In a survey, the planning value for the population proportion is p^* = 0.10. How large a sample should be taken to provide a 91% confidence interval with a margin of error of 0.02? Round your answer up to the next whole number.

Answer:
Given:
The population proportion $(p^{\ast })=0.10$
The margin of error $(e)=0.02$
The confidence interval level $=91\mathrm{\%}$

Solution:
→ The significance level at 91% confidence interval:
$(\alpha )=1-0.91=0.09$

→ The critical value at 0.09 significance level $(z_c)=1.70$

$\Rightarrow$ The sample size $(n):$
→ Here we need to use a margin of error formula to find out the sample size

$\therefore e=z_c\cdot \sqrt{\frac{p(1–p)}{n}}\therefore 0.02=1.70\cdot \sqrt{\frac{0.10(1–0.10)}{n}}\therefore n=650.25\approx 650$

Final answer:
The sample size $(n)=650$