Construct a 90 percent confidence interval for the proportio

Construct a 90 percent confidence interval for the proportion of all depositors who ask for cash back.(Round your answers to 4 decimal places.)

Solution

Confidence Interval For Proportion
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)=16
Sample Size(n)=48
Sample proportion = x/n =0.3333
Confidence Interval = [ 0.3333 ±Z a/2 ( Sqrt ( 0.3333*0.6667) /48)]
= [ 0.3333 - 1.645* Sqrt(0.0046) , 0.3333 + 1.65* Sqrt(0.0046) ]
= [ 0.2214,0.4452