A researcher wishes to estimate, with 99% confidence, the population proportion of adults who are confident with their country's banking system. His estimate must be accurate within 2% of the population proportion.

a. No preliminary estimate is available. Find the minimum sample size needed.
b. Find the minimum sample size needed, using a prior study that found that 24% of the respondents said they are confident with their country's banking system.
c. Compare the results from parts (a) and (b).

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.

a. No preliminary estimates is available. Assume that $p =0.5$.

Given that margin of error $E =0.02$. The confidence coefficient is $0.99$.

The critical value of $Z$ is $Z_{\alpha/2} = 2.58$.

z-critical0.01
z-critical 0.01

The minimum sample size required to estimate the proportion is

$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.5(1-0.5)\bigg(\frac{2.58}{0.02}\bigg)^2\\ &=4160.25\\ &\approx 4161. \end{aligned} $$

Thus, the sample of size $n=4161$ will ensure that the $99$% confidence interval for the proportion will have a margin of error $0.02$.

b. Given that $p = 0.24$.

Given that margin of error $E =0.025$. The confidence coefficient is $0.99$.

The critical value of $Z$ is $Z_{\alpha/2} = 2.58$.

The minimum sample size required to estimate the proportion is

$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.24(1-0.24)\bigg(\frac{2.58}{0.025}\bigg)^2\\ &=1942.6038\\ &\approx 1943. \end{aligned} $$

Thus, the sample of size $n=1943$ will ensure that the $99$% confidence interval for the proportion will have a margin of error $0.025$.

c. Having an estimate of the population proportion reduces the minimum sample size needed.

Further Reading