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

Answer:

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)

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 level of $1-\alpha$, 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:

$n = 760, \pi = \frac{1}{5} = 0.2$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 - 1.645\sqrt{\frac{0.2*0.8}{760}} = 0.176$

The upper limit of this interval is:

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

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)