Very frequently asked question in statistical consulting is, how large should the sample size be to estimate a mean?

The answer will depend on the variation associated with the random variable under observation. The statistician could correctly respond, only one item is needed, provided that the standard deviation of the distribution is zero. That is, if σ σ is equal zero, then the value of that one item would necessarily equal the unknown mean of the distribution. This is the extreme case and one that is not met in practice. However, the smaller the variance, the smaller the sample size needed to achieve a given degree of accuracy.

The type of statistic we see most often in newspaper and magazines is an estimate of a proportion p. We might, for example, want to know the percentage of the labor force that is unemployed or the percentage of voters favoring a certain candidate. Sometimes extremely important decisions are made on the basis of these estimates. If this is the case, we would most certainly desire short confidence intervals for p with large confidence coefficients. We recognize that these conditions will require a large sample size. On the other hand, if the fraction p being estimated is not too important, an estimate associated with a longer confidence interval with a smaller confidence coefficients is satisfactory; and thus a smaller sample size can be used.

In general, to find the required sample size to estimate p, recall that the point estimate of p is p ^ = z α/2 p ^ ( 1− p ^ ) n . p ^ = z α/2 p ^ ( 1− p ^ ) n .

Suppose we want an estimate of p that is within ε ε of the unknown p with 100( 1−α )% 100( 1−α )% confidence where ε= z α/2 p ^ ( 1− p ^ )/n ε= z α/2 p ^ ( 1− p ^ )/n is the maximum error of the point estimate p ^ =y/n p ^ =y/n . Since p ^ p ^ is unknown before the experiment is run, we cannot use the value of p ^ p ^ in our determination of n. However, if it is known that p is about equal to p * p * , the necessary sample size n is the solution of ε= z α/2 p ∗ ( 1− p ∗ ) n . ε= z α/2 p ∗ ( 1− p ∗ ) n . That is, n= z α/2 2 p ∗ ( 1− p ∗ ) ε 2 . n= z α/2 2 p ∗ ( 1− p ∗ ) ε 2 .