A gigantic warehouse located in Tampa, Florida, stores approximately 60 million empty aluminum beer and soda cans. Recently, a fire occurred at the warehouse. The smoke from the fire contaminated many of the cans with black spot, rendering them unusable. A University of South Florida statistician was hired by the insurance company to estimate the true proportion of cans in the warehouse that were contaminated by the fire. How many aluminum cans should be randomly sampled to estimate the true proportion to within .02 with 90% confidence?


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 $E =0.02$. The confidence coefficient is $0.9$. Assume that the proportion is $p =0.5$.

Z-critical value
Z-critical value

The critical value of $Z$ is $Z_{\alpha/2} =Z_{0.05}= 1.645$.

The minimum sample size required to estimate the proportion is

$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.5(1-0.5)\bigg(\frac{1.645}{0.02}\bigg)^2\\ &=1691.2656\\ &\approx 1692. \end{aligned} $$

Thus, the sample of size $n=1692$ will ensure that the $90$% confidence interval for the proportion will have a margin of error $0.02$.

Further Reading