Respuesta :
Using the margin of error, it is found that he must survey 601 adults in order to be 95% confidence that his estimate is within 4 percentage points of the true population percentage.
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]\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]\frac{1+\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confident, hence:
[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
We have no estimate, hence, [tex]\pi = 0.5[/tex] is used.
Within 4%, hence, it is needed to find n for which M = 0.04.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96(0.5)[/tex]
[tex]\sqrt{n} = \frac{1.96(0.5)}{0.04}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96(0.5)}{0.04}\right)^2[/tex]
[tex]n = 600.25[/tex]
Rounding up, 601 adults must be sampled.
A similar problem is given at https://brainly.com/question/15133177
