To calculate the required sample size for a survey, the formula used is:

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

Where:

• $Z$ is the critical value corresponding to the confidence level (for 95% confidence, $Z \approx 1.96$),
• $p$ is the assumed proportion (0.5 in this case),
• $E$ is the margin of error (in decimal form).

Part (a): Margin of Error = 5% (0.05)

Plugging in the values: $n = \frac{1.96^2 \cdot 0.5 \cdot (1 - 0.5)}{0.05^2}$

$n = \frac{3.8416 \cdot 0.25}{0.0025}$

$n = \frac{0.9604}{0.0025} = 384.16$

Since sample size must be a whole number, round up to the nearest whole number: $n = 385$

Part (b): Margin of Error = 2% (0.02)

Plugging in the values: $n = \frac{1.96^2 \cdot 0.5 \cdot (1 - 0.5)}{0.02^2}$

$n = \frac{3.8416 \cdot 0.25}{0.0004}$

$n = \frac{0.9604}{0.0004} = 2401$

Since sample size must be a whole number, round up to the nearest whole number: $n = 2401$

Final Answers:

(a) To have a margin of error no greater than 5%: 385 students.
(b) To have a margin of error no greater than 2%: 2401 students.

---

Do you need detailed steps or clarification?

Here are 5 related questions to expand your understanding:

1. What happens to the required sample size if the confidence level increases to 99%?
2. How does changing the assumed proportion ($p$) affect the sample size?
3. Can the sample size formula be applied to smaller populations?
4. What if the margin of error were reduced to 1%—how large would the sample size need to be?
5. How is the $Z$-value determined for various confidence levels?

Tip: When the true proportion is unknown, using $p = 0.5$ provides the most conservative (largest) sample size estimate.