In a survey, the planning value for the population proportion is p^* = 0.35. How large a sample should be taken to provide a 92% confidence interval with a margin of error of 0.06? Round your answer up to the next whole number.

Answer:
Given:
The population proportion $(p^*)=0.35$
The margin of error $(e)=0.06$
The confidence interval level $=92%$

Solution:
→ The significance level at 92% confidence interval:
$(\alpha)=1-0.92=0.08$

→ The critical value at 0.08 significance level $(z_c)=1.75$

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

$$ \therefore e = z_c \cdot \sqrt{\frac{p(1 – p)}{n}} \quad $$ $$ \therefore 0.06 = 1.75 \cdot \sqrt{\frac{0.35(1 – 0.35)}{n}} \quad $$ $$ \therefore n = 193.533 $$ $$ \approx 194 $$

Final answer:
The sample size $(n) = 194 $