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}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which:
is the sample proportion.
In this problem, the parameters are:
![z = 2.575, n = 2000, \pi = 0.77](https://tex.z-dn.net/?f=z%20%3D%202.575%2C%20n%20%3D%202000%2C%20%5Cpi%20%3D%200.77)
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](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.77%20-%202.575%5Csqrt%7B%5Cfrac%7B0.77%280.23%29%7D%7B2000%7D%7D%20%3D%200.7458)
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](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.77%20%2B%202.575%5Csqrt%7B%5Cfrac%7B0.77%280.23%29%7D%7B2000%7D%7D%20%3D%200.7942)
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