I am working on the following problem.
"A CEO wants to know what proportion of the company’s employees are routinely late for work. What is the most conservative estimate of the sample size that would limit the margin of error to within 0.01 of the population proportion for a 90% confidence interval?"
This is a proportion CI so I know that the margin of error $m$ is
$$m = z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$
In this problem my understanding is that $m=.01$, $z_{\alpha/2} =1.96$ and $\hat{p}=.5$ to maximize the standard error value.
Now here is my issue, if I am not mistaken the value of $n$ must be at least $6766$ and I find that to be way beyond what I have expected.
Was there a mistake in calculation or assumption here?
I would appreciate your help.