Question

A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 533 randomly selected Americans surveyed, 351 were in favor of the initiative. Round answers to 4 decimal places where possible.

Answer 1

The 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative is (0.6247, 0.6923).

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)`.

Of the 533 randomly selected Americans surveyed, 351 were in favor of the initiative.

This means that
`n = 533, \pi = (351)/(533) = 0.6585`

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)} = 0.6585 - 1.645\sqrt{(0.6585*0.3415)/(533)} = 0.6247`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.6585 + 1.645\sqrt{(0.6585*0.3415)/(533)} = 0.6923`

The 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative is (0.6247, 0.6923).