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 $(p^*)=0.10$
The margin of error $(e)=0.02$
The confidence interval level $=91\%$

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

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