Using the z-distribution, it is found that the 99% confidence interval is (0.041, 0.2256), and it means that we are 99% sure that the population proportion is in this interval.
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
In this problem, we have a 99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.575.
The other parameters are given by:
[tex]n = 90, \pi = \frac{12}{90} = 0.1333[/tex]
Then, the bounds of the interval are found as follows:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1333 - 2.575\sqrt{\frac{0.1333(0.8667)}{90}} = 0.041[/tex]
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1333 + 2.575\sqrt{\frac{0.1333(0.8667)}{90}} = 0.2256[/tex]
The 99% confidence interval is (0.041, 0.2256), and it means that we are 99% sure that the population proportion is in this interval.
More can be learned about the z-distribution at https://brainly.com/question/25890103
