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Using the z-distribution, it is found that the smallest sample size required to obtain the desired margin of error is of 2401.
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.
[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 of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence level, hence[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].
In this problem, there is no prior estimate, hence [tex]\pi = 0.5[/tex] is used.
The smallest sample size needed is n for which M = 0.02, hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.96(0.5)[/tex]
[tex]\sqrt{n} = \frac{1.96(0.5)}{0.02}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96(0.5)}{0.02}\right)^2[/tex]
[tex]n = 2401[/tex]
The smallest sample size required to obtain the desired margin of error is of 2401.
You can learn more about the z-distribution at https://brainly.com/question/25844663
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