Respuesta :
Answer:
We must sample at least 1068 individuals.
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 half-width of the interval is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
They desire the interval to have a half-width of no more than 0.03. How many individuals must they sample?
We do not know the true proportion, so we use [tex]\pi = 0.5[/tex], which is the proportion for which we are going to need the largest sample size.
They must sample at least n individuals, in which N is found when [tex]M = 0.03[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96*0.5[/tex]
[tex]0.03\sqrt{n} = 0.98[/tex]
[tex]\sqrt{n} = \frac{0.98}{0.03}[/tex]
[tex](\sqrt{n})^{2} = (\frac{0.98}{0.03})^{2}[/tex]
[tex]n = 1067.1[/tex]
Rounding up
We must sample at least 1068 individuals.
