A recent poll of 2350 US voters found that 1034 of them curr

A recent poll of 2350 U.S. voters found that 1034 of them currently approve of Obama’s performance as president. Make a 99% confidence interval for the proportion of all U.S. voters that approve of Obama’s performance. Describe the meaning of the interval you found.

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)=1034
Sample Size(n)=2350
Sample proportion = x/n =0.44
Confidence Interval = [ 0.44 ±Z a/2 ( Sqrt ( 0.44*0.56) /2350)]
= [ 0.44 - 2.576* Sqrt(0) , 0.44 + 2.58* Sqrt(0) ]
= [ 0.414,0.466] ~ [ 41.4%, 46.6% ]
Interpretations:
1) We are 99% confidence that the interval [ 41.4%, 46.6% ] contains the true population
proportion Po

2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contain the contains the true population
proportion Po