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 $0.04$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
