Respuesta :
Using the z-distribution, as we are working with a proportion, it is found that the margin of error for a 95% confidence interval would be of 0.0008.
What is a confidence interval of proportions?
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] is the sample proportion.
- z is the critical value.
- n is the sample size.
The margin of error is given by:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.
Considering the 99% confidence interval, which has z = 2.575, the estimate and the sample size are given by:
[tex]\pi = \frac{0.67 + 0.74}{2} = 0.705[/tex]
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.74 - 0.705 = 2.575\sqrt{\frac{0.705(0.295)}{n}}[/tex]
[tex]0.035 = 2.575\sqrt{\frac{0.705(0.295)}{n}}[/tex]
[tex]0.035\sqrt{n} = 2.575\sqrt{0.705(0.295)}[/tex]
[tex]\sqrt{n} = \frac{2.575\sqrt{0.705(0.295)}}{0.035}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.705(0.295)}}{0.035}\right)^2[/tex]
[tex]n = 1126[/tex]
Hence, the margin of error is given by:
[tex]M = 1.96\sqrt{\frac{0.705(0.295)}{1126}} = 0.0008[/tex]
More can be learned about the z-distribution at https://brainly.com/question/25890103
