Respuesta :
Answer:
We need a sample size of at least 369.
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 margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
For this problem, we have that:
[tex]\pi = 0.32[/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 large of a sample size is required?
We need a sample size of at least n.
n is found when [tex]M = 0.04[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.645\sqrt{\frac{0.32*0.68}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.645\sqrt{0.32*0.68}[/tex]
[tex]\sqrt{n} = \frac{1.645\sqrt{0.32*0.68}}{0.04}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645\sqrt{0.32*0.68}}{0.04})^{2}[/tex]
[tex]n = 368.02[/tex]
Rounding up
We need a sample size of at least 369.
