To determine the required sample size $n$ for a 99.5% confidence interval with a margin of error of 0.07, we can use the following formula for estimating proportions:

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

where:

• $z$ is the z-score corresponding to the desired confidence level,
• $p$ is the estimated proportion (0.67 in this case),
• $E$ is the margin of error (0.07).

Step 1: Find the z-score

For a 99.5% confidence level, the z-score $z$ is approximately 2.807.

Step 2: Plug in the values

Given:

• $p = 0.67$
• $E = 0.07$
• $z = 2.807$

$n = \frac{(2.807)^2 \cdot 0.67 \cdot (1 - 0.67)}{(0.07)^2}$

Step 3: Calculate the sample size

1. Calculate $z^2$: $(2.807)^2 = 7.8784$
2. Calculate $p(1 - p)$: $0.67 \times (1 - 0.67) = 0.67 \times 0.33 = 0.2211$
3. Substitute into the formula:

$n = \frac{7.8784 \cdot 0.2211}{0.0049} = \frac{1.7410}{0.0049} \approx 355.31$

Since the sample size must be a whole number, we round up to the nearest whole number:

$n \approx 356$

Answer

A sample of 356 elementary school children is needed to obtain a 99.5% confidence interval with a margin of error of 0.07.

Would you like more details or have any questions?

Related Questions:

1. How do we interpret a 99.5% confidence interval in this context?
2. Why do we round up the sample size instead of rounding to the nearest integer?
3. How would the sample size change if the desired margin of error were smaller, such as 0.05?
4. What if the confidence level were reduced to 95%—how would that affect the sample size?
5. Why is it important to have an accurate estimate for the proportion $p$ when calculating sample size?

Tip:

When calculating sample sizes, always ensure the confidence level and margin of error align with the study's goals, as a higher confidence level or smaller margin of error will generally require a larger sample.