Respuesta :
Answer:
A sample of 1068 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]
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].
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?
We need a sample of n.
n is found when M = 0.03.
We have no prior estimate of [tex]\pi[/tex], so we use the worst case scenario, which is [tex]\pi = 0.5[/tex]
Then
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.03}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}[/tex]
[tex]n = 1067.11[/tex]
Rounding up
A sample of 1068 is needed.
