Question

A random sample of 1500 home owners in a particular city found 465 home owners who had a swimming pool in their backyard. Find a 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard. Round your final answer to three decimal places.

Answer 1

The 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard is (0.287, 0.333).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`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:

A random sample of 1500 home owners in a particular city found 465 home owners who had a swimming pool in their backyard. This means that
`n = 1500` and
`\pi = (465)/(1500) = 0.31`.

Find a 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard.

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2)` = 0.975, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.31 - 1.96\sqrt{(0.31*0.69)/(1500)} = 0.287`

The upper limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.31 + 1.96\sqrt{(0.31*0.69)/(1500)} = 0.333`

The 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard is (0.287, 0.333).