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.