# What happens to the sample standard deviation when the sample size is increased?

Jan 22, 2017

It depends on the actual data added to the sample, but generally, the sample S.D. will approach the actual population S.D.

#### Explanation:

The formula for sample standard deviation is

$s = \sqrt{\frac{{\sum}_{i = 1}^{n} {\left({x}_{i} - \overline{x}\right)}^{2}}{n - 1}}$

while the formula for the population standard deviation is

$\sigma = \sqrt{\frac{{\sum}_{i = 1}^{N} {\left({x}_{i} - \mu\right)}^{2}}{N - 1}}$

where

• $n$ is the sample size,
• $N$ is the population size,
• $\overline{x}$ is the sample mean, and
• $\mu$ is the population mean.

As $n$ increases towards $N$, the sample mean $\overline{x}$ will approach the population mean $\mu$, and so the formula for $s$ gets closer to the formula for $\sigma$.

Thus, as $n \to N , s \to \sigma$.

## Note:

When $n$ is small compared to $N$, the sample mean $\overline{x}$ may behave very erratically, darting around $\mu$ like an archer's aim at a target very far away. Adding a single new data point is like a single step forward for the archer—his aim should technically be better, but he could still be off by a wide margin. Thus, incrementing $n$ by 1 may shift $\overline{x}$ enough that $s$ may actually get further away from $\sigma$.

It is only over time, as the archer keeps stepping forward—and as we continue adding data points to our sample—that our aim gets better, and the accuracy of $\overline{x}$ increases, to the point where $s$ should stabilize very close to $\sigma$.