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

Answer:

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).

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

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 $\pi = \frac{465}{1500} = 0.31$.

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 $1 - \frac{0.05}{2}$ = 0.975, so $Z = 1.96$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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).