Respuesta :
Answer:
We need a sample of size at least 13.
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]
90% confidence interval: (0.438, 0.642).
The proportion estimate is the halfway point of these two bounds. So
[tex]\pi = \frac{0.438 + 0.642}{2} = 0.54[/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].
Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?
We need a sample of size at least n.
n is found when M = 0.08. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.08 = 1.96\sqrt{\frac{0.54*0.46}{n}}[/tex]
[tex]0.08\sqrt{n} = 1.96\sqrt{0.54*0.46}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.54*0.46}}{0.08}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96\sqrt{0.54*0.46}}{0.08})^{2}[/tex]
[tex]n = 12.21[/tex]
Rounding up
We need a sample of size at least 13.
