People with children

Question

Out of 300 people sampled, 255 had children. Based on this, construct a 95% confidence interval for the true population proportion of people with children.  What is the 90% confidence interval to estimate the population proportion? Round your answer to three decimal places.

Answer 1

Hi Bailey,

First, we must verify that pn and (1-p)n both exceed 10. To do this, we compute p, which some books will call p-hat for p predicted--I'll stick with p to avoid confusion:

p= 255/300

p=0.85

Now, we need to compute np and n(1-p), where

n=sample size=300

np=300*.85=255

n(1-p)=300*.15=45

We can proceed with the z-confidence interval:

CI=p +/- z*sqrt[(p(1-p)/n)]

p=proportion we are testing

z*=z-critical value depending on confidence level

n=sample size

So, for this problem:

p=0.85

n=300

z*=1.645

Note: 1.645 is the z-critical value for all 90% 2-sided z confidence intervals, so you might want to memorize it even though it can be derived from the z-table. Proceeding:

CI=0.85 +/- 1.645*sqrt((0.85*0.15)/300)

CI=0.85 +/- 0.034

CI=(0.816, 0.884)

I notice you have many questions posted here--let me know via Wyzant if you want to meet online and talk through some of these. I hope this helps.