Question

Sleep apnea is a disorder in which there are pauses in breathing during sleep. People with this condition must wake up frequently to breathe. In a sample of 433 people aged 65 and over, 107 of them had sleep apnea.

(b) Construct a 99.5% confidence interval for the proportion of those aged 65 and over who have sleep apnea. Round the answer to three decimal places.

Answer 1

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.

1. **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).

2. **Calculate the standard error (SE)** for the proportion:
\[
SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.247(1 - 0.247)}{433}} \approx \sqrt{\frac{0.247 \times 0.753}{433}} \approx \sqrt{\frac{0.186801}{433}} \approx \sqrt{0.000431} \approx 0.0208
\]

3. **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.

4. **Calculate the margin of error (ME)**:
\[
ME = z \cdot SE = 2.576 \cdot 0.0208 \approx 0.0537
\]

5. **Construct the confidence interval**:
\[
\text{CI} = \left( p - ME, p + ME \right) = \left( 0.247 - 0.0537, 0.247 + 0.0537 \right) = \left( 0.1933, 0.3007 \right)
\]

6. **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:

\[
\boxed{(0.193, 0.301)}
\]