Standard Error

The standard deviation of the sample mean

The mean of a random sample of n values, X̄, has variance Var$\left(\bar{X}\right)=\frac{{\sigma }^2}{n}$(X̄) = σ²/n, where $\sigma$σ is the standard deviation of the population. Take the square root to get its standard deviation: $\sqrt{\frac{{\sigma }^2}{n}}=\frac{\sigma }{\sqrt{n}}$√(σ²/n) = σ/√n. The n comes out as a root, because $\sqrt{n}$√n is the square root of n.

The standard deviation of the sample mean is called the standard error, often written SE. It measures how far a sample mean typically lands from the population mean $\mu$μ.

Suppose adult heights have standard deviation $\sigma =12$σ = 12 cm, and you measure n = 36 people. The standard error is $12÷\sqrt{36}=12÷6=2$12 ÷ √36 = 12 ÷ 6 = 2 cm. Single heights are spread out by about 12 cm, but the mean of 36 heights by only about 2 cm.

Heights with mean 170 cm and $\sigma =12$σ = 12 cm. The two dashed curves are one height, the lowest and widest, and the mean of 4 heights, with standard error 12 ÷ 2 = 6 cm. The tall gold curve is the mean of 36, with standard error 2 cm.

Fast at first, then slowly

With $\sigma =12$σ = 12, the standard error is $\frac{12}{\sqrt{n}}$12/√n. One reading gives 12. Four readings give 12 ÷ 2 = 6, already half. Sixteen give 12 ÷ 4 = 3, thirty-six give 2, and sixty-four give 12 ÷ 8 = 1.5.

The first few readings make a large difference: going from 1 reading to 4 removes 6 cm of standard error. Going from 64 readings to 100 removes only 1.5 − 1.2 = 0.3 cm, although it adds 36 readings.

The standard error $\frac{12}{\sqrt{n}}$12/√n against the sample size n. It falls steeply from 12 at n = 1 to 6 at n = 4, and then flattens: 3 at n = 16, 2 at n = 36, 1.5 at n = 64.

Four times the readings for half the error

To halve the standard error, $\sqrt{n}$√n must double, and that needs four times as many readings, since $\sqrt{4n}=2\sqrt{n}$√(4n) = 2√n. The heights of 36 people give a standard error of 2 cm. For 1 cm, measure 4 × 36 = 144 people: $\sqrt{144}=12$√144 = 12, and 12 ÷ 12 = 1.

For 0.5 cm, halve it again with 4 × 144 = 576 people: $\sqrt{576}=24$√576 = 24, and 12 ÷ 24 = 0.5. Each halving costs four times the readings, so precision grows expensive quickly. To divide the standard error by 3, you need 9 times as many readings, and to divide it by 10, a hundred times as many.

Six hundred sample means, each from a sample of n values drawn from a population with mean 1 and standard deviation 1, piled up by size. At n = 16 the standard error is $1÷\sqrt{16}=0.25$1 ÷ √16 = 0.25. Drag n to 64 and it is 1 ÷ 8 = 0.125, and the pile closes in around 1, half as wide.

When $\sigma$σ is not known

In practice the population standard deviation is rarely known. It is estimated by the standard deviation of the sample, s, and the standard error is estimated by $\frac{s}{\sqrt{n}}$s/√n.

A machine fills bags of rice. A sample of 50 bags has standard deviation s = 3.5 g, so the standard error of their mean mass is about $3.5÷\sqrt{50}\approx 3.5/7.071\approx 0.495\text{g}$3.5 ÷ √50 ≈ 3.5 / 7.071 ≈ 0.495 g.

The standard error of a proportion

A proportion is a mean too. Score each person 1 if they say yes and 0 if they say no; the mean of the scores is the proportion who said yes. If the population proportion is p, each score has variance p(1 − p), so the standard error of the sample proportion is $\sqrt{\frac{p\left(1-p\right)}{n}}$√(p(1 − p)/n).

If half of all voters favor a plan, p = 0.5, and a poll asks 100 voters, the standard error of the poll’s proportion is $\sqrt{0.5\times 0.5÷100}=\sqrt{0.0025}=0.05$√(0.5 × 0.5 ÷ 100) = √0.0025 = 0.05, or 5 percentage points. With 400 voters it is $\sqrt{0.25÷400}=0.025$√(0.25 ÷ 400) = 0.025.

The usual mistakes

Dividing $\sigma$σ by n. It is the variance that is divided by n; $\sigma$σ is divided by $\sqrt{n}$√n. With $\sigma =12$σ = 12 and n = 36, the standard error is 12 ÷ 6 = 2, not $12÷36=\frac{1}{3}$12 ÷ 36 = 1/3.

Using $\sigma$σ for the mean. $\sigma =12$σ = 12 is the spread of one reading; the mean of 36 readings has standard error 2.

Doubling the sample to halve the error. Doubling n divides the standard error by $\sqrt{2}\approx 1.41$√2 ≈ 1.41, so 72 people give $12÷\sqrt{72}\approx 1.41$12 ÷ √72 ≈ 1.41 cm, not 1 cm.