Asimple random sample of 1350 registered voters shows that 58% favor candidate a over candidate b. which is the 99% confidence interval for the percent of the population of all registered voters who prefer candidate a over candidate b.

Given:
n = 1350, sample size
p = 58% = 0.58, proportion that faors candidate A over candidate B
Confidence level = 99%

The 99% confidence interval is

where
z* = 2.58 from tables.


Therefore the confidence interval is
(0.58 - 0.0347, 0.58+0.0347)
 = (0.5453, 0.6147)
= (54.5%, 61.5%) nearest tenth


The 99% confidence interval is (54.5%, 61.5%), or approximately (55%, 62%)