may be assessed based on the power of a hypothesis test. For example, if we are comparing the support for a certain political candidate among women with the support for that candidate among men, we may wish to have 80% power to detect a difference in the support levels of 0.04 units. Estimation Estimation of a proportion A relatively simple situation is estimation of a proportion. For example, we may wish to estimate the proportion of residents in a community who are at least 65 years old. The estimator of a proportion is , where X is the number of 'positive' e.g., the number of people out of the n sampled people who are at least 65 years old). When the observations are independent, this estimator has a (scaled) binomial distribution (and is also the sample mean of data from a Bernoulli distribution). The maximum variance of this distribution is 0.25, which occurs when the true parameter is p = 0.5. In practice, since p is unknown, the maximum variance is often used for sample size assessments. If a reasonable estimate for p is known the quantity may be used in place of 0.25. For sufficiently large n, the distribution of will be closely approximated by a normal distribution. Using this and the Wald method for the binomial distribution, yields a confidence interval of the form , where Z is a standard Z-score for the desired level of confidence (1.96 for a 95% confidence interval). If we wish to have a confidence interval that is W units total in width (W/2 on each side of the sample mean), we will solve for n, yielding the sample size , in the case of using .5 as the most conservative estimate of the proportion. (Note: W/2 = margin of error.) In the figure below one can observe how sample sizes for binomial proportions change given different confidence levels and margins of error. Otherwise, the formula would be , which yields . For example, if we are interested in estimating the proportion of the US population who supports a particular presidential candidate, and we want the width of 95% confidence interval to be at most 2 percentage points (0.02), then we would need a sample size of (1.96)2/ (0.022) = 9604. It is reasonable to use the 0.5 estimate for p in this case because the presidential races are often close to 50/50,