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 422 people aged 65 and over, 116 of them had sleep apnea.

a) Find a point estimate for the population proportion of those aged 65 and over who have sleep apnea. Round the answer to three decimal places.

use a ti-84 calculator

Answer 1

(b) Construct a 99% confidence interval for the proportion of those aged 65 and over who have sleep apnea. Round the answer to three decimal places.

Answer 2

To find the point estimate for the population proportion of those aged 65 and over who have sleep apnea, you can use the following formula:

\[
\hat{p} = \frac{x}{n}
\]

where:
- \( \hat{p} \) is the point estimate of the population proportion,
- \( x \) is the number of people with sleep apnea,
- \( n \) is the total number of people in the sample.

In this case:
- \( x = 116 \) (the number of people with sleep apnea),
- \( n = 422 \) (the total number of people in the sample).

Now, plug in the values:

\[
\hat{p} = \frac{116}{422}
\]

Calculating this gives:

\[
\hat{p} \approx 0.275
\]

Now rounding to three decimal places, the point estimate is:

\[
\hat{p} \approx 0.275
\]

**Final answer: 0.275**

Answer 3

To construct a 99% confidence interval for the proportion of those aged 65 and over who have sleep apnea, 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:
- \( \hat{p} \) is the sample proportion,
- \( z \) is the z-score corresponding to the desired confidence level,
- \( n \) is the sample size.

### Step 1: Find the sample proportion (\( \hat{p} \))
From the previous calculation, we have:

\[
\hat{p} = \frac{116}{422} \approx 0.275
\]

### Step 2: Determine the z-score for a 99% confidence interval
For a 99% confidence level, the z-score (from a z-table) is approximately:

\[
z \approx 2.576
\]

### Step 3: Calculate the standard error (SE)
Calculate the standard error of the proportion:

\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.275(1 - 0.275)}{422}}
\]

Calculating \( SE \):

\[
SE = \sqrt{\frac{0.275 \times 0.725}{422}} \approx \sqrt{\frac{0.199875}{422}} \approx \sqrt{0.000473} \approx 0.0217
\]

### Step 4: Calculate the margin of error (ME)
Now calculate the margin of error:

\[
ME = z \cdot SE = 2.576 \cdot 0.0217 \approx 0.0559
\]

### Step 5: Construct the confidence interval
Now we can construct the confidence interval:

\[
\hat{p} \pm ME = 0.275 \pm 0.0559
\]

Calculating the lower and upper bounds:

- Lower bound: \( 0.275 - 0.0559 \approx 0.2191 \)
- Upper bound: \( 0.275 + 0.0559 \approx 0.3309 \)

### Step 6: Round to three decimal places
The 99% confidence interval for the proportion of those aged 65 and over who have sleep apnea is:

\[
(0.219, 0.331)
\]

**Final answer: (0.219, 0.331)**