Respuesta :
Using the z-distribution, it is found that a sample size of 3,385 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, we have a 98% confidence level, hence[tex]\alpha = 0.98[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.98}{2} = 0.99[/tex], so the critical value is z = 2.327.
We have no prior estimate, hence [tex]\pi = 0.5[/tex] is used, which is when the largest sample size is needed. To find the sample size, we solve the margin of error expression for n when M = 0.02, hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 2.327\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.02\sqrt{n} = 2.327 \times 0.5[/tex]
[tex]\sqrt{n} = \left(\frac{2.327 \times 0.5}{0.02}\right)[/tex]
[tex](\sqrt{n})^2 = \left(\frac{2.327 \times 0.5}{0.02}\right)^2[/tex]
n = 3,385.
A sample size of 3,385 is required.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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