Respuesta :
Using the z-distribution, as we are working with a proportion, it is found that samples of 937 should be taken.
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 that:
- The estimate is of [tex]\pi = 0.675[/tex].
- The margin of error is of M = 0.03.
- 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 [tex]z = 1.96[/tex].
Then, we solve for n to find the minimum sample size.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.675(0.325)}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.675(0.325)}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.675(0.325)}}{0.03}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.675(0.325)}}{0.03}\right)^2[/tex]
[tex]n = 936.4[/tex]
Rounding up, it is found that samples of 937 should be taken.
More can be learned about the z-distribution at https://brainly.com/question/25890103
