Question

A random sample of 150 mortgages in the state of Florida was randomly selected. From this sample, 17 were found to be delinquent on their current payment. The 98% confidence interval for the proportion based on this sample is ________.

Answer 1

The 98% confidence interval for the proportion based on this sample is (0.0531, 0.1735).

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 = 150, \pi = (17)/(150) = 0.1133`

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.327`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.1133 - 2.327\sqrt{(0.1133*0.8867)/(150)} = 0.0531`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1133 + 2.327\sqrt{(0.1133*0.8867)/(150)} = 0.1735`

The 98% confidence interval for the proportion based on this sample is (0.0531, 0.1735).