Assignment two:

IIlustrative Example $1\mathrm{.}1$

Assuming $N_0$ represents the sample size; $T=$ tolerance error $)=(0\mathrm{.}05$ prevalence from a previous study $)=(18\mathrm{.}5\%$; and, $Z=1\mathrm{.}96,$ is the level of significance $($or standard normal deviate$)$ which corresponds to $95\%$ confidence level.

Substitute into Eq$.(1\mathrm{.}3)$:

$N_0=\frac{{1\mathrm{.}96}^2(0\mathrm{.}185)(1-0\mathrm{.}185)}{{0\mathrm{.}05}^2}$~~$232$ persons Remove $18\mathrm{.}5\%$ for P and take it as unknown and sample size is $300.$ Solve for P$.$