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Sagot :
Using the z-distribution, it is found that 1068 students have to be polled to be 95% confident of the outcome within +/- 3% of the vote.
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 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.
We have no estimate for the true proportion, hence we use [tex]\pi = 0.5[/tex], and solve for n when M = 0.03. Then:
[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(0.5)[/tex]
[tex]\sqrt{n} = \frac{1.96(0.5)}{0.03}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96(0.5)}{0.03}\right)^2[/tex]
n = 1067.11.
Rounding up, 1068 students have to be polled to be 95% confident of the outcome within +/- 3% of the vote.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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