Respuesta :
Using the z-distribution, as we are working with a proportion, the minimum sample size required to ensure a 90% confidence interval for p to within 0.02 is of 1692.
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 of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In this problem:
- No prior estimate, hence the proportion is [tex]\pi = 0.5[/tex].
- Margin of error of M = 0.02.
- 90% confidence level, hence[tex]\alpha = 0.9[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so [tex]z = 1.645[/tex].
Then, solving for n:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.645\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.645(0.5)[/tex]
[tex]\sqrt{n} = \frac{1.645(0.5)}{0.02}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.645(0.5)}{0.02}\right)^2[/tex]
[tex]n = 1691.3[/tex]
Rounding up, a sample size of 1692 is needed.
More can be learned about the z-distribution at https://brainly.com/question/25890103
