Inference about a Population Proportion - Margin of Error and Sample Size

Last one :)


An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by the local dealer.The customer relations department will survey a random sample of customers and compute a 99% confidence interval for the proportion who are not satisfied.

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if themargin of error of the confidence interval is to be about 0.015.

(b) When the sample is actually contacted, 10% of the sample say they are not satisfied. What is the margin of error of the 99% confidence interval?

a) 4719

b) 0.0113

Yes? No? Close?!!

Thanks in advance!
I used the method of finding a sample size that was in the talkstats example (it is different than the one we were taught) and I got the same answer so I'm pretty confident that question a) = 4719

However, I don't see an example to find out the margin or error similar to the method I was taught in class.

m = margin or error
z* = Confidence level C from Table C (99%)
p = p(hat) or estimated sample p (0.10)
n = sample size (4719)

I don't know how to type it, but p(1-p)/n in the formula is under the 'square root' sign.

m = z*(p(1-p)/n)

So when I plug in my numbers I get the following:

m = 2.576(0.1(1-0.1)/4719)
= 2.576(0.1(0.9)/4719)
= 2.576(0.09/4719)
= 2.576(0.00001907)
then did the square root of 0.00001907
= 2.576(0.0044)
= 0.0113

Am I incorrect?

Is the margin or error NOT 0.0113, and if so, can you please tell me where I'm going wrong!?!

Thanks in advance!