Question

****PLEASE SHOW HOW TO SOLVE IN EXCEL NOT HANDWRITTEN***

1. If I have a sample of 200 randomly selected people in a city of whom 120 are Christians, what would be a 95% confidence interval for the proportion of Christians in that city? If I want my confidence interval with a margin of error of 1% or less, how many people should I have in my sample?

Answer 1

Ans:

sample size(n)=200
x=120
sample proportion=120/200=0.6
confidence level=0.95
alpha=1-0.95=0.05
alpha/2=0.025
critical z values=normsinv(0.025) or normsinv(1-0.025)=+/-1.96
Margin of error=1.96*sqrt(0.6*(1-0.6)/200)=0.068
Point estimate=0.6
Lower limit=0.6-0.068=0.532
Upper limit=0.6+0.0679=0.668
95% Confidence interval for proportion of Christians in the city
=(0.532 ,   0.668)
Now,when margin of error=0.01
Sample size required,n=1.96^2*0.6*(1-0.6)/0.01^2=9220