To construct a 99% confidence interval for the true population proportion of people with kids, we can use the formula for the confidence interval for a proportion:
\[ \hat{p} \pm Z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]
Where:
- \(\hat{p}\) is the sample proportion
- \(Z\) is the z-score corresponding to the desired confidence level
- \(n\) is the sample size
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Calculate the sample proportion (\(\hat{p}\)): \[ \hat{p} = \frac{x}{n} = \frac{70}{100} = 0.7 \]
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Find the z-score for a 99% confidence level. The z-score for 99% confidence is approximately \(2.576\).
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Calculate the standard error (SE): \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.7(1 - 0.7)}{100}} = \sqrt{\frac{0.7 \times 0.3}{100}} = \sqrt{\frac{0.21}{100}} = \sqrt{0.0021} \approx 0.04583 \]
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Calculate the margin of error (ME): \[ ME = Z \times SE = 2.576 \times 0.04583 \approx 0.1187 \]
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Construct the confidence interval: \[ \hat{p} \pm ME = 0.7 \pm 0.1187 \] This gives us:
- Lower limit: \(0.7 - 0.1187 \approx 0.5813\)
- Upper limit: \(0.7 + 0.1187 \approx 0.8187\)
Thus, the 99% confidence interval for the true population proportion of people with kids is:
\[ 0.581 < p < 0.819 \]
Final answer:
\[ 0.581 < p < 0.819 \]