Respuesta :
Answer:
a) We need a sample size of at least 3109.
b) We need a sample size of at least 4145.
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].
(a) he uses a previous estimate of 25%?
we need a sample of size at least n.
n is found when [tex]M = 0.02, \pi = 0.25[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 2.575\sqrt{\frac{0.25*0.75}{n}}[/tex]
[tex]0.02\sqrt{n} = 2.575\sqrt{0.25*0.75}[/tex]
[tex]\sqrt{n} = \frac{2.575\sqrt{0.25*0.75}}{0.02}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575\sqrt{0.25*0.75}}{0.02})^{2}[/tex]
[tex]n = 3108.1[/tex]
We need a sample size of at least 3109.
(b) he does not use any prior estimates?
When we do not use any prior estimate, 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\sqrt{0.5*0.5}[/tex]
[tex]\sqrt{n} = \frac{2.575\sqrt{0.5*0.5}}{0.02}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575\sqrt{0.5*0.5}}{0.02})^{2}[/tex]
[tex]n = 4144.1[/tex]
Rounding up
We need a sample size of at least 4145.
