The standard error of the proportion is calculated using the formula:

$\sigma_p = \sqrt{\frac{p(1-p)}{n}}$

Where:

• $p$ is the population proportion.
• $n$ is the sample size.

Given:

$p = 0.38$, $1-p = 0.62$

Part a: $n = 30$

$\sigma_p = \sqrt{\frac{0.38 \cdot 0.62}{30}} = \sqrt{\frac{0.2356}{30}} = \sqrt{0.007853333} \approx 0.0886$

Part b: $n = 60$

$\sigma_p = \sqrt{\frac{0.38 \cdot 0.62}{60}} = \sqrt{\frac{0.2356}{60}} = \sqrt{0.003926667} \approx 0.0627$

Part c: $n = 90$

$\sigma_p = \sqrt{\frac{0.38 \cdot 0.62}{90}} = \sqrt{\frac{0.2356}{90}} = \sqrt{0.002617778} \approx 0.0512$

Final Answers:

a) $\sigma_p = 0.0886$
b) $\sigma_p = 0.0627$
c) $\sigma_p = 0.0512$

Let me know if you need further clarification or assistance!

Related Questions:

1. What does the standard error of a proportion represent?
2. How does increasing the sample size affect the standard error?
3. Why is it important to round to four decimal places in statistical calculations?
4. What are some applications of the standard error in real-world data analysis?
5. How would the standard error change if the proportion $p$ were closer to 0 or 1?

Tip:

When calculating the standard error, always double-check your values for $p$ and $n$, as small errors can significantly affect the results.