Respuesta :
Answer:
A sample size of 2166 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 of a confidence interval is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
98% confidence level
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].
Solving
We need to find n when M = 0.025.
We dont know the proportion, so we use [tex]\pi = 0.5[/tex], which is when we are going to need the largest sample size for this estimate.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.025 = 2.327\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.025\sqrt{n} = 2.327*0.5[/tex]
[tex]\sqrt{n} = \frac{2.327*0.5}{0.025}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.327*0.5}{0.025})^{2}[/tex]
[tex]n = 2166[/tex]
A sample size of 2166 is needed.
