Construct a 99 confidence intervsl to estimate the populatio

Construct a 99% confidence intervsl to estimate the population proportion with a sample proportion equal to .80 and a sample size equal to 300.

a 99% confidence interval estimates that the population proportion is between a lower limit of ___ and an upper limit of ____.

(round to three decimal places as needed.)

Use \"cumulative probabilities for the standard normal distribution.\"

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)=2.4
Sample Size(n)=300
Sample proportion = x/n =0.008
Confidence Interval = [ 0.008 ±Z a/2 ( Sqrt ( 0.008*0.992) /300)]
= [ 0.008 - 2.58* Sqrt(0) , 0.008 + 2.58* Sqrt(0) ]
= [ -0.005,0.021]