Using the z-distribution, as we are working with a proportion, it is found that the 99% confidence interval for the proportion of all U. S. Adults who would include the 9/11 attacks on their list of 10 historic events is (0.7458, 0.7942). It means that we are 99% sure that the true proportion for all U.S. adults is between these two bounds.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, the parameters are:

$z = 2.575, n = 2000, \pi = 0.77$

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7458$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7942$

The 99% confidence interval for the proportion of all U. S. Adults who would include the 9/11 attacks on their list of 10 historic events is (0.7458, 0.7942). It means that we are 99% sure that the true proportion for all U.S. adults is between these two bounds.

More can be learned about the z-distribution at brainly.com/question/25890103

7 0

1 year ago

X− 4 ≤ 3 explain how to solve

rusak2 [61]

Answer:

x ≤ 7

Step-by-step explanation:

1. add 4 to both sides x-4+4 ≤ 3+4

2. simplify x ≤ 7

7 0

2 years ago