Question

Out of 100 people sampled, 70 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

< p <

Answer 1

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

1. **Calculate the sample proportion (\(\hat{p}\))**:
\[
\hat{p} = \frac{x}{n} = \frac{70}{100} = 0.7
\]

2. **Find the z-score for a 99% confidence level**. The z-score for 99% confidence is approximately \(2.576\).

3. **Calculate the standard error (SE)**:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.7(1 - 0.7)}{100}} = \sqrt{\frac{0.7 \times 0.3}{100}} = \sqrt{\frac{0.21}{100}} = \sqrt{0.0021} \approx 0.04583
\]

4. **Calculate the margin of error (ME)**:
\[
ME = Z \times SE = 2.576 \times 0.04583 \approx 0.1187
\]

5. **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:

\[
0.581 < p < 0.819
\]