Answer to Question #188905 in Calculus for Alice

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,

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

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

$CI = 0.43 \pm 0.028$