Using the z-distribution, it is found that a sample size of 1066 is required.
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 .
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
From the previous study, the estimate is of 0.23, hence [tex]\pi = 0.23[/tex].
98% confidence level, hence[tex]\alpha = 0.98[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.98}{2} = 0.99[/tex], so [tex]z = 2.327[/tex].
The sample size is n for which M = 0.03, hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 2.327\sqrt{\frac{0.23(0.77)}{n}}[/tex]
[tex]0.03\sqrt{n} = 2.327\sqrt{0.23(0.77)}[/tex]
[tex]\sqrt{n} = \left(\frac{2.327\sqrt{0.23(0.77)}}{0.03}\right)[/tex]
[tex](\sqrt{n})^2 = \left(\frac{2.327\sqrt{0.23(0.77)}}{0.03}\right)^2[/tex]
[tex]n = 1065.5[/tex]
Rounding up, a sample size of 1066 is required.
A similar problem is given at https://brainly.com/question/12517818
