Question

Out of 300 people sampled, 162 had children. Based on this, construct a 95% 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=. ROUND ANSWER TO 1 DECIMAL PLACE!.
   3.)  nˆp(1−ˆp)=  

4.) Confidence Interval: What is the 95% confidence interval to estimate the population proportion? ROUND ANSWER TO 3 DECIMAL PLACES!

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

6.) Interpret: Interpret the confidence interval in context of estimating the population proportion of people having children. ????????

Answer 1

3)
300 * 0.54 *(1-0.54) = 74.5

4)

sample proportion, = 0.54
sample size, n = 300
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.54 * (1 - 0.54)/300) = 0.0288

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0288
ME = 0.0564

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.54 - 1.96 * 0.0288 , 0.54 + 1.96 * 0.0288)
CI = (0.484 , 0.596)

5)

0.484 < p < 0.596

6)

we are 95% confident that the population proportion of people having children. is between 0.484 and 0.596