Imagine you have a random sample with n observations x_1, …, x_n, independently and identically distributed according to some distribution with mean mu and variance s^2.
The sample mean is sum(x_i)/n (the expected value is mu as one would hope). Doing some manipulations we find that this has variance s^2/n, i.e. a large n means a small variance, so larger samples are more tightly clustered around mu.
That’s not such a rigorous answer:
Imagine you have a random sample with
nobservationsx_1, …,x_n, independently and identically distributed according to some distribution with meanmuand variances^2.The sample mean is
sum(x_i)/n(the expected value ismuas one would hope). Doing some manipulations we find that this has variances^2/n, i.e. a largenmeans a small variance, so larger samples are more tightly clustered aroundmu.