Question

Out of 600 people sampled, 438 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places

Answer 1

p = 438 / 600 = 0.73

n = 600

Z for 90% confidence interval = Z_0.05 = 1.645

Confidence interval = (p + Z_0.05 * sqrt(p * (1 - p) / n))

                                = (0.73 + 1.645 * sqrt(0.73 * (1 - 0.73) / 600))

                                = (0.73 + 0.03)

                                = (0.7 , 0.760)