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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

User Arnyeinstein
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1 Answer

19 votes
19 votes

1. Find the sample proportion:


(212)/(400)=0.53

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):


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

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

User RamiReddy
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