To solve this problem, we can use the formula for estimating the sample size needed to estimate a population proportion when no preliminary estimate is available. The formula is:

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

Where:

• $Z$ is the Z-score corresponding to the desired confidence level.
• $p$ is the estimated population proportion (since no estimate is given, we use $p = 0.5$, which maximizes the sample size).
• $E$ is the margin of error (in this case, 2% or 0.02).

For a 90% confidence level, the Z-score is approximately 1.645.

Now, we can plug in the values:

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

First, calculate each part:

$1.645^2 = 2.706025$

$0.5(1 - 0.5) = 0.25$

$0.02^2 = 0.0004$

Now, plug them in:

$n = \left( \frac{2.706025 \times 0.25}{0.0004} \right) = \left( \frac{0.67650625}{0.0004} \right) = 1691.27$

Therefore, the required sample size is 1692 (since sample size must be a whole number, we round up).

Let me know if you'd like further clarification! Would you like additional details or have any questions?

Here are five related questions:

1. What if the confidence level was increased to 95%? How would the sample size change?
2. How does the margin of error affect the sample size?
3. What if we had a preliminary estimate of the population proportion? How would that change the formula?
4. How does the Z-score change for different confidence levels?
5. How do we calculate the Z-score for a given confidence level?

Tip: Always round up your sample size result, even if the decimal part is small. This ensures you have a sufficient sample.