To determine the sample size needed for a poll with a specified margin of error and confidence level, we can follow a step-by-step process using standard statistical formulas for sample size calculations. Here’s the detailed solution:
### Step-by-Step Solution:
1. Identify the given parameters:
- Margin of error (E): 3.73 percentage points, which we convert to decimal form, [tex]\( E = 0.0373 \)[/tex].
- Confidence level: 99%, which corresponds to a critical value:
- From the provided table, the critical value for a 99% confidence level ([tex]\( z^* \)[/tex]) is 2.576.
2. Assume the proportion (p) of the population:
- Often, we assume [tex]\( p = 0.5 \)[/tex] because it provides the maximum sample size (most conservative estimate).
3. Use the margin of error formula for proportions:
The margin of error for a proportion can be given by:
[tex]\[
E = z^* \times \sqrt{\frac{p(1-p)}{n}}
\][/tex]
Where [tex]\( n \)[/tex] is the sample size we need to solve for.
4. Rearrange the formula to solve for [tex]\( n \)[/tex]:
[tex]\[
n = \left( \frac{z^* \times \sqrt{p(1-p)}}{E} \right)^2
\][/tex]
Plugging in the known values:
- [tex]\( z^* = 2.576 \)[/tex]
- [tex]\( p = 0.5 \)[/tex]
- [tex]\( E = 0.0373 \)[/tex]
5. Calculate the sample size:
[tex]\[
n = \left( \frac{2.576 \times \sqrt{0.5 \times 0.5}}{0.0373} \right)^2
\][/tex]
[tex]\[
n = \left( \frac{2.576 \times \sqrt{0.25}}{0.0373} \right)^2
\][/tex]
[tex]\[
n = \left( \frac{2.576 \times 0.5}{0.0373} \right)^2
\][/tex]
[tex]\[
n = \left( \frac{1.288}{0.0373} \right)^2
\][/tex]
[tex]\[
n = \left( 34.5403 \right)^2
\][/tex]
[tex]\[
n \approx 1192.9
\][/tex]
6. Round up to the nearest whole number:
Since sample size must be a whole number and we typically round up to ensure our sample is large enough:
[tex]\[
n \approx 1193
\][/tex]
Therefore, to achieve a 99% confidence interval with a margin of error of 3.73 percentage points, at least [tex]\( \boxed{1193} \)[/tex] voters should be sampled.
