Respuesta :
Using the margin of error for the z-distribution, the sample sizes are given as follows:
a) 822.
b) 1068.
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 given by:
[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.
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.
Item a:
The estimate is of [tex]\pi = 0.26[/tex], hence we solve for n when M = 0.03.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.26(0.74)}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.26(0.74)}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.26(0.74)}}{0.03}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.26(0.74)}}{0.03}\right)^2[/tex]
n = 821.2
A sample of 822 is needed.
Item b:
No prior estimate, hence [tex]\pi = 0.5[/tex].
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.5(0.5)}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.03}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.03}\right)^2[/tex]
n = 1067.11
A sample of 1068 is needed.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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