Out of 400 people sampled, 212 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

Respuesta :

1. Find the sample proportion:

[tex]\frac{212}{400}=0.53[/tex]

2. Find the z-critical value for 95% confidence:

For 95% confidence interval the z-critical value is 1.96

3. Use the next formula to find the Confidence interval (CI):

[tex]CI=Sample\text{ }proportion\pm z-critical*\sqrt{\frac{sample\text{ }proportion*(1-sample\text{ }proportion)}{sample\text{ }size}}[/tex][tex]\begin{gathered} CI=0.53\pm1.96*\sqrt{\frac{0.53(1-0.53)}{400}} \\ \\ CI=0.53\pm1.96*\sqrt{\frac{0.53*0.47}{400}} \\ \\ CI=0.53\pm0.049 \\ \\ CI=(0.481,0.579) \end{gathered}[/tex]

Then, the confidence interval is: (0.481, 0.579)