To determine the required sample size to estimate a population proportion when you have no preliminary estimate of the population proportion, you can use the following formula for sample size calculation:

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

Where:

• $Z$ is the z-score corresponding to the desired confidence level.
• $p$ is the estimated population proportion (if unknown, use 0.5 as it maximizes the sample size).
• $E$ is the margin of error (in decimal form).

Step 1: Identify the parameters

• Confidence level = 98%, so the corresponding z-score $Z = 2.33$ (you can find this from z-tables or standard normal distribution).
• Margin of error $E = 1.5\% = 0.015$.
• Since no preliminary estimate of the population proportion is available, use $p = 0.5$.

Step 2: Plug the values into the formula

$n = \frac{{(2.33)^2 \cdot 0.5(1 - 0.5)}}{(0.015)^2}$

Step 3: Calculate

$n = \frac{{(5.4289) \cdot 0.5(0.5)}}{0.000225}$ $n = \frac{{5.4289 \cdot 0.25}}{0.000225}$ $n = \frac{{1.357225}}{0.000225}$ $n \approx 6021.0$

Thus, the required sample size is approximately 6021.

Conclusion:

To estimate the population proportion with 98% confidence and a margin of error of 1.5%, you would need a sample size of 6021.

Would you like more details or explanations on any part of this solution?

Here are some related questions for further exploration:

1. What if the margin of error was reduced to 1%? How would it affect the sample size?
2. How is the z-score for different confidence levels determined?
3. What if you had a preliminary estimate of the population proportion? How would the calculation change?
4. How would a larger or smaller confidence level affect the required sample size?
5. Why do we use $p = 0.5$ when no prior estimate is available?

Tip: Always round up your sample size to the next whole number, as partial samples are not possible in real-world scenarios.