To calculate the margin of error $E$ at a 99% confidence level, we use the following formula:

$E = z_{\alpha/2} \times \frac{s}{\sqrt{n}}$

Where:

• $n = 20$ (sample size),
• $\bar{x} = 49$ (sample mean, though it's not used for the margin of error),
• $s = 5$ (sample standard deviation),
• $z_{\alpha/2}$ is the z-value corresponding to a 99% confidence level.

The z-value for a 99% confidence level is approximately 2.576.

Now, applying the formula:

$E = 2.576 \times \frac{5}{\sqrt{20}} = 2.576 \times \frac{5}{4.4721} \approx 2.576 \times 1.118$

$E \approx 2.879$

Thus, the margin of error is approximately 2.88 (rounded to two decimal places).

Would you like more details or have any questions?

Related Questions:

1. How do we interpret the margin of error in this context?
2. What is the importance of the confidence level in determining the margin of error?
3. How would the margin of error change if the sample size increased?
4. What z-value corresponds to other confidence levels (e.g., 90%, 95%)?
5. How can we use this margin of error to construct a confidence interval?

Tip: The larger the sample size, the smaller the margin of error, as the sample better approximates the population.