How many people should be surveyed if we want to estimate the proportion of students who prefer using the test center? Assume we want 90% confidence we are within 5% of the true population proportion and that a prior study showed 53% preferred it.
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 margin of error $E =0.05$. The confidence coefficient is $1-\alpha=0.9$. The sample proportion is $p =0.53$.
The critical value of $Z$ is $Z_{\alpha/2} =Z_{0.05}= 1.64$.
The minimum sample size required to estimate the proportion is
$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.53(1-0.53)\bigg(\frac{1.64}{0.05}\bigg)^2\\ &=267.992\\ &\approx 268. \end{aligned} $$
Thus, the sample of size $n=268$ will ensure that the $90$% confidence interval for the proportion will have a margin of error $0.05$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
