In: Statistics and Probability
In a random sample of 600 U.S. high school students, 11% (or 66 out of 600) said they do not plan on having children.
What sample size would allow us to create a 99% confidence interval for the true proportion of U.S. high school students who do not plan on having children with a margin of error of only 0.02?
Group of answer choices
1083
459
220
1625
882
Solution :
Given that,
= 0.11
1 - = 1 - 0.11 = 0.89
margin of error = E = 0.02
At 99% confidence level the z is,
= 1 - 99%
= 1 - 0.99 = 0.01
/2 = 0.005
Z/2 = 2.576 ( Using z table )
Sample size = n = (Z/2 / E)2 * * (1 - )
= (2.576 / 0.02)2 * 0.11 * 0.89
=1625
Sample size = 1625