Using the z-distribution, it is found that the lower bound of the 99% confidence interval is given by:
d. 68.39%.
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 sample size and estimate are given by:
[tex]n = 2002, \pi = 0.71[/tex]
Hence, the lower bound is given by:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.71 - 2.575\sqrt{\frac{0.71(0.29)}{2002}} = 0.6839[/tex]
Hence the lower bound is of 68.39%, which means that option D is correct.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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