To calculate the number of voters needed for a $95\%$ confidence interval with a margin of error of $1\%$, we use the following formula for sample size $n$:

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

Where:

• $Z$ is the critical value for the desired confidence level ($1.96$ for $95\%$),
• $p$ is the proportion (approval rating, expressed as a decimal: $0.36$),
• $E$ is the margin of error (expressed as a decimal: $0.01$).

Step-by-Step Calculation:

1. $Z = 1.96$,
2. $p = 0.36$,
3. $1 - p = 0.64$,
4. $E = 0.01$.

Substitute these into the formula:

$n = \frac{1.96^2 \cdot 0.36 \cdot 0.64}{0.01^2}$

Simplify:

1. $1.96^2 = 3.8416$,
2. $0.36 \cdot 0.64 = 0.2304$,
3. $0.01^2 = 0.0001$.

Thus:

$n = \frac{3.8416 \cdot 0.2304}{0.0001}$

$n = \frac{0.8845056}{0.0001}$

$n = 8845.056$

Final Answer:

The required sample size is approximately 8,846 voters (rounded up to the nearest whole number).

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Would you like to explore more details or have any questions?

Related Questions:

1. How does changing the confidence level affect the sample size?
2. What happens to the sample size if the margin of error is reduced?
3. Can you calculate the sample size for a $90\%$ confidence interval with the same margin of error?
4. How is the critical value $Z$ determined for different confidence levels?
5. Why is rounding up necessary when calculating sample size?

Tip:

Always round up when calculating sample size, as even one voter less might not meet the desired margin of error.