Question

How a large sample should be surveyed to estimate the true proportion of college students who do a laundry once a week within 3% with 95% confidence?

Answer 1

$ME=z_{0.025}\sqrt{\frac{p(1-p)}{n}}$

$n=(\frac{z_{0.025}}{ME})^2p(1-p)=(\frac{1.96}{0.03})^20.5(1-0.5)=1068.$