Sample mean and the central limit theorem: VCE Specialist Mathematics Unit 4

What this dot point is asking

VCAA wants you to understand that the sample mean $\bar{X}$ is itself a random variable with its own distribution, to know its mean and standard deviation (the standard error), to see how increasing the sample size sharpens that distribution, and to state and use the central limit theorem that makes $\bar{X}$ approximately normal for large samples. This underpins confidence intervals and hypothesis testing.

The sample mean is a random variable

Take a random sample $X_1, X_2, \dots, X_n$ of size $n$ from a population. The sample mean

depends on which sample is drawn, so it varies from sample to sample: it is a random variable with its own distribution, called the sampling distribution of the mean.

Mean and standard error

Suppose the population has mean $\mu$ and standard deviation $\sigma$, and the observations are independent. Using the rules for linear combinations of independent random variables:

The mean of $\bar{X}$ equals the population mean, so $\bar{X}$ is an unbiased estimator of $\mu$. Its standard deviation, called the standard error, is $\frac{\sigma}{\sqrt{n}}$, smaller than the population standard deviation by a factor of $\sqrt{n}$.

The effect of sample size

Because the standard error is $\frac{\sigma}{\sqrt{n}}$, larger samples give a sample mean clustered more tightly around $\mu$. The shrinking is by $\frac{1}{\sqrt{n}}$, not $\frac{1}{n}$: to halve the standard error you must quadruple the sample size. This diminishing return is why very precise estimates need large samples.

The central limit theorem

The central limit theorem states that, for a sufficiently large sample size $n$, the distribution of the sample mean is approximately normal:

regardless of the shape of the population distribution. So even if the population is skewed or discrete, the sample mean of a large sample behaves normally. This is what lets us use normal-based methods (confidence intervals, hypothesis tests) for the mean. If the population is itself normal, $\bar{X}$ is exactly normal for any $n$.

Examples in context

Example 1. Quadrupling $n$ from $25$ to $100$ changes the standard error from $\frac{\sigma}{5}$ to $\frac{\sigma}{10}$, halving it.

Example 2. For a normal population $\bar{X}$ is exactly normal for any $n$, so no large-sample approximation is needed.

Try this

Q1. State the mean and standard error of $\bar{X}$ for $\mu = 20$, $\sigma = 4$, $n = 16$. [2 marks]

• Cue. Mean $20$, standard error $\frac{4}{\sqrt{16}} = 1$.

Q2. By what factor must $n$ increase to halve the standard error? [1 mark]

• Cue. A factor of $4$.

Q3. State the central limit theorem in one sentence. [2 marks]

• Cue. For large $n$, $\bar{X}$ is approximately normal with mean $\mu$ and standard error $\frac{\sigma}{\sqrt{n}}$, whatever the population shape.