Respuesta :
Using the z-distribution, as we are working with a proportion, it is found that a sample size of 1701 is required.
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]
The margin of error is:
[tex]M = 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.
In this problem, the parameters are:
- 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].
- Estimate of [tex]\pi = 0.23[/tex].
- Margin of error of M = 0.02.
Hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{\frac{0.23(0.77)}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.96\sqrt{0.23(0.77)}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.23(0.77)}}{0.02}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.23(0.77)}}{0.02}\right)^2[/tex]
[tex]n = 1700.8[/tex]
Rounding up, a sample size of 1701 is required.
More can be learned about the z-distribution at https://brainly.com/question/25890103
