2.4 Is the sample mean centred on the truth?

The sample means cluster around 7. Is that a coincidence of this simulation?

No individual sample produced 7. Yet the average of the thousand sample means is 6.932 — essentially the population value.

It is not a coincidence, and it does not depend on the deck. It follows from the definition of the sample mean and one line of algebra.

\[ \begin{aligned} E(\bar{X}) &= E\!\left(\frac{X_1 + X_2 + \cdots + X_n}{n}\right) \\[4pt] &= \frac{1}{n}\Big(E(X_1) + E(X_2) + \cdots + E(X_n)\Big) \\[4pt] &= \frac{1}{n}\underbrace{(\mu + \mu + \cdots + \mu)}_{n \text{ times}} \\[4pt] &= \frac{n\mu}{n} \;=\; \mu \end{aligned} \]

The third line is the one that matters. It replaces every \(E(X_i)\) with the same \(\mu\), and it is entitled to do so only because the observations are identically distributed — each drawn from the population we are asking about. Averaging \(n\) of them then leaves \(\mu\) unchanged.

Look back at Survey B. Its observations were identically distributed among themselves, but not drawn from the population of interest, so the \(\mu\) substituted in that third line was the urban mean rather than the national one. The algebra was never wrong; it answered a question nobody had asked.

This is the first of the two properties promised earlier, and it now has a name.

An estimator is unbiased if its expected value equals the parameter it estimates. The sample mean is unbiased for the population mean:

\[E(\bar{X}) = \mu\]

Unbiasedness is a statement about the procedure, not about any particular estimate. It says the rule is not systematically too high or too low. It does not say your \(\bar{x}\) is close to \(\mu\) — every one of the thousand samples above missed.