Q.The mean and standard deviation of a set of $n_1$ observations are $\overline{x}_1$ and $s_1$, respectively while the mean and standard deviation of another set of $n_2$ observations are $\overline{x}_2$ and $s_2$, respectively. Show that the standard deviation of the combined set of $(n_1 + n_2)$ observations is given by

[!FORMULA] $\text{S.D.} = \sqrt{\frac{n_1(s_1)^2 + n_2(s_2)^2}{n_1 + n_2} + \frac{n_1 n_2 (\overline{x}_1 - \overline{x}_2)^2}{(n_1 + n_2)^2}}$

Concept: Combined Standard Deviation via Variance Decomposition

The variance of a combined dataset depends on both the within-group variances and the separation between group means.

Step 1. The combined mean is $\overline{x} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2}$.

Step 2. Variance of the first group about the combined mean:

The cross-term vanishes since $\sum(x_i - \overline{x}_1) = 0$. Similarly for the second group: $s_2^2 + (\overline{x}_2 - \overline{x})^2$.

Step 3. Combined variance:

$\sigma^2 = \frac{n_1[s_1^2 + (\overline{x}_1 - \overline{x})^2] + n_2[s_2^2 + (\overline{x}_2 - \overline{x})^2]}{n_1 + n_2}$ …

Why combining datasets isn't straightforward

When you merge two groups of observations, you can't simply average their standard deviations. The combined variance depends not only on how spread out each group is internally, but also on how far apart the two group means are from each other. If the groups have very different means, the combined dataset will show additional spread beyond what either group had individually.

The key insight: variance measures the average squared distance from the mean. When we pool data, we need a new combined mean, and every observation's distance must be recalculated from this new center.

Derivation

Let the first set have observations $x_1, x_2, \ldots, x_{n_1}$ and the second set have $y_1, y_2, \ldots, y_{n_2}$.

1. Find the combined mean

The combined mean $\overline{x}$ of all $(n_1 + n_2)$ observations is the weighted average:

This follows because the sum of all observations equals $n_1 \overline{x}_1 + n_2 \overline{x}_2$.

2. Express the combined variance

The variance of the combined set is:

We need to relate each term to the known quantities $s_1^2$, $s_2^2$, $\overline{x}_1$, and $\overline{x}_2$.

3. Rewrite the first sum using a clever algebraic trick

For any observation $x_i$ from the first group:

Squaring both sides:

Summing over all $n_1$ observations:

where we used $s_1^2 = \frac{1}{n_1} \sum_{i=1}^{n_1} (x_i - \overline{x}_1)^2$.

4. Apply the same logic to the second group

By identical reasoning:

5. Combine and simplify

The combined variance becomes:

6. Simplify the second term

Substitute $\overline{x} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2}$: …