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 $(p^*)=0.45$
The margin of error $(e)=0.03$
The confidence interval level $=98\%$

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

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