A university is conducting a survey in order to estimate the results of an upcoming local election. They intend to take a simple random sample of individuals from the population of eligible voters. Each selected individual will be asked whether they will vote for Candidate A, yes or no. The university intends to use the results to create a 95% CI for the population proportion of voters who will vote for Candidate A. They desire the interval to have a half-width of no more than 0.03. How many individuals must they sample? (Hint: the half-width of a CI for a population proportion is maximized when the sample proportion is set to p = 0.5.)

Answer:We must sample at least 1068 individuals.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 whichz is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].The half-width of the interval is:[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]95% confidence levelSo [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].They desire the interval to have a half-width of no more than 0.03. How many individuals must they sample?We do not know the true proportion, so we use [tex]\pi = 0.5[/tex], which is the proportion for which we are going to need the largest sample size.They must sample at least n individuals, in which N is found when [tex]M = 0.03[/tex]. So[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]0.03\sqrt{n} = 0.98[/tex][tex]\sqrt{n} = \frac{0.98}{0.03}[/tex][tex](\sqrt{n})^{2} = (\frac{0.98}{0.03})^{2}[/tex][tex]n = 1067.1[/tex]Rounding upWe must sample at least 1068 individuals.