Accepted Answer

Hi there!

The confidence interval for a proportion goes from:

p - z*sqrt(p(1-p)/N) to p + z*sqrt(p(1-p)/N)

p is the proportion that we are given, z is the z value for the 95% interval (1.96, from a table), and N is the sample size

N = 86991
z = 1.96
p = 1143/86991 = 0.013139
1-p = 0.986861

0.013139 - 1.96*sqrt(0.013139*0.986861/86991) to 0.013139 + 1.96*sqrt(0.013139*0.986861/86991)

Do the math:
0.01238 to 0.01390


We are allowed to do this question, even though P is small, because Np and N(1-p) are both reasonably sized (more than 10):
Np = 1143
N(1-p) = 85848

We have enough "hits" of both negative and positive results. Our sample is large enough (greater than 1000) that even with a tiny "p" value, we can make the normality assumption.


Let me know if you have any questions. If not, thanks for pressing "Accept".

-Scott

Expert:	Scott
Pos. Feedback:	100.0 %
Accepts:
Answered:	5/26/2008

MIT Graduate

College degree in math... proficient in all levels -- from algebra to calculus