The sampling distribution of the sample mean (WACE Year 12 Mathematics Specialist Unit 4)

What this dot point is asking

SCSA wants you to treat the sample mean as a random variable, state its mean and standard error, and explain the effect of sample size, before bringing in the central limit theorem.

Why we study the sample mean

The sample mean is the natural estimator of the population mean: it is the single number a survey or experiment reports to summarise a sample. But because each sample is a different random draw, the reported value would change if the sample were repeated, and statistics is the study of how to reason reliably despite that variation. The sampling distribution captures exactly this variation, telling us how far a single sample mean is likely to fall from the true population mean, and it is the foundation on which the central limit theorem and confidence intervals are built later in the unit.

The sample mean as a random variable

Take a random sample of size $n$ from a population with mean $\mu$ and standard deviation $\sigma$. The sample mean $\bar{x}$ depends on which units happen to be drawn, so it changes from sample to sample. Thinking of all possible samples, the sample mean is a random variable $\bar{X}$ with its own distribution, called the sampling distribution.

Its mean and standard error

The first says the sample mean is unbiased: on average it equals the population mean. The second, the standard error, measures how much the sample mean typically deviates from $\mu$. Crucially it has $\sqrt{n}$, not $n$, in the denominator.

Why the mean and standard error take these values

The two results follow from the rules for the mean and variance of a sum of independent random variables. The sample mean is $\bar{X} = \dfrac{1}{n}(X_1 + X_2 + \dots + X_n)$ where each $X_i$ is an independent draw from the population. Since each $E(X_i) = \mu$, linearity of expectation gives $E(\bar{X}) = \dfrac{1}{n}(n\mu) = \mu$. For the variance, independence lets the variances add, so $\text{Var}(X_1 + \dots + X_n) = n\sigma^2$, and dividing by $n$ scales the variance by $\dfrac{1}{n^2}$, giving $\text{Var}(\bar{X}) = \dfrac{n\sigma^2}{n^2} = \dfrac{\sigma^2}{n}$. Taking the square root yields the standard error $\dfrac{\sigma}{\sqrt{n}}$. This derivation explains why the $\sqrt{n}$, not $n$, appears.

Why precision improves with sample size

This explains why large samples give sample means tightly clustered around $\mu$, while small samples scatter widely. The population spread $\sigma$ is fixed; only the standard error shrinks.

The shape of the sampling distribution

The mean and standard error describe the centre and spread of the sampling distribution, but not its shape. The shape depends on the population. If the population is itself normal, then $\bar{X}$ is exactly normal for every sample size, however small, because a sum of independent normal variables is normal. If the population is not normal, $\bar{X}$ is only approximately normal, and only for large $n$; that approximation is the content of the central limit theorem, which builds directly on this dot point. Distinguishing the exact-normal case (normal population) from the approximate-normal case (any population, large $n$) is a common SCSA discrimination, so always note which situation a question describes before quoting a distribution for $\bar{X}$.