Describe the distribution of the sample mean, use the Central Limit Theorem, and find unbiased estimates of the population mean and variance from a sample

What this dot point is asking

SEAB wants you to describe the distribution of the sample mean, state and use the Central Limit Theorem, compute the standard error, and find unbiased estimates of the population mean and variance from sample data.

The answer

The sampling distribution of the mean

If samples of size $n$ are drawn from a population with mean $\mu$ and variance $\sigma^2$, the sample mean $\bar{X}$ has

The mean of the sampling distribution equals the population mean (so $\bar{X}$ is unbiased), and its spread shrinks as $n$ grows. The standard error is $\dfrac{\sigma}{\sqrt{n}}$.

The Central Limit Theorem

The Central Limit Theorem (CLT) states that for a sufficiently large sample size $n$, the sample mean is approximately normally distributed,

regardless of the population's distribution. This is what lets us use normal-based methods even when the population is not normal, provided $n$ is large (commonly $n \geq 30$).

Unbiased estimators

From a sample, the unbiased estimate of the population mean is the sample mean $\bar{x} = \dfrac{\sum x}{n}$. The unbiased estimate of the population variance uses the $n - 1$ divisor:

Dividing by $n - 1$ rather than $n$ corrects the tendency of the sample to underestimate the spread.

Why the standard error matters

Because $\operatorname{Var}(\bar{X}) = \dfrac{\sigma^2}{n}$ decreases with $n$, larger samples give more precise estimates of $\mu$. This is the quantitative reason bigger samples are better.

Examples in context

Example 1. Quality monitoring. A production line checks samples of $40$ items; by the CLT the sample mean weight is approximately normal even though individual weights are not, letting an operator flag drift using normal control limits.

Example 2. Survey precision. Quadrupling a survey's sample size halves the standard error (since it scales as $\frac{1}{\sqrt{n}}$), the quantitative trade-off pollsters weigh between cost and precision.

Try this

Q1. A population has $\sigma = 10$. Find the standard error of the mean for a sample of size $25$. [2 marks]

• Cue. $\dfrac{10}{\sqrt{25}} = 2$.

Q2. State what the Central Limit Theorem guarantees about the sample mean. [2 marks]

• Cue. For large $n$, $\bar{X}$ is approximately normal with mean $\mu$ and variance $\frac{\sigma^2}{n}$, regardless of the population distribution.

Q3. Why is the population variance estimated with an $n - 1$ divisor? [1 mark]

• Cue. To make the estimator unbiased, correcting the sample's tendency to underestimate the spread.