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 mean
The width of the confidence interval = μ + E - (μ - E) = μ + E - μ + E = 2E
The 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.
