Respuesta :
Answer:
A sample of 801 is needed.
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]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
25% of U.S. homes have a direct satellite television receiver.
This means that [tex]\pi = 0.25[/tex]
How large a sample is necessary to estimate the true population of homes which do with 95% confidence and within 3 percentage points?
This is n for which M = 0.03. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.25*0.75}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.25*0.75}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.25*0.75}}{0.03}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.25*0.75}}{0.03})^2[/tex]
[tex]n = 800.3[/tex]
Rounding up:
A sample of 801 is needed.
