suppose That in a random sample of 40 Americans 95 prefer ch

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)=36
Sample Size(n)=40
Sample proportion = x/n =0.9
Confidence Interval = [ 0.9 ±Z a/2 ( Sqrt ( 0.9*0.1) /40)]
= [ 0.9 - 2.576* Sqrt(0.002) , 0.9 + 2.58* Sqrt(0.002) ]
= [ 0.778,1.022]