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To calculate the margin of error given a standard error of 4.29 and a 98% confidence level, we use the following formula:

\[
\text{Margin of Error} = z_{\alpha/2} \times \text{Standard Error}
\]

1. **Standard error** is given as 4.29.
2. **Confidence level** is 98%. 

The value of \( z_{\alpha/2} \) for a 98% confidence level is approximately **2.33** (this is the critical value from the standard normal distribution for 98% confidence).

Now, using the formula:

\[
\text{Margin of Error} = 2.33 \times 4.29
\]

\[
\text{Margin of Error} = 9.9957 \approx 10.00
\]

So, the margin of error is approximately **10.00**.

Would you like further details or clarifications?

### Related Questions:
1. How is the \( z_{\alpha/2} \) value derived for different confidence levels?
2. What is the relationship between confidence level and margin of error?
3. How do we interpret the margin of error in the context of confidence intervals?
4. What happens to the margin of error if the standard error increases or decreases?
5. How does a change in sample size affect the margin of error?

#### Tip:
Remember, as the confidence level increases, the margin of error also increases, because higher confidence requires a wider interval.

Standard error = 4.29, confidence level 98%. The margin of error is ?

Learn how to calculate the margin of error with a standard error of 4.29 and a 98% confidence level. This problem involves statistical concepts such as confidence intervals and normal distribution (Z-scores). Suitable for students in Grades 10-12.

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