Answer:
Given:
The population proportion $(p^*)=0.30$
The margin of error $(e)=0.08$
The confidence interval level $=99%$
Solution: โ The significance level at 99% confidence interval: $(\alpha)=1-0.99=0.01$
โ The critical value at 0.01 significance level $(z_c)=2.576$
$ \Rightarrow $ The sample size $(n):$ โ Here we need to use a margin of error formula to find out the sample size
$$ \therefore e = z_c \cdot \sqrt{\frac{p(1 – p)}{n}} \quad $$ $$ \therefore 0.08 = 2.576 \cdot \sqrt{\frac{0.30(1 – 0.30)}{n}} \quad $$ $$ \therefore n = 218.413 $$ $$ \approx 218 $$
Final answer: The sample size $(n) = 218 $
