To solve this problem, we need to find the margin of error (ME) for a 99% confidence level.

The formula for the margin of error is:

$ME = z^* \times \frac{s}{\sqrt{n}}$

Where:

• $z^*$ is the z-score corresponding to the confidence level,
• $s$ is the sample standard deviation,
• $n$ is the sample size.

Step 1: Identify the values from the problem

• $n = 32$ (sample size),
• $\bar{x} = 38$ (sample mean, but not needed for margin of error calculation),
• $s = 8$ (sample standard deviation),
• Confidence level = 99%.

Step 2: Find the critical value $z^*$

For a 99% confidence level, the critical value $z^*$ is approximately 2.576 (from the standard normal distribution table).

Step 3: Apply the formula

Now, substitute the known values into the margin of error formula:

$ME = 2.576 \times \frac{8}{\sqrt{32}}$

$ME = 2.576 \times \frac{8}{5.6569}$

$ME = 2.576 \times 1.4142 \approx 3.644$

Final answer:

The margin of error is approximately 3.64 (rounded to two decimal places).

Would you like further explanation on any of the steps?

Here are 5 related questions:

1. What is the significance of using a 99% confidence level?
2. How would the margin of error change if the sample size increased?
3. What is the formula to calculate margin of error for proportions?
4. How do you interpret the confidence interval in real-world terms?
5. Why is the sample mean $\bar{x}$ not necessary to calculate the margin of error?

Tip: A larger sample size reduces the margin of error, improving the precision of your estimate.