Why does the margin of error decrease as the sample size $n$ $n$ increases?

Question

4788 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-10T08:30:40

Margin of error = $C \cdot \frac{s}{\sqrt{n}}$ $C * \frac{s}{\sqrt{n}}$

where C = critical value and $s =$ $s =$ standard deviation, and $n =$ $n =$ sample size.

Because of this inverse relationship, as sample size increases, the margin of error decreases.

Why does the margin of error decrease as the sample size nnn increases?