Respuesta :
Answer:
A sample size of 664 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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
If the candidate only wants a 5% margin of error at a 99% confidence level, what size of sample is needed
A sample size of n is needed.
n is found when M = 0.05.
We have no estimate, so we use [tex]\pi = 0.05[/tex]. Then
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 2.575\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.05\sqrt{n} = 2.575*0.5[/tex]
[tex]\sqrt{n} = \frac{2.575*0.5}{0.05}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575*0.5}{0.05})^{2}[/tex]
[tex]n = 663.1[/tex]
Rounding up
A sample size of 664 is needed.
