Question

A Gallup survey of 2322 adults (at least 18 years old) in the U.S. found that 408 of them have donated blood in the past two years. Construct a 90% confidence interval for the population proportion of adults in the U.S. who have donated blood in the past two years.

Answer 1

The 90% confidence interval for the population proportion of adults in the U.S. who have donated blood in the past two years is (0.1627, 0.1887).

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)`.

A Gallup survey of 2322 adults (at least 18 years old) in the U.S. found that 408 of them have donated blood in the past two years.

This means that
`n = 2322, \pi = (408)/(2322) = 0.1757`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a p-value of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.1757 - 1.645\sqrt{(0.1757*0.8243)/(2322)} = 0.1627`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1757 + 1.645\sqrt{(0.1757*0.8243)/(2322)} = 0.1887`

The 90% confidence interval for the population proportion of adults in the U.S. who have donated blood in the past two years is (0.1627, 0.1887).