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