In a survey, the planning value for the population proportion is p^* = 0.45. How large a sample should be taken to provide a 98% confidence interval with a margin of error of 0.03? Round your answer up to the next whole number.

Answer:
Given:
The population proportion $(p^{\ast })=0.45$
The margin of error $(e)=0.03$
The confidence interval level $=98\mathrm{\%}$

Solution:
→ The significance level at 98% confidence interval:
$(\alpha )=1-0.98=0.02$

→ The critical value at 0.02 significance level $(z_c)=2.33$

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

$\therefore e=z_c\cdot \sqrt{\frac{p(1–p)}{n}}\therefore 0.03=2.33\cdot \sqrt{\frac{0.45(1–0.45)}{n}}\therefore n=1492.948\approx 1493$

Final answer:
The sample size $(n)=1493$