Respuesta :
Answer:
A sample of 796 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 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]
In this question, we have that:
[tex]\pi = 0.53[/tex]
91% confidence level
So [tex]\alpha = 0.09[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.09}{2} = 0.955[/tex], so [tex]Z = 1.695[/tex].
How large should a sample be if the margin of error is .03 for a 91% confidence interval
We need a sample of n, which is found when [tex]M = 0.03[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.695\sqrt{\frac{0.53*0.47}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.695\sqrt{0.53*0.47}[/tex]
[tex]\sqrt{n} = \frac{1.695\sqrt{0.53*0.47}}{0.03}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.695\sqrt{0.53*0.47}}{0.03})^{2}[/tex]
[tex]n = 795.2[/tex]
Rounding up
A sample of 796 is needed.
