Question

Statistics - Finding a lower bound for a confidence interval help

Answer 1

http://puu.sh/lCPTC/b58a0fffba.png

So when I attempted this problem I found the mean value to be 11/152 = .07236
The variance = (.07236)(1-.07236)=.06713 Sd = sqrt(.06713) = .259.
Standard deviation accounting for population = .259/sqrt(152) = .021

To get a lower boundary of 95% we need the z score -1.645
Therefore, I would think the answer would be:
.0723-1.645(.021) = .03775

However, webassign says this is wrong and should be .045.

Where am I making a mistake?