Respuesta :
Answer:
655 people would need to 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].
In this question, we have that:
[tex]\pi = 0.68[/tex]
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
How many people would need to be surveyed for a 90% confidence interval to ensure the margin of error would be less than 3%?
We need to survey n adults.
n is found when M = 0.03. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.645\sqrt{\frac{0.68*0.32}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.645\sqrt{0.68*0.32}[/tex]
[tex]\sqrt{n} = \frac{1.645\sqrt{0.68*0.32}}{0.03}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645\sqrt{0.68*0.32}}{0.03})^{2}[/tex]
[tex]n = 654.3[/tex]
Rounding up
655 people would need to be surveyed.
