Respuesta :
Answer:
Standard deviation of the students = 0.408
The students would you have to poll to be 95% confident of the outcome within /- 2% of the vote
= 0.408 X 1600
= 652.8
Step-by-step explanation:
Step(i):-
Given sample size 'n' = 1600
95% confidence interval of the margin error is determined by
[tex]M.E = \frac{Z_{0.05} S.D}{\sqrt{n} }[/tex]
Level of significance = 0.05
Z₀.₀₅ = 1.96
Given Margin of error = 2% = 0.02
[tex]M.E = \frac{Z_{0.05} S.D}{\sqrt{n} }[/tex]
[tex]0.02 = \frac{1.96 X S.D}{\sqrt{1600} }[/tex]
0.02 X √1600 = 1.96 X S.D
[tex]S.D = \frac{0.02X\sqrt{1600} }{1.96} = 0.408[/tex]
Standard deviation of the students = 0.408
The students would you have to poll to be 95% confident of the outcome within /- 2% of the vote
= 0.408 X 1600
= 652.8
Using the z-distribution, it is found that you would have to poll 2401 students.
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which z is the z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
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].
There is no prior estimate, hence [tex]\pi = 0.5[/tex] is used.
The sample size is n for which M = 0.02, hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.96(0.5)[/tex]
[tex]\sqrt{n} = \left(\frac{1.96(0.5)}{0.02}\right)[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96(0.5)}{0.02}\right)^2[/tex]
[tex]n = 2401[/tex]
You would have to poll 2401 students.
To learn more about the z-distribution, you can take a look at https://brainly.com/question/12517818
