Respuesta :
Answer:
358 randomly selected air passengers must be surveyed.
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 given by:
[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].
Assume that a prior survey suggests that about 22​% of air passengers prefer an aisle seat.
This means that [tex]\pi = 0.22[/tex]
How many randomly selected air passengers must you​ survey?
We need a sample of n, and n is found for which [tex]M = 0.051[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.051 = 2.327\sqrt{\frac{0.22*0.78}{n}}[/tex]
[tex]0.051\sqrt{n} = 2.327\sqrt{0.22*0.78}[/tex]
[tex]\sqrt{n} = \frac{2.327\sqrt{0.22*0.78}}{0.051}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.327\sqrt{0.22*0.78}}{0.051})^2[/tex]
[tex]n = 357.2[/tex]
Rounding up
358 randomly selected air passengers must be surveyed.
