacement) to ensure that her estimates \(\hat
\) are within \(\epsilon=0.04\) of the true proportion \(p\)?
Now that we know the correct formula for the confidence interval for \(p\) of a small population, we can follow the same procedure we did for determining the sample size for estimating a proportion \(p\) of a large population. The researcher's goal is to estimate \(p\) so that the error is no larger than 0.04. That is, the goal is to calculate a 95% confidence interval such that:
\(\hat
\pm \epsilon=\hat
\pm 0.04\)
Now, we know the formula for an approximate \((1-\alpha)100\%\) confidence interval for a proportion \(p\) of a small population is:
So, again, we should proceed by equating the terms appearing after each of the above \(\pm\) signs, and solving for \(n\). That is, equate:
and solve for \(n\). Doing the algebra yields:
That looks simply dreadful! Let's make it look a little more friendly to the eyes:
where \(m\) is defined as the sample size necessary for estimating the proportion \(p\) for a large population, that is, when a correction for the population being small and finite is not made. That is:
Now, before we make the calculation for our particular example, let's take a step back and summarize what we have just learned.
Estimating a population proportion \(p\) of a small finite populationThe sample size necessary for estimating a population proportion \(p\) of a small finite population with \((1-\alpha)100\%\) confidence and error no larger than \(\epsilon\) is:
is the sample size necessary for estimating the proportion \(p\) for a large population.
