Using the z-distribution, as we are working with a proportion, the 99% confidence interval is given by:
a) (0.733, 0.927).
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 p = 0.83, n = 100, hence the bounds of the interval are given as follows.
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.83 - 2.575\sqrt{\frac{0.83(0.17)}{100}} = 0.733[/tex]
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.83 + 2.575\sqrt{\frac{0.83(0.17)}{100}} = 0.927[/tex]
Hence option a is correct.
More can be learned about the z-distribution at https://brainly.com/question/25890103
