A statistic is a random variable

◈ 5 cards

A sample mean varies from sample to sample, so it has a distribution of its own — the sampling distribution — built here by listing every sample of size 2 from four cheques.

The number you compute is itself random

Module 2 drew a sample and Module 3 computed its mean. Draw a different sample and the mean changes. The sample mean $\overset{x}{ˉ}$ is therefore a random variable — its value depends on which sample chance handed you — and like every random variable it has a distribution: the sampling distribution of $\overset{ˉ}{X}$ , the distribution of the sample mean over all the samples that could have been drawn.

This is the most confused sentence in the course, so fix it now. The distribution of $X$ describes one cheque: how long a single cheque takes to clear. The sampling distribution of $\overset{ˉ}{X}$ describes the average of a sample of cheques. Same population, different random variable, different distribution. Every exam question says which it is asking about.

Worked example — every sample of two cheques

Harbourview Clearing has four cheques in its queue, with clearing times of 2, 4, 6 and 8 days. Treat these four as the population: $μ = (2 + 4 + 6 + 8) /4 = 5$ days. Draw a sample of $n = 2$ cheques without replacement. There are $(2 4) = 6$ possible samples, each equally likely:

Sample	$\overset{x}{ˉ}$
{2, 4}	3
{2, 6}	4
{2, 8}	5
{4, 6}	5
{4, 8}	6
{6, 8}	7

The six sample means are 3, 4, 5, 5, 6, 7. Collecting them into a pmf — Module 6's table — gives the sampling distribution:

$\overset{x}{ˉ}$	3	4	5	6	7
$P (\overset{ˉ}{X} = \overset{x}{ˉ})$	1/6	1/6	2/6	1/6	1/6

It is a distribution like any other, and it answers probability questions: $P (\overset{ˉ}{X} \geq 6) = 2/6 = 0.3333$ — one sample in three has an average clearing time of six days or more, though only one cheque in four actually takes that long.

Two things the table shows

Its mean is $μ$ . The average of the six sample means is $(3 + 4 + 5 + 5 + 6 + 7) /6 = 30/6 = 5 = μ$ . The sample mean is neither systematically too high nor too low: it is an unbiased estimator of $μ$ . Unbiased does not mean right — four of the six samples miss $μ$ — it means the misses average out to zero.

Its spread is smaller than the population's. The population values range from 2 to 8; the sample means only from 3 to 7, and they crowd around 5. Averaging cancels some of the variation. L8.2 puts a formula on exactly how much.

Why you never do this again

Enumeration works for four cheques. For a population of 40,000 invoices and samples of 36 there are more samples than atoms in the galaxy, and no table can be built. Modules 8.2–8.4 replace the table with two formulas and a theorem that give the sampling distribution's mean, spread and shape without listing anything. But the meaning is fixed by this table: the sampling distribution is what you would see if you drew every possible sample and computed $\overset{x}{ˉ}$ for each.

Choosing between estimators

If two statistics both estimate $μ$ without bias, prefer the one whose sampling distribution is tighter — the smaller standard error, the name for the SD of a sampling distribution. The sample mean beats the sample median on this criterion for normal data, which is why the paper's inference is built on $\overset{x}{ˉ}$ .

The sampling distribution of x̄ built by hand. Its mean is 5 = μ (x̄ is unbiased); its range 3–7 is narrower than the population’s 2–8.