Variance and standard deviation

Spread, in the units of the data

Two branches can share a mean and behave completely differently. The variance measures how far values sit from the mean, on average and squared; the standard deviation takes the square root so the answer is back in the data's units.

$$s^2=\frac{\sum (x_i-\bar{x}{)}^2}{n-1},s=\sqrt{s^2}$$

The shortcut form avoids computing every deviation:

$$s^2=\frac{\sum x_i^2-n{\bar{x}}^2}{n-1}$$

It needs only two sums — $\sum x$ and $\sum x^2$ — and those two sums are the sums-table habit that Module 4 (covariance, the slope) and Module 13 reuse. Build the table once and every later formula reads from it.

Worked example — Bramble & Voss audit hours

A Guelph audit practice, Bramble & Voss, logged its monthly audit-hour totals for six months: 42, 38, 51, 45, 39, 49.

The sums. $\sum x=264$, $n=6$, $\bar{x}=264/6=44$. $\sum x^2=1,764+1,444+2,601+2,025+1,521+2,401=11,756$.

Shortcut. $s^2=\frac{11,756-6\times 44^2}{5}=\frac{11,756-11,616}{5}=\frac{140}{5}=28$, so $s=\sqrt{28}=5.29$ hours.

Deviation table, as a check. The deviations are −2, −6, 7, 1, −5, 5 (they sum to 0, as they must); squared, 4, 36, 49, 1, 25, 25, which total 140. Same numerator, same 28. When the two disagree, the arithmetic is wrong somewhere — usually a squared deviation.

Read it: a typical month is about 5.3 hours from the 44-hour mean.

Why n − 1

The deviations $x_i-\bar{x}$ are not six free numbers. Because $\bar{x}$ was computed from the same data, they are forced to sum to zero — once five are known, the sixth is fixed. Only $n-1$ of them carry independent information, and that is the degrees of freedom the divisor counts. The second justification is the one Module 8 will make precise: dividing by $n$ produces a variance that is too small on average — the sample's values are closer to their own mean than to the population mean — and $n-1$ corrects the bias exactly.

Every tool in this course agrees: statistics.stdev, R's sd() and var(), and the cheat sheet all divide by $n-1$. The one that does not is statistics.pstdev (and numpy's default std), which divides by $n$ and is for a population — the whole set of units, not a sample of them. On the audit hours it prints 4.83 where 5.29 is expected; a 4.83 on the paper is this mistake and loses the mark.

A second run, by deviations

Five values: 10, 12, 15, 11, 17. $\bar{x}=65/5=13$. Deviations −3, −1, 2, −2, 4; squared 9, 1, 4, 4, 16; total 34. $s^2=34/4=8.5$, $s=2.915$. The shortcut gives the same: $\sum x^2=879$, $879-5\times 169=34$.