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 25% of the respondents said they are confident with their country's banking system.

Solution

The formula for the minimum 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 estimate is available. Thus we assume that $p = 0.5$.

Given that margin of error $E =0.02$. The confidence coefficient is $1-\alpha = 0.99$. Thus $\alpha= 0.01$.

Z Critical value 0.01
Z Critical value 0.01

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

The minimum sample size required to estimate the proportion is

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

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. A prior study found that 25% of the respondents said they are confident with their country's banking system. Thus $p=0.25$.

Given that margin of error $E =0.02$. The confidence coefficient is $1-\alpha = 0.99$. That is $\alpha = 0.01$.

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

The minimum sample size required to estimate the proportion is

$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.25(1-0.25)\bigg(\frac{2.58}{0.02}\bigg)^2\\ &=3120.1875\\ &\approx 3121. \end{aligned} $$

Thus, the sample of size $n=3121$ will ensure that the $99$% confidence interval for the proportion will have a margin of error $0.02$.

Further Reading