## Elementary Statistics (12th Edition)

If $\hat{p}$ is known:$n=\frac{z^2_{\frac{\alpha}{2}}\cdot \hat{p}\cdot (1-\hat{p})}{E^2}.$ If $\hat{p}$ is unknown:$n=\frac{z^2_{\frac{\alpha}{2}}\cdot0.25}{E^2}.$ a) Here, $\hat{p}$ is unknown, hence $n=\frac{2.575^2\cdot0.25}{0.01^2}=16577.$ b)Here, $\hat{p}$ is known, it is 90%=0.9, hence $n=\frac{1.96^2\cdot(0.9)\cdot(1-0.9)}{0.01^2}=5968.$ c)Yes, because it reduces the result from a) by more than 10000, which is about 60-70% of the value, which is a lot.