To construct a confidence interval for the proportion of left-handed golfers, we use the given dataTo construct a confidence interval for the proportion of left-handed golfers, we use the given

• Sample proportion ($\hat{p}$): $0.15$ (since 15% of 2480 golfers are left-handed)
• Margin of error (ME): $0.03$ (given as 3%)

The confidence interval (CI) is calculated as follows:

$\text{CI} = \left( \hat{p} - \text{ME}, \hat{p} + \text{ME} \right)$

Substituting the values:

$\text{CI} = \left( 0.15 - 0.03, 0.15 + 0.03 \right) = \left( 0.12, 0.18 \right)$

So, the correct answer is:

D. $(0.12, 0.18)$

Would you like more details or have any questions?

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Here are some additional questions related to confidence intervals:

1. What is the interpretation of a 95% confidence interval in statistics?
2. How does sample size affect the margin of error in a confidence interval?
3. What is the formula for calculating the margin of error in a proportion confidence interval?
4. How does increasing the confidence level affect the width of the confidence interval?
5. What assumptions must be met for constructing a confidence interval for a proportion?
6. How would the confidence interval change if the margin of error was reduced?
7. Can you explain the difference between a confidence interval and a prediction interval?
8. How would you calculate the necessary sample size to achieve a specific margin of error?

Tip: Remember that the confidence interval gives a range within which we expect the true population proportion to lie, with a certain level of confidence (e.g., 95%).