Respuesta :
Answer:
a) n = 1037.
b) n = 1026.
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) Assume that nothing is known about the percentage to be estimated.
We need to find n when M = 0.04.
We dont know the percentage to be estimated, so we use [tex]\pi = 0.5[/tex], which is when we are going to need the largest sample size.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 2.575\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.04\sqrt{n} = 2.575*0.5[/tex]
[tex](\sqrt{n}) = \frac{2.575*0.5}{0.04}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575*0.5}{0.04})^{2}[/tex]
[tex]n = 1036.03[/tex]
Rounding up
n = 1037.
(b) Assume prior studies have shown that about 55% of full-time students earn bachelor's degrees in four years or less.
[tex]\pi = 0.55[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 2.575\sqrt{\frac{0.55*0.45}{n}}[/tex]
[tex]0.04\sqrt{n} = 2.575*\sqrt{0.55*0.45}[/tex]
[tex](\sqrt{n}) = \frac{2.575*\sqrt{0.55*0.45}}{0.04}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575*\sqrt{0.55*0.45}}{0.04})^{2}[/tex]
[tex]n = 1025.7[/tex]
Rounding up
n = 1026.
