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.
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
In this problem, the parameters are:
[tex]z = 2.575, n = 2000, \pi = 0.77[/tex]
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7458[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 2.575\sqrt{\frac{0.77(0.23)}{2000}} = 0.7942[/tex]
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 https://brainly.com/question/25890103