Out of 400 people sampled 20 had kids Based on this construc

Out of 400 people sampled, 20 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. As in the reading, in your calculations: Give your answers to three decimals Give your answers as decimals, to three 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)=20
Sample Size(n)=400
Sample proportion = x/n =0.05
Confidence Interval = [ 0.05 ±Z a/2 ( Sqrt ( 0.05*0.95) /400)]
= [ 0.05 - 2.576* Sqrt(0.000118) , 0.05 + 2.58* Sqrt(0.000118) ]
= [ 0.022,0.078]