Question

A philanthropic organisation sent free mailing labels and greeting cards to a random sample of 100 comma 000 potential donors on their mailing list and received 5066 donations. What is the 99% confidence interval?

Answer 1

The 99% confidence interval for the proportion of donors who donated is (0.0547, 0.0585).

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

For this problem, we have that:

`n = 100000, \pi = (5066)/(100000) = 0.0566`

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`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.0566 - 2.575\sqrt{(0.0566*0.9434)/(100000)} = 0.0547`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.0566 + 2.575\sqrt{(0.0566*0.9434)/(100000)} = 0.0585`

The 99% confidence interval for the proportion of donors who donated is (0.0547, 0.0585).