confidence interval

數學神童

The results of a survey showed that 3600 out of 10000 families regularly purchased a specific weekly magazine .
a)Find the approximate 95% confidence limits for the proportion of families buying the magazine.
b)Estimate the additional number of families to be contacted if the probability that the estimated proportion is in error by more than 0.01 is to be at most 1%

Answer: a)0.36+-9.408x10^-3
            b)n=5277

Can anyone show me that working?

puangenlun

x~B(n,p)
alpha=0.05
n=10000,
p=3600/10000=0.36
q=1-p=0.64

u=np=3600
sigma=sqrt(npq)

1-alpha=P(|x-u|<1.96*sigma/sqrt(n))
|x-u|<1.96*sigma/sqrt(n)
x between u(+-)1.96*sigma/sqrt(n)=3600(+-)0.9408

average =x/n=0.36(+-)0.9408*10(-3)