# Solved: What sample size would you need to construct a 96 percent confidence interval around the proportion for voter turnout during the next election

#### ByDr. Raju Chaudhari

Feb 27, 2021

What sample size would you need to construct a 96 percent confidence interval around the proportion for voter turnout during the next election that would provide a margin of error of 4 percent? Assume the population proportion has been estimated at 55 percent.

#### Solution

The formula to estimate the sample size required to estimate the proportion is

 $$n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2$$
where $p$ is the proportion of success, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

Given that the proportion is $p =0.55$, margin of error $E =0.04$. The confidence coefficient is $1-\alpha = 0.96$. The critical value of $Z$ is $Z_{\alpha/2} =Z_{0.02}= 2.0537$.

The minimum sample size required to estimate the proportion is

 \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.55(1-0.55)\bigg(\frac{2.0537}{0.04}\bigg)^2\\ &=652.4229\\ &\approx 653. \end{aligned}

Thus, the sample of size $n=653$ will ensure that the $0.96$\% confidence interval for the proportion will have a margin of error $0.04$.