Respuesta :
Answer:
They need to survey 4145 residents.
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 given by:
[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 p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
At least how many residents do they need to survey if they want to be at least 99% confident that the sample proportion is within 0.02 of the true proportion?
This is n for which [tex]M = 0.02[/tex]. As we have no estimate for the proportion, we use [tex]\pi = 0.5[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 2.575\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.02\sqrt{n} = 2.575*0.5[/tex]
[tex]\sqrt{n} = \frac{2.575*0.5}{0.02}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575*0.5}{0.02})^2[/tex]
[tex]n = 4144.1[/tex]
Rounding up:
They need to survey 4145 residents.
