To calculate the minimum sample size required to estimate the population proportion when the margin of error, confidence level, and the values of $p$ and $q$ are unknown, we use the following formula for the sample size $n$:

$n = \frac{Z^2 \cdot p \cdot q}{E^2}$

Where:

• $Z$ is the Z-value corresponding to the given confidence level,
• $p$ is the estimated population proportion (if unknown, we assume $p = 0.5$),
• $q = 1 - p$,
• $E$ is the margin of error.

Step 1: Z-value for 99% confidence level

For a 99% confidence level, the Z-value corresponding to the middle 99% of the normal distribution is approximately 2.576.

Step 2: Set $p = 0.5$ and $q = 0.5$

Since $p$ and $q$ are unknown, we assume $p = 0.5$ and $q = 1 - p = 0.5$. This maximizes the required sample size and is the most conservative assumption.

Step 3: Use the margin of error

Given the margin of error $E = 0.028$, we can now plug the values into the formula.

$n = \frac{(2.576)^2 \cdot (0.5) \cdot (0.5)}{(0.028)^2}$

Step 4: Perform the calculation

Let’s calculate the value of $n$.The minimum sample size required to estimate the population proportion, given a margin of error of 0.028 and a 99% confidence level, is 2116.

Do you want more details on the calculation or have any questions?

Here are 5 related questions for further exploration:

1. How would the sample size change if the margin of error is reduced to 0.02?
2. What happens to the sample size if the confidence level is lowered to 95%?
3. Why do we assume $p = 0.5$ when $p$ and $q$ are unknown?
4. How would using a known population proportion impact the sample size?
5. How does the sample size formula differ for estimating means instead of proportions?

Tip: When $p$ and $q$ are unknown, assuming $p = 0.5$ is a conservative choice that ensures the largest sample size needed.