Question

Out of 600 people sampled, 42 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids.

Use GeoGebra to calculate! Give your answers as decimals, to three places

Answer 1

So we can be 95% confident that the true proportion of people with kids in the population is between 0.044 and 0.096.

Explanation:

To construct a confidence interval, we need to use the formula:

CI = p ± zsqrt((p(1-p))/n)

Where:

CI = confidence interval

p = sample proportion (in this case, 42/600 = 0.07)

z = the z-score corresponding to the desired level of confidence (in this case, 1.96 for a 95% confidence interval)

n = sample size (in this case, 600)

Plugging in the numbers, we get:

CI = 0.07 ± 1.96sqrt((0.07(1-0.07))/600)

Simplifying this, we get:

CI = 0.07 ± 0.026

Therefore, the 95% confidence interval for the true population proportion of people with kids is:

0.044 ≤ p ≤ 0.096