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To solve this problem, we first need to calculate the margin of error and the confidence interval based on the information provided. Here's how we proceed:

### Step 1: Gather Information
- Sample proportion (\(\hat{p}\)): 0.58 (since 58% of employers were likely to require higher contributions).
- Sample size (\(n\)): 743.
- Confidence level: 99%.

### Step 2: Find the Z-Score for a 99% Confidence Level
A 99% confidence level corresponds to a Z-score of approximately 2.576 (using the standard Z-table).

### Step 3: Calculate the Standard Error (SE)
The standard error for a proportion is given by:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
Substituting the values:
\[
SE = \sqrt{\frac{0.58 \times (1 - 0.58)}{743}}
\]
\[
SE = \sqrt{\frac{0.58 \times 0.42}{743}} \approx \sqrt{\frac{0.2436}{743}} \approx 0.0181
\]

### Step 4: Calculate the Margin of Error (ME)
\[
ME = Z \times SE
\]
\[
ME = 2.576 \times 0.0181 \approx 0.0466
\]

### Step 5: Calculate the Confidence Interval
The confidence interval is given by:
\[
\text{Confidence Interval} = \hat{p} \pm ME
\]
\[
\text{Lower Bound} = 0.58 - 0.0466 = 0.5334
\]
\[
\text{Upper Bound} = 0.58 + 0.0466 = 0.6266
\]

### Final Answers
- **Margin of Error**: 0.0466 (rounded to four decimal places)
- **99% Confidence Interval**: (0.5334, 0.6266) (rounded to four decimal places)

Let me know if you'd like further details or have any questions.

### Additional Questions:
1. What is the formula for calculating the margin of error for a proportion?
2. How does the sample size affect the margin of error and confidence interval?
3. Why is the Z-score different for varying confidence levels?
4. How would the margin of error change if the confidence level was 95% instead of 99%?
5. What assumptions are made when calculating confidence intervals for proportions?

**Tip**: Always ensure the sample size is large enough when calculating confidence intervals for proportions to meet the conditions for normal approximation.

A recent survey showed that 58% of U.S. employers were likely to require higher employee contributions for healthcare coverage in the upcoming year. The sample size is 743, and we are asked to compute the margin of error and the 99% confidence interval.

Learn how to calculate the 99% confidence interval for the proportion of U.S. employers likely to require higher healthcare contributions. This problem involves understanding margin of error, sample size, and the Z-score for 99% confidence level.

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