In a survey, the planning value for the population proportion is p^* = 0.50. How large a sample should be taken to provide a 95% confidence interval with a margin of error of 0.04? Round your answer up to the next whole number.

Answer:
Given:
The population proportion $(p^*)=0.50$
The margin of error $(e)=0.04$
The confidence interval level $=95\%$

Solution:
→ The significance level at 95% confidence interval:
$(\alpha)=1-0.95=0.05$

→ The critical value at 0.05 significance level $(z_c)=1.96$

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

$$ \therefore e = z_c \cdot \sqrt{\frac{p(1 – p)}{n}} \quad $$ $$ \therefore 0.04 = 1.96 \cdot \sqrt{\frac{0.50(1 – 0.50)}{n}} \quad $$ $$ \therefore n = 600.25 $$ $$ \approx 601 $$

Final answer:
The sample size $(n) = 601 $