Construct a confidence interval of the population proportion

Construct a confidence interval of the population proportion at the given level of confidence.

x=240, n=300, 99% confidence

The 99% cofidence interval is (___,___) (Use ascending order. Round to three decimal places as needed)

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)=240
Sample Size(n)=300
Sample proportion = x/n =0.8
Confidence Interval = [ 0.8 ±Z a/2 ( Sqrt ( 0.8*0.2) /300)]
= [ 0.8 - 2.58* Sqrt(0.001) , 0.8 + 2.58* Sqrt(0.001) ]
= [ 0.74,0.86]