statistics question

Question

Out of 400 people sampled, 284 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

Answer 1

Use the formula CI = p̂ ± z√[p̂(1-p̂)/n] where p̂=284/400=0.71, n=400, and z=1.96 (for a 95% confidence level):

CI = p̂ ± z√[p̂(1-p̂)/n] = 0.71 ± 1.96√[(0.71)(1-0.71)/400] ≈ {0.666,0.754}

Thus, we are 95% confident that the true population proportion of people with kids is between 0.666 and 0.754

Hope this helped!