In a survey of 10000 American adults 25 said they believed i

In a survey of 10000 American adults, 25% said they believed in astrology. Please calculate a 99% confidence interval for the proportion of all adults Americans who believe in astrology.

Solution

CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=2500
Sample Size(n)=10000
Sample proportion = x/n =0.25
Confidence Interval = [ 0.25 ±Z a/2 ( Sqrt ( 0.25*0.75) /10000)]
= [ 0.25 - 2.58* Sqrt(0) , 0.25 + 2.58* Sqrt(0) ]
= [ 0.239,0.261]