# Combined Data Set Standard Deviation

Can the sample deviation of a combined data set be lower than the sample deviations of each separate data set? So, at first thought, my answer was no and that it can be equal to that of each separate data set at the lowest. However, when I tried to show this, I supposed that the mean, sample standard deviations, and sample size of both data sets are equal.

Fooling around with some examples such as: $$\text{data set 1 = data set 2: 1,2,3,4,5} \quad N=5, \mu=3, s\approx 1.58 \Rightarrow$$ $$\text{combined data set: 1,1,2,2,3,3,4,4,5,5} \quad N=10,\mu=3 s\approx 1.49$$ So I'm thinking that for all sample data sets in which the sample sizes, sample standard deviations, and mean are equal, the sample standard deviation of the combined data set will be lower, while it will approach an equal value as $N$ goes to infinity.

How can I show this mathematically?

Usually the sample variance is taken to be the unbiased estimator of $\sigma_X^2$:$$s^2\equiv \frac{1}{N-1} \sum_{i = 1}^{N} (x_i - \bar{X})^2$$

So when you combine two identical data sets, the sample size doubles $N \to 2N$, the sample mean $\bar{X}$ is unchanged, the sum of squares $\sum_{i = 1}^{N} (x_i - \bar{X})^2$ also doubles, and you divide it by a number that is slightly less than doubled $2N -1$.

When $N \to \infty$, the sample size approaches "being doubled" $(2N-1)/N \to 2$ and $s^2 \to \sigma_X^2$, approaching what you call "equal value".

• You should use the standard notation: sample size (small) $n$, sample mean $\bar X$, sample variance $s^2,$ and sample standard deviation $s.$ It is customary to use $N$ and $\mu$ for populations. – BruceET Oct 4 '16 at 0:04
• @BruceET yeah you're right. – Lee David Chung Lin Oct 4 '16 at 0:14
• Intended for OP, put in wrong place. – BruceET Oct 6 '16 at 4:57

Start with the sample means $\bar X_1 = T_1/n = \frac{\sum_{i=1}^n X_i}{n}$ of the first sample and $\bar X_2 = T_2/m= \frac{\sum_{i=n+1}^{n+m} X_i}{m}$ of the second. Then the sample mean of the combined sample is $X_c = T_c/(n+m) = (T_1 + T_2)/(n+m) = \frac{\sum_{i-1}^{n+m} X_i} {n+m}.$ So if you know the sample sizes $n$ and $n+m$ and the sample means $\bar X_1$, $\bar X_c,$ you can easily solve for $\bar X_2.$

Now consider that the variance $s^2$ of a sample can be written in terms of the 'computational formula' $$s^2 = \frac{\sum_{i=1}^n X_i^2 - n\bar X^2}{n-1} = \frac{Q - n\bar X^2}{n-1}.$$

Now if the sample sizes, sample sum, and sample sums of squares of two samples are known, it is possible to combine the sample sizes, find $T_c = T_1 + T_2$, and find $Q_c = Q_1 + Q_2,$ and then use these quantities to find $s_c^2$ from $s_1^2$ and $s_2^2.$

Further, knowing the sample variance, mean, and total, it is possible to solve for the sample sum of squares. It follows that sample sizes, means, and variances for individual samples can be used to find sample sizes, means, and variances for combined samples.

Now you can use sample size 5, mean 3, and SD 1.58 for sample 1 and sample size 10, mean 3, and SD 1.49 to find exact sample size, mean, and SD for sample 2. No guess work about it.

Finally, if all samples are from the same population with mean $\mu$ and standard devation $\sigma$, we expect sample means and SDs to get closer to $\mu$ and $\sigma$ as sample size increases.