Respuesta :
Using the z-distribution for the margin of error, it is found that 355 samples should be taken.
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 given by:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
The estimate is of 18%, hence [tex]\pi = 0.18[/tex].
95% confidence, 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].
The minimum sample size is n for which M = 0.04, hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.18(0.82)}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96\sqrt{0.18(0.82)}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.18(0.82)}}{0.04}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.18(0.82)}}{0.04}\right)^2[/tex]
[tex]n = 354.4[/tex]
Rounding up, 355 samples should be taken.
A similar problem is given at https://brainly.com/question/14936818
