To construct a confidence interval for the proportion of people aged 65 and over who have sleep apnea, we can use the formula for a confidence interval for a population proportion.
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Determine the sample proportion (\( p \)): \[ p = \frac{x}{n} = \frac{107}{433} \approx 0.247 \] where \( x \) is the number of individuals with sleep apnea (107), and \( n \) is the total sample size (433).
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Calculate the standard error (SE) for the proportion: \[ SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.247(1 - 0.247)}{433}} \approx \sqrt{\frac{0.247 \times 0.753}{433}} \approx \sqrt{\frac{0.186801}{433}} \approx \sqrt{0.000431} \approx 0.0208 \]
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Determine the z-score for a 99.5% confidence interval: The z-value corresponding to a 99.5% confidence level (which corresponds to 0.5% in each tail) is approximately 2.576.
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Calculate the margin of error (ME): \[ ME = z \cdot SE = 2.576 \cdot 0.0208 \approx 0.0537 \]
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Construct the confidence interval: \[ \text{CI} = \left( p - ME, p + ME \right) = \left( 0.247 - 0.0537, 0.247 + 0.0537 \right) = \left( 0.1933, 0.3007 \right) \]
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Round to three decimal places:
- Lower limit: \( 0.193 \)
- Upper limit: \( 0.301 \)
Thus, the 99.5% confidence interval for the proportion of those aged 65 and over who have sleep apnea is:
\[ \boxed{(0.193, 0.301)} \]
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