Using the z-distribution, it is found that the minimum sample size required is of 42,190.
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
The margin of error is given by:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In this problem, we have a 96% confidence level, hence[tex]\alpha = 0.96[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.96}{2} = 0.98[/tex], so the critical value is z = 2.054.
As for the other parameters, we have that M = 0.005 and [tex]\pi = 0.5[/tex], as there are no estimates of the proportion. Hence, we solve for n to find the minimum sample size.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.005 = 2.054\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.005\sqrt{n} = 2.054(0.5)[/tex]
[tex]\sqrt{n} = \frac{2.054(0.5)}{0.005}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{2.054(0.5)}{0.005}\right)^2[/tex]
[tex]n = 42189.16[/tex]
Rounding up, the minimum sample size required is of 42,190.
More can be learned about the z-distribution at https://brainly.com/question/25890103
