In a survey, the planning value for the population proportion is p^* = 0.15. How large a sample should be taken to provide a 94% confidence interval with a margin of error of 0.09? Round your answer up to the next whole number.

Answer:
Given:
The population proportion $(p^*)=0.15$
The margin of error $(e)=0.09$
The confidence interval level $=94%$

Solution:
→ The significance level at 94% confidence interval:
$(\alpha)=1-0.94=0.06$

→ The critical value at 0.06 significance level $(z_c)=1.88$

$ \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 0.09 = 1.88 \cdot \sqrt{\frac{0.15(1 – 0.15)}{n}} \quad \therefore n = 55.634 \approx 56 $

Final answer:
The sample size $(n) = 56 $