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We are tasked with estimating the proportion of teenagers who want more family discussions about school and calculating a 99% confidence interval based on the data provided. Let's break down the information:

### Given:
- Sample size \( n = 536 \)
- Proportion of teenagers wanting more discussion about school \( p = 0.37 \) (37%)
- Confidence level = 99%

### Formula:
For a confidence interval for a population proportion, we use the formula:

\[
CI = p \pm Z \times \sqrt{\frac{p(1-p)}{n}}
\]

Where:
- \( p \) is the sample proportion (0.37)
- \( n \) is the sample size (536)
- \( Z \) is the Z-score corresponding to the confidence level (for 99%, \( Z = 2.576 \))

### Step 1: Calculate the margin of error
First, calculate the standard error (SE):

\[
SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37(1 - 0.37)}{536}} = \sqrt{\frac{0.37 \times 0.63}{536}} \approx 0.02095
\]

Now, calculate the margin of error (ME):

\[
ME = Z \times SE = 2.576 \times 0.02095 \approx 0.054
\]

### Step 2: Confidence interval
Now, construct the confidence interval:

\[
CI = 0.37 \pm 0.054
\]

This gives us:

\[
CI = (0.37 - 0.054, 0.37 + 0.054) = (0.316, 0.424)
\]

### Conclusion:
The correct answer is \( \mathbf{0.37 \pm 0.054} \).

Would you like further details or explanations on any of the steps? 

Here are some related questions you might find helpful:

1. How is the Z-score determined for a given confidence level?
2. Why do we use the square root in the standard error formula?
3. What are the implications of using a 99% confidence level compared to a 95% level?
4. How would the confidence interval change if the sample size were smaller?
5. How does the proportion \( p \) affect the width of the confidence interval?

**Tip:** The higher the confidence level, the wider the confidence interval becomes due to the larger Z-score used.

A newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the results of a poll, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk about religion. These and other percentages were based on a national sampling of 536 teenagers. Estimate the proportion of all teenagers who want more family discussions about school. Use a 99% confidence level.

Learn how to estimate the proportion of teenagers wanting more discussions about school using a 99% confidence interval. The problem uses a sample of 536 teenagers and highlights the formulas for calculating standard error and margin of error. This topic is suitable for Grades 11-12.

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