Questions: Out of 400 people sampled, 268 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places <p<

Solution Steps

The sample proportion of people with kids is calculated as follows:

\[ \hat{p} = \frac{x}{n} = \frac{268}{400} = 0.67 \]

To construct a \(95\%\) confidence interval for the population proportion, we use the formula:

\[ \hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

Where:

• \(\hat{p} = 0.67\) (sample proportion)
• \(n = 400\) (sample size)
• \(z \approx 1.96\) (z-value for \(95\%\) confidence level)

Substituting the values into the formula:

\[ 0.67 \pm 1.96 \cdot \sqrt{\frac{0.67(1 - 0.67)}{400}} \]

Calculating the margin of error:

\[ \sqrt{\frac{0.67(0.33)}{400}} = \sqrt{\frac{0.2211}{400}} \approx \sqrt{0.00055275} \approx 0.0235 \]

Thus, the margin of error is:

\[ 1.96 \cdot 0.0235 \approx 0.0460 \]

Now, we can find the confidence interval:

\[ 0.67 - 0.0460 \approx 0.624 \] \[ 0.67 + 0.0460 \approx 0.716 \]

The \(95\%\) confidence interval for the true population proportion of people with kids is:

\[ (0.624, 0.716) \]

Final Answer