Answer:
A sample size of 752 is needed.
Step-by-step explanation:
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]1-\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]1 - \frac{\alpha}{2}[/tex].
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
If the candidate wants a 3% margin of error at a 90% confidence level, what size of sample is needed?
We have no estimate of the proportion, so we use [tex]\pi = 0.5[/tex].
The sample size is n for which M = 0.03. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.645*0.5[/tex]
[tex]\sqrt{n} = \frac{1.645*0.5}{0.03}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.645*0.5}{0.03})^2[/tex]
[tex]n = 751.67[/tex]
Rounding up:
A sample size of 752 is needed.
