According to the information given, using the z-distribution, it is found that a sample of 1066 is needed.
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]\alpha[/tex], we have the following confidence interval of proportions, according to the z-distribution.
[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]\frac{1+\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In this problem, the estimate is of 52%, hence [tex]\pi = 0.52[/tex].
95% confidence level
So [tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
The minimum sample size is n for which M = 0.03, hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.52(0.48)}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.52(0.48)}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.52(0.48)}}{0.03}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.52(0.48)}}{0.03})^2[/tex]
[tex]n = 1065.4[/tex]
Rounding up, a sample of 1066 is needed.
A similar problem is given at https://brainly.com/question/14515126
