Question

A simple 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.

Answer 1

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

`p\pm z*\sqrt{ (p(1-p))/(n) }`
where
z* = 2.58 from tables.


`z* \sqrt{ (p(1-p))/(n) } = 2.58 \sqrt{ (0.58*(1-0.58))/(1350) }=0.0347`
Therefore the confidence interval is
(0.58 - 0.0347, 0.58+0.0347)
= (0.5453, 0.6147)
= (54.5%, 61.5%) nearest tenth

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