Linear combinations and statistical inference: VCE Specialist Mathematics Unit 4

What this dot point is asking

VCAA wants you to find the mean and variance of linear combinations of independent random variables, to use the resulting distribution of the sample mean $\bar{X}$, to construct a confidence interval for a population mean, and to carry out a hypothesis test for a mean by computing a test statistic and a $p$ value. This is the statistical inference strand of Specialist, distinct from the proportion-based inference in Mathematical Methods.

Mean and variance of linear combinations

For any random variables, expectation is linear:

whether or not $X$ and $Y$ are independent. Variance, however, behaves differently. For independent $X$ and $Y$,

The coefficients are squared, and the cross term vanishes because independence makes the covariance zero. Note that even for a difference, variances add: $\mathrm{Var}(X - Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)$, since $(-1)^2 = 1$.

The distribution of the sample mean

Take a random sample $X_1, X_2, \dots, X_n$ of independent observations from a population with mean $\mu$ and standard deviation $\sigma$. The sample mean is $\bar{X} = \frac{1}{n}\sum X_i$. Applying the linear-combination rules:

So the sample mean is unbiased for $\mu$, and its spread shrinks as $n$ grows. By the central limit theorem, for large $n$ the distribution of $\bar{X}$ is approximately normal, $\bar{X} \sim N\!\left(\mu, \frac{\sigma^2}{n}\right)$, regardless of the population's shape. The quantity $\frac{\sigma}{\sqrt{n}}$ is the standard error of the mean.

Confidence intervals for a population mean

An approximate $C\%$ confidence interval for $\mu$, when $\sigma$ is known and $n$ is large, is

where $z$ is the standard normal value capturing the central $C\%$. For a 95% interval, $z \approx 1.96$. The interval is a range of plausible values for $\mu$; the confidence level refers to the long-run proportion of such intervals that would contain the true mean, not the probability that $\mu$ lies in this one fixed interval.

Hypothesis testing for a mean

A test of the mean compares a null hypothesis $H_0: \mu = \mu_0$ against an alternative. The steps:

1. State $H_0: \mu = \mu_0$ and the alternative $H_1$ (one-sided, $\mu > \mu_0$ or $\mu < \mu_0$, or two-sided $\mu \neq \mu_0$).
2. Compute the test statistic $z = \dfrac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$, which measures how many standard errors the observed mean sits from $\mu_0$.
3. Find the $p$ value: the probability, assuming $H_0$ is true, of observing a sample mean at least as extreme as $\bar{x}$.
4. Compare the $p$ value with the significance level $\alpha$ (commonly $0.05$). If $p < \alpha$, reject $H_0$; otherwise do not reject it.

Why independence and squaring matter

The single most error-prone step is the variance rule. Means combine linearly with their coefficients, but variances combine with the squares of the coefficients, and only when the variables are independent. This is why doubling a measurement ($2X$) quadruples its variance, and why averaging $n$ independent values divides the variance by $n$ rather than by $\sqrt{n}$. Keeping variance and standard deviation distinct (one is the square of the other) prevents most slips in this strand.

Examples in context

Example 1. Variance of a sum. If $X$ and $Y$ are independent with $\mathrm{Var}(X) = 4$ and $\mathrm{Var}(Y) = 9$, then $\mathrm{Var}(2X + Y) = 4(4) + 1(9) = 25$, so $\mathrm{sd} = 5$.

Example 2. Confidence interval. A sample of $n = 25$ with $\bar{x} = 50$ and known $\sigma = 10$ gives standard error $\frac{10}{5} = 2$, so a 95% interval is $50 \pm 1.96(2) = 50 \pm 3.92$, that is $(46.08, 53.92)$.

Try this

Q1. $X$ and $Y$ are independent with $\mathrm{Var}(X) = 5$, $\mathrm{Var}(Y) = 3$. Find $\mathrm{Var}(X - Y)$. [2 marks]

• Cue. Variances add: $5 + 3 = 8$.

Q2. A sample of $n = 36$ from a population with $\sigma = 12$ has mean $\bar{x} = 70$. State the standard error of the mean. [1 mark]

• Cue. $\frac{12}{\sqrt{36}} = \frac{12}{6} = 2$.

Q3. For a test of $H_0: \mu = 100$ against $H_1: \mu \neq 100$, the test statistic is $z = 2$. State the $p$ value. [2 marks]

• Cue. Two-sided: $p = 2\,P(Z > 2) \approx 2(0.0228) = 0.0456$.