Compute unbiased estimates of a population mean and variance and construct and interpret confidence intervals for a population mean

What this dot point is asking

SEAB wants you to estimate population parameters from sample data: to compute unbiased estimates of the population mean and variance (the latter using the $n - 1$ divisor), and to construct and interpret a confidence interval for a population mean. The interpretation of a confidence interval, in terms of the long-run capture rate, is examined as carefully as the calculation.

The answer

Estimators and unbiasedness

A statistic computed from a sample is an estimator of a population parameter. It is unbiased if its expected value equals the parameter it estimates. The sample mean is an unbiased estimator of the population mean:

The unbiased estimate of variance

Dividing the sum of squared deviations by $n$ underestimates the population variance, because the deviations are taken about the sample mean. The unbiased estimate uses $n - 1$:

The divisor $n - 1$ (the degrees of freedom) is what makes $s^2$ unbiased.

The distribution of the sample mean

For a sample of size $n$ from a population with mean $\mu$ and variance $\sigma^2$, the sample mean has

By the Central Limit Theorem, for large $n$ the sample mean is approximately normally distributed, which underpins the confidence interval.

Constructing a confidence interval for the mean

A confidence interval gives a range of plausible values for $\mu$. With the population standard deviation $\sigma$ known (or a large sample),

where $z^*$ is the critical value for the chosen confidence level ($1.96$ for $95\%$, $2.576$ for $99\%$). The term $\dfrac{\sigma}{\sqrt{n}}$ is the standard error and $z^*\dfrac{\sigma}{\sqrt{n}}$ the margin of error.

Interpreting a confidence interval

A $95\%$ confidence interval does not mean there is a $95\%$ probability the true mean lies in this particular interval. It means the procedure produces an interval that captures the true mean in $95\%$ of repeated samples. Wider confidence (say $99\%$) gives a wider interval; a larger sample narrows it.

Examples in context

Example 1. Polling. A political poll reports a percentage with a margin of error; that margin is $z^*\dfrac{\sigma}{\sqrt{n}}$, and the "95% confidence" caveat is exactly the long-run capture interpretation, which is why larger polls report tighter margins.

Example 2. Quality assurance. A factory estimates the mean fill of bottles from a sample and reports a confidence interval; if the target value lies outside the interval, the process is flagged, linking estimation directly to hypothesis testing.

Try this

Q1. Why does the unbiased estimate of variance divide by $n - 1$? [2 marks]

• Cue. Deviations are measured about the sample mean, which understates spread; the $n - 1$ divisor (degrees of freedom) corrects the bias.

Q2. State the $95\%$ confidence interval formula for a mean with known $\sigma$. [1 mark]

• Cue. $\bar{x} \pm 1.96\dfrac{\sigma}{\sqrt{n}}$.

Q3. Does a $95\%$ confidence interval mean a $95\%$ chance the true mean is inside it? [1 mark]

• Cue. No; it means $95\%$ of intervals from repeated samples would contain the true mean.