Question

3. In a random sample of 600 families in the city of Ottawa 360 live in a rental housing. Find a 95% confidence interval for the proportion of the renters in Ottawa.

Answer 1

sample success x =	360
sample size          n=	600
sample proportion p̂ =x/n=360/600=	0.6000
std error se= √(p*(1-p)/n) =sqrt(0.6*0.4/600)=	0.0200
for 95 % CI value of z=	1.960
margin of error E=z*std error   =	0.0392
lower bound=p̂ -E =0.6-0.0392= =	0.5608
Upper bound=p̂ +E=0.6+0.0392 =	0.6392

from above 95% confidence interval for population proportion =(0.5608 , 0.6392)