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Let's solve the problem step by step.

The question asks for the margin of error for a sample mean at a 98% confidence level. The formula for the margin of error \( E \) is:

\[
E = z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}
\]

Where:
- \( z_{\alpha/2} \) is the critical value of the z-distribution for a 98% confidence level.
- \( s = 8 \) is the sample standard deviation.
- \( n = 17 \) is the sample size.

### Step 1: Find the critical value (\( z_{\alpha/2} \))
For a 98% confidence level, the critical value \( z_{\alpha/2} \) corresponds to a cumulative probability of \( 0.99 \) (since \( 1 - \frac{\alpha}{2} = 0.99 \)).

Looking up this value in a standard normal distribution table or using a calculator gives:
\[
z_{\alpha/2} \approx 2.326
\]

### Step 2: Apply the formula
Now, substitute the values into the formula for margin of error:

\[
E = 2.326 \cdot \frac{8}{\sqrt{17}}
\]

Let's calculate this.The margin of error at a 98% confidence level is approximately **4.51**.

Would you like more details on this calculation or have any other questions?

Here are five related questions you could explore:
1. How does increasing the sample size affect the margin of error?
2. What is the difference between margin of error and confidence interval?
3. How do you calculate the confidence interval using the margin of error?
4. What critical value would be used for a 95% confidence level?
5. How would the margin of error change if the standard deviation were higher?

**Tip:** A larger sample size reduces the margin of error, leading to more precise estimates.

Assuming the population has an approximate normal distribution, if a sample size n = 17 has a sample mean x̄ = 47 with a sample standard deviation s = 8, find the margin of error at a 98% confidence level. Round the answer to two decimal places.

Learn how to calculate the margin of error for a sample mean when given a sample size of 17, a sample mean of 47, and a standard deviation of 8. This problem involves using the critical value for a 98% confidence level. The solution is suitable for college-level statistics.

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