Question

Solve for the confidence interval estimate of the population proportion.

Mikay conducted poll survey in which 520 of 1200 randomly selected voters indicated their preference for a certain candidate. What is the true population proportion of voters who prefer the candidate using 95% confidence interval?

Answer 1

We have given that,

$p = \dfrac{520}{1200} = 0.43$

Confidence interval can be given as,

$CI = p \pm z \sqrt{\dfrac{p(1-p)}{n}}$

$CI = 0.43 \pm 1.96 \sqrt{\dfrac{0.43 \times 0.57}{1200}}$

$CI = 0.43 \pm 0.028$