To calculate the required sample size for this poll, we can use the formula for sample size in proportion estimation:

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

where:

• $Z$ is the Z-score corresponding to the desired confidence level (for 99% confidence, $Z = 2.576$),
• $p$ is the estimated proportion of people supporting the candidate. If we don't have a prior estimate, we use $p = 0.5$ to maximize the sample size,
• $E$ is the margin of error, which is 1% or $0.01$ in this case.

Step-by-Step Calculation:

1. Identify the Z-score: For a 99% confidence level, $Z = 2.576$.
2. Assume proportion (p): We use $p = 0.5$ as no prior data is given.
3. Set margin of error (E): $E = 0.01$.

Plugging in these values:

$n = \frac{(2.576)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.01)^2}$

Calculating each component:

$n = \frac{6.635 \cdot 0.5 \cdot 0.5}{0.0001}$ $n = \frac{1.65875}{0.0001}$ $n \approx 16587.5$

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

$n = 16588$

Answer:

The required sample size is 16,588 people.

Would you like more details on each step or further explanation?

Here are five related questions that expand on this concept:

1. How would the required sample size change if the confidence level were 95% instead of 99%?
2. What would happen to the sample size if we had a prior estimate that $p = 0.6$?
3. How does the margin of error influence the sample size requirement?
4. Why do we use $p = 0.5$ when there is no prior data?
5. How is the Z-score determined for different confidence levels?

Tip: When estimating proportions, using $p = 0.5$ provides the largest possible sample size, ensuring the estimate remains valid even in the worst case.