Section 8.3:Confidence Intervals (Proportions)

Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 83 had kids. Based on this, construct a 90% 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

90% confidence interval for a proportion

x=83

n=100

phat = 0.83

phat +/- z(α/2) * sqrt(phat(1-phat)/n)

0.83 +/- 1.645*sqrt(0.83*(1-0.83)/100)

0.83 +/- 0.06179

0.768 < p < 0.892