In a survey, the planning value for the population proportion is p^* = 0.40. How large a sample should be taken to provide a 90% confidence interval with a margin of error of 0.05? Round your answer up to the next whole number.

Answer:
Given:
The population proportion $(p^*)=0.40$
The margin of error $(e)=0.05$
The confidence interval level $=90%$

Solution:
→ The significance level at 90% confidence interval:
$(\alpha)=1-0.90=0.10$

→ The critical value at 0.10 significance level $(z_c)=1.645$

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

$$ \therefore e = z_c \cdot \sqrt{\frac{p(1 – p)}{n}} \quad $$ $$ \therefore 0.05 = 1.645 \cdot \sqrt{\frac{0.40(1 – 0.40)}{n}} \quad $$ $$ \therefore n = 258.202 $$ $$ \approx 258$$

Final answer:
The sample size $(n) = 258 $