Respuesta :
Answer:
A sample size of 474 is required.
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]
Based on previous evidence, you believe the population proportion is approximately 73%.
This means that [tex]\pi = 0.73[/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].
How large of a sample size is required?
A sample size of n is required, and n is found when M = 0.04. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.73*0.27}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96\sqrt{0.73*0.27}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.73*0.27}}{0.04}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.73*0.27}}{0.04})^2[/tex]
[tex]n = 473.24[/tex]
Rounding up:
A sample size of 474 is required.
