Question

in a survey of 1002 people, 701 said that they voted in a recent presidential election. voting records show that 61% of eligible voters actually did vote. find a 99% confidence interval estimate of the proportion of people who say that they voted

Answer 1

The 99% confidence interval estimate of the proportion of people who say that they voted is (0.6623, 0.7369).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

In a survey of 1002 people, 701 said that they voted in a recent presidential election.

This means that
`n = 1002, \pi = (701)/(1002) = 0.6996`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a p-value of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.6996 - 2.575\sqrt{(0.6996*0.3004)/(1002)} = 0.6623`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.6996 + 2.575\sqrt{(0.6996*0.3004)/(1002)} = 0.7369`

The 99% confidence interval estimate of the proportion of people who say that they voted is (0.6623, 0.7369).