To construct a 99% confidence interval for the population proportion $p$, we can use the formula for the confidence interval of a proportion:

$\hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

where:

• $\hat{p}$ is the sample proportion.
• $Z_{\alpha/2}$ is the critical value for the 99% confidence level (for 99%, $Z_{\alpha/2} \approx 2.576$).
• $n$ is the sample size.

Step 1: Calculate the sample proportion ($\hat{p}$)

$\hat{p} = \frac{33}{250} = 0.132$

Step 2: Find the critical value $Z_{\alpha/2}$

For a 99% confidence level, the critical value is: $Z_{\alpha/2} = 2.576$

Step 3: Compute the standard error

$\text{Standard Error} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.132(1 - 0.132)}{250}}$ $\text{Standard Error} = \sqrt{\frac{0.132 \times 0.868}{250}} = \sqrt{\frac{0.114576}{250}} = \sqrt{0.0004583} = 0.02142$

Step 4: Calculate the margin of error

$\text{Margin of Error} = Z_{\alpha/2} \times \text{Standard Error} = 2.576 \times 0.02142 = 0.0552$

Step 5: Find the confidence interval

$\hat{p} \pm \text{Margin of Error} = 0.132 \pm 0.0552$

• Lower bound: $0.132 - 0.0552 = 0.0768$
• Upper bound: $0.132 + 0.0552 = 0.1872$

Final Answer

• Lower bound: $0.077$ (rounded to three decimal places)
• Upper bound: $0.187$ (rounded to three decimal places)

Let me know if you have further questions or need more details!

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Here are 5 follow-up questions to deepen understanding:

1. How does changing the confidence level affect the width of the interval?
2. What is the interpretation of this 99% confidence interval?
3. How would a larger sample size affect the confidence interval?
4. What assumptions are made when constructing this confidence interval?
5. How can we determine the required sample size for a desired margin of error?

Tip: The critical value $Z_{\alpha/2}$ depends on the desired confidence level. Make sure you use the correct value for your calculations!