$n \ge \left(\frac{z_{\alpha/2}}{2}\right)^2p(1-p) = \left(\frac{1.96}{0.03}\right)^2 \frac{19}{20}\frac{1}{20} =202.75$?

Question

A polling firm proposes to estimate proportions in a population by the corresponding proportions in the samples so that the margin of error is at most $3$%, and this $19$ times out of $20$ on average.

What is the minimum sample size here? The answer is $1068$ here.

For me that answer is wrong because

$$n \ge \left(\frac{z_{\alpha/2}}{\text{margin_error}}\right)^2p(1-p) = \left(\frac{1.96}{0.03}\right)^2 \frac{19}{20}\frac{1}{20} =202.75.$$

So the minimum size must be $203$. Where am I wrong?

Answer 1

The wording of the question is not especially clear, but I understand the 19 out of 20 as giving you only the value of $\alpha$. Consequently the problem does not explicitly tell you what value of $p$ to use to estimate the standard deviation. This would imply that the one asking you the question doesn't have any idea about $p$ and that thus you as the analyst must make the worst possible assumption about it (the one that maximizes the standard deviation).

Thankfully in the case of the binomial distribution this is actually possible: you just estimate $p$ as $1/2$. Doing that seems to get you the desired answer here.