To construct a 99% confidence interval for the true population proportion of people with kids, we will use the formula for the confidence interval for a proportion:
\[ \hat{p} \pm z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]
where:
- \(\hat{p}\) = sample proportion of successes (people with kids)
- \(z\) = z-value corresponding to the desired confidence level
- \(n\) = sample size
Step 1: Calculate the sample proportion \(\hat{p}\)
\[ \hat{p} = \frac{x}{n} = \frac{14}{200} = 0.07 \]
Step 2: Find the z-value for a 99% confidence interval
For a 99% confidence interval, the z-value can be found from a standard normal distribution table or a calculator. The z-value for 99% confidence is approximately:
\[ z \approx 2.576 \]
Step 3: Calculate the standard error (SE)
\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.07(1 - 0.07)}{200}} = \sqrt{\frac{0.07 \times 0.93}{200}} = \sqrt{\frac{0.0651}{200}} \approx \sqrt{0.0003255} \approx 0.0180 \]
Step 4: Calculate the margin of error (ME)
\[ ME = z \cdot SE = 2.576 \cdot 0.0180 \approx 0.0464 \]
Step 5: Construct the confidence interval
\[ \hat{p} - ME \quad \text{to} \quad \hat{p} + ME \]
Calculating the endpoints:
\[ 0.07 - 0.0464 \approx 0.0236 \] \[ 0.07 + 0.0464 \approx 0.1164 \]
Final Confidence Interval
The 99% confidence interval for the true population proportion is:
\[ (0.024, 0.116) \]
Thus, rounding to three decimal places:
\[ \text{Confidence Interval: } (0.024, 0.116) \]
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