Q.The population standard deviation of the monthly sales (in Rs. '000) of a chain's outlets is known to be Rs. 15,000 (i.e., sigma = 15 in Rs. '000). Find the standard error of the sample mean when

Since SE depends on the square root of the sample size, increasing n by a factor of 9 (from 25 to 225) should shrink SE by a factor of exactly 3, not 9 - this numerically demonstrates why a much larger sample is needed to meaningfully tighten an estimate. …

Given $\sigma = 15$ (Rs. '000).

1. $n = 25$:

   $SE = \dfrac{\sigma}{\sqrt{n}} = \dfrac{15}{\sqrt{25}} = \dfrac{15}{5} = 3$

2. $n = 225$:

   $SE = \dfrac{15}{\sqrt{225}} = \dfrac{15}{15} = 1$

   Check: $\sqrt{25} = 5$ and $\sqrt{225} = 15$ exactly (since $15^2 = 225$); both divisions ($15\div5=3$, $15\div15=1$) are confirmed by direct calculation. …

A very common mistake is to assume SE falls in direct (linear) proportion to n - e.g., expecting a 9-fold increase in n to re …