Answer:
The minimum sample size should be of 381.
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 given by:
[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].
Without any prior knowledge of the proportion, what should be our minimum sample size
We dont know the population proportion, so we use [tex]\pi = 0.5[/tex], which is when the largest sample size is needed. We have to find n for which [tex]M = 0.05[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 1.95\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.05\sqrt{n} = 1.95*0.5[/tex]
[tex]\sqrt{n} = \frac{1.95*0.5}{0.05}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.95*0.5}{0.05})^2[/tex]
[tex]n = 380.25[/tex]
Rounding up
The minimum sample size should be of 381.