Question

In a recent poll of 760 homeowners in the United States, one in five homeowners reports having a home equity loan that or she is currently paying off. Using a confidence coefficient of 0.9, derive the interval estimate for the proportion of all homeowners in the United States that hold a home equity loan. (Keep three decimal places)

Answer 1

The 90% confidence interval for the proportion of all homeowners in the United States that hold a home equity loan is (0.176, 0.224)

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 = 760, \pi = (1)/(5) = 0.2`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue 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)} = 0.2 - 1.645\sqrt{(0.2*0.8)/(760)} = 0.176`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.2 + 1.645\sqrt{(0.2*0.8)/(760)} = 0.224`

The 90% confidence interval for the proportion of all homeowners in the United States that hold a home equity loan is (0.176, 0.224)