An analyst takes a random sample of 25 firms in the telecommunications industry and constructs a confidence interval for the mean return for the prior year. Holding all else constant, if he increased the sample size to 30 firms, how are the standard error of the mean and the width of the confidence interval affected? (A) Standard error of the mean increases, Width of confidence interval becomes wider.(B) Standard error of the mean increases, width of confidence intervale becomes narrower.(C) Standard error of the mean Decreases, width of confidence interval becomes wider.(D) Standard error of the mean decreases, width of confidence interval becomes narrower.(E) Cannot be determined.

Answer:The standard error decreases and the width of the confidence interval also decreases.Step-by-step explanation:The standard error of a distribution (E) is given as:[tex]E=z_{\frac{\alpha}{2} }*\frac{\sigma}{\sqrt{n} }[/tex]Where n is the sample size, [tex]z_{\frac{\alpha}{2} }[/tex] is the z score of he confidence and [tex]\sigma[/tex] is the standard deviation.The sample size is inversely proportional to the standard error. If the sample size is increased and everything else is constant, the standard would decrease since they are inversely proportional to each other. The confidence interval = μ ± E = (μ - E, μ + E). μ is the meanThe width of the confidence interval = μ + E - (μ - E) = μ + E - μ + E = 2EThe width of the confidence interval is 2E, therefore as the sample size increase, the margin of error decrease and since the width of the confidence interval is directly proportional to the margin of error, the width of the confidence interval also decreases.