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Now suppose $X$ is a binary variable measured on a population in which an unknown proportion $p$ of the population meets some condition of interest, with $X = 1$ if the subject meets that condition and $X = 0$ otherwise. Thus:

$$X \sim B(1,p)$$

Let $X_1,...,X_n$ denote planned measurements of $X$ on a random sample of size $n$ from the population. Then:

$$S = X_1 + \cdot\cdot\cdot + X_n \sim B(n,p)$$

Let $\hat{p} = \cfrac{S}{n}$. We saw previously that:

$$E(\hat{p}) = p \,\,\,\,\,\,\,\,\,\,\,\,\,\, \text{and} \,\,\,\,\,\,\,\,\,\,\,\,\,\, V(\hat{p}) = \cfrac{p(1 - p)}{n}$$

Moreover, we learned previously that when $n$ is large, the Central Limit Theorem implies that $\hat{p}$ has an approximate

$$N\bigg(p,\cfrac{p(1-p)}{n}\bigg)$$ distribution.

Unfortunately, we cannot use this distribution to make a CI for $p$, because the variance of $\hat{p}$ depends on the unknown value of $p$.

But now consider $\tilde{p} = \cfrac{S + 2}{n + 4}$.

It has been found (Agresti and Coull, 1998) that when $n$ is large, the distribution of $\tilde{p}$ is well-approximated by the

$$N\bigg(p,\cfrac{\tilde{p}(1 - \tilde{p})}{n + 4}\bigg)$$ distribution.

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If you are planning a study and you prefer a CI for $p$ to have a margin of error no larger than some $K > 0$ while maintaining the same confidence level $1 - \alpha$, you must determine the minimum sample size $n$ necessary to achieve this goal.

However, the variance of $\tilde{p}$ depends on $\tilde{p}$, which is unknown to you before data are collected. But this variance is maximized when $\tilde{p}(1-\tilde{p})$ is maximized, that is, when $\tilde{p} = 0.5$.

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