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)

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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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 760, \pi = \frac{1}{5} = 0.2[/tex]

90% confidence level

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

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 - 1.645\sqrt{\frac{0.2*0.8}{760}} = 0.176[/tex]

The upper limit of this interval is:

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

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)