Respuesta :
Answer:
The expression is [tex]n = (\frac{1.645*0.5}{0.03})^2[/tex]
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is of:
[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 p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
What expression would give the smallest sample size that will result in a margin of error of no more than 3 percentage points?
We have to find n for which M = 0.03.
We have no prior estimate for the proportion, so we use [tex]\pi = 0.5[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.645*0.5[/tex]
[tex]\sqrt{n} = \frac{1.645*0.5}{0.03}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.645*0.5}{0.03})^2[/tex]
[tex]n = (\frac{1.645*0.5}{0.03})^2[/tex]
The expression is [tex]n = (\frac{1.645*0.5}{0.03})^2[/tex]
