To determine the required sample size ($n$) for estimating a population proportion with a given margin of error, confidence level, and no preliminary estimate for the proportion, we use the following formula:

$n = \frac{{Z^2 \cdot p(1 - p)}}{{E^2}}$

Where:

• $Z$ is the Z-score corresponding to the confidence level.
• $p$ is the estimated population proportion (since there is no prior estimate, we use $p = 0.5$, which maximizes the sample size).
• $E$ is the margin of error.

Step 1: Identify the values

• Confidence level = 90% → $Z$-score for 90% confidence is 1.645.
• Margin of error $E$ = 2% or 0.02.
• Since we have no preliminary estimate for the population proportion, we use $p = 0.5$.

Step 2: Apply the formula

$n = \frac{{(1.645)^2 \cdot 0.5(1 - 0.5)}}{{(0.02)^2}}$

$n = \frac{{(2.706025) \cdot (0.25)}}{{0.0004}}$

$n = \frac{{0.67650625}}{{0.0004}}$

$n = 1691.265625$

Step 3: Final Answer

Since the sample size must be a whole number, round up to the nearest whole number:

$n = 1692$

Therefore, you would need a sample size of 1,692.

Would you like any further explanation or more details? Here are some related questions:

1. How does increasing the confidence level affect the sample size?
2. What happens to the required sample size if we have a prior estimate for $p$?
3. How do we calculate the Z-score for different confidence levels?
4. What impact does changing the margin of error have on sample size?
5. How is this formula derived from the margin of error formula?

Tip: Always round the sample size up to ensure the margin of error stays within the desired range.