Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 146 had kids. Based on this, construct a 95% confidence interval for the proportion p of adult residents who are parents in this county.

Express your answer in tri-inequality form. Give your answers as decimals, to three places.

< p <

Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

p =      ±

Answer 1

Sample proportion = 146 / 200 = 0.73

95% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.73 - 1.96 * sqrt [ 0.73 ( 1 - 073) / 200 ] < p < 0.73 + 1.96 * sqrt [ 0.73 ( 1 - 073) / 200 ]

0.668 < p < 0.792

Confidence interval in terms of E is

Z/2 * sqrt [ ( 1 - ) / n ]

0.73 1.96 * sqrt [ 0.73 ( 1 - 073) / 200 ]

0.73 0.062