To determine the required sample size $n$ for a 90% confidence level with a 4.74% margin of error and a reported support proportion of 44%, we’ll use the formula for sample size in a proportion:

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

where:

• $Z$ is the z-value for the confidence level (90% confidence corresponds to $Z = 1.645$),
• $\sigma = \sqrt{p(1 - p)}$, with $p = 0.44$,
• $E$ is the margin of error (4.74% or 0.0474).

Step-by-Step Calculation

1. Calculate the Standard Deviation, $\sigma$: $\sigma = \sqrt{0.44 \times (1 - 0.44)} = \sqrt{0.44 \times 0.56} = \sqrt{0.2464} \approx 0.4964$

2. Plug Values into the Formula: $n = \left( \frac{1.645 \times 0.4964}{0.0474} \right)^2$

3. Calculate the Sample Size:

   • Compute the numerator: $1.645 \times 0.4964 \approx 0.8165$
   • Divide by the margin of error: $\frac{0.8165}{0.0474} \approx 17.23$
   • Square the result: $n \approx (17.23)^2 \approx 296.88$

Rounding up to the nearest whole number (since sample size must be an integer), we get: $n = 297$

Answer:

The required sample size is approximately 297 voters.

Would you like more details on the steps, or do you have any other questions?

Related Questions:

1. How does changing the confidence level affect the sample size needed?
2. What happens to the sample size if the margin of error is decreased?
3. How is the sample size formula derived for population proportions?
4. Why do we round up sample sizes in confidence interval calculations?
5. How would this calculation differ with a larger margin of error?

Tip:

Always round up the sample size to ensure the margin of error is met within the desired confidence level.