Question

Out of 200 people sampled, 94 had children. Based on this, construct a 99% confidence interval for the true population proportion of people with children.

Preliminary:

1. Is it safe to assume that n≤5% of all people with children?
   • No
   • Yes
   
   
2. Verify nˆp(1−ˆp)≥10. YOU MUST ROUND YOUR ANSWER TO ONE DECIMAL PLACE.
3. nˆp(1−ˆp)=

4.Confidence Interval: What is the 99% confidence interval to estimate the population proportion? YOU MUST ROUND ANSWER TO THREE DECIMALS PLACES.

5. ?????? <p< ????

Answer 1

yes

nˆp(1−ˆp)= 200*(94/200)*(106/200) = 49.8

4)
sample proportion, = 0.47
sample size, n = 200
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.47 * (1 - 0.47)/200) = 0.0353

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.47 - 2.58 * 0.0353 , 0.47 + 2.58 * 0.0353)
CI = (0.379 , 0.561)

0.379 < p < 0.561