Question

A newspaper article about the results of a poll states: "In theory, the results of such a poll, in 99 cases out of 100 should differ by no more than 2 percentage points in either direction from what would have been obtained by interviewing all voters in the United States." Find the sample size suggested by this statement.

Answer 1

The sample size suggested by this statement is of at least 4145.

Explanation:

This statement states that the 99% confidence interval has a margin of error of 2 percentage points.

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 zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

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

99% confidence level:

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

Find the sample size suggested by this statement.

The sample size is at least n.

n is found when M = 0.02.

We don't know the true proportion, so we use
`\pi = 0.5`, which is when the largest sample size will be needed.

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

`0.02 = 2.575\sqrt{(0.5*0.5)/(n)}`

`0.02√(n) = 2.575*0.5`

`√(n) = (2.575*0.5)/(0.02)`

`(√(n))^(2) = ((2.575*0.5)/(0.02))^(2)`

`n = 4144.14`

Rounding up

The sample size suggested by this statement is of at least 4145.