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Statistical inference (WACE Year 12 Mathematics Specialist Unit 4)

Syllabus dot point

“Use the distribution of the sample mean and the central limit theorem to build confidence intervals for a population mean”

9 min read

Quick answer

The sample mean Xˉ\bar{X} from a sample of size nn has mean μ\mu and standard deviation (standard error) σn\frac{\sigma}{\sqrt{n}}; by the central limit theorem Xˉ\bar{X} is approximately normal for large nn, so an approximate confidence interval for the population mean is xˉ±zσn\bar{x} \pm z\frac{\sigma}{\sqrt{n}}, with z≈1.96z \approx 1.96 for 95%95\% confidence.

Jump to a section
  1. What this dot point is asking
  2. From sample to population
  3. The sampling distribution of the sample mean
  4. The central limit theorem
  5. Confidence intervals for a population mean
  6. Effect of sample size and confidence level
  7. Exam-style questions

What this dot point is asking

SCSA Unit 4 statistical inference is about reasoning from a sample to a population. You must know how the sample mean behaves across repeated samples, state and use the central limit theorem, and construct and interpret confidence intervals for a population mean.

From sample to population

Inference is the logic of reasoning backwards: a sample is observed and a statement about the whole population is made. We cannot measure every member of a population, so we draw a random sample and use its mean as an estimate of the population mean. The difficulty is that a different sample would give a different estimate, so a single number is not enough; we need to quantify the uncertainty around it. This dot point develops the machinery, the sampling distribution, the central limit theorem and the confidence interval, that turns a single sample mean into a defensible statement about the population, complete with a stated level of confidence.

The sampling distribution of the sample mean

Take repeated random samples of size nn from a population with mean μ\mu and standard deviation σ\sigma. Each sample gives a sample mean xˉ\bar{x}, and these vary from sample to sample. The random variable Xˉ\bar{X} has:

Mean and standard error of the sample mean

E(Xˉ)=μ,sd(Xˉ)=σn.E(\bar{X}) = \mu, \qquad \text{sd}(\bar{X}) = \frac{\sigma}{\sqrt{n}}.

The quantity σn\dfrac{\sigma}{\sqrt{n}} is the standard error. It shrinks as nn grows, so larger samples give sample means clustered more tightly around the true mean μ\mu.

The central limit theorem

Central limit theorem

For a sufficiently large sample size nn, the distribution of the sample mean Xˉ\bar{X} is approximately normal, Xˉ∼N ⁣(μ,σ2n)\bar{X} \sim N\!\left(\mu, \dfrac{\sigma^2}{n}\right), regardless of the shape of the original population distribution.

This is what makes inference possible: even if the underlying population is skewed, the sample mean behaves normally for large nn, so we can use normal-distribution probabilities and zz-values.

Confidence intervals for a population mean

An approximate confidence interval estimates μ\mu from one sample. With known (or large-sample estimated) population standard deviation σ\sigma, the interval is

xˉ±z σn,\bar{x} \pm z\,\frac{\sigma}{\sqrt{n}},

where zz is the critical value for the chosen confidence level: z≈1.645z \approx 1.645 for 90%90\%, z≈1.96z \approx 1.96 for 95%95\%, and z≈2.576z \approx 2.576 for 99%99\%. The term zσnz\dfrac{\sigma}{\sqrt{n}} is the margin of error.

The correct interpretation of a 95%95\% confidence interval: if we repeated the sampling many times and built an interval each time, about 95%95\% of those intervals would contain the true mean μ\mu. It is not a 95%95\% probability that μ\mu lies in this one interval.

Constructing a 95% confidence interval

A sample of n=64n = 64 has mean xˉ=50.4\bar{x} = 50.4 with population standard deviation σ=8\sigma = 8. Construct a 95% confidence interval for the population mean.

Step 1: Compute the standard error

The standard error measures how much the sample mean is likely to vary from sample to sample. Dividing σ\sigma by n\sqrt{n} accounts for the fact that averaging over more observations reduces spread.

σn=864=88=1.\frac{\sigma}{\sqrt{n}} = \frac{8}{\sqrt{64}} = \frac{8}{8} = 1.

Step 2: Determine the margin of error

For a 95% confidence level we use the critical value z=1.96z = 1.96, which corresponds to the central 95% of the standard normal distribution. Multiplying zz by the standard error gives the half-width of the interval.

margin of error=1.96×1=1.96.\text{margin of error} = 1.96 \times 1 = 1.96.

Step 3: Construct the confidence interval

We add and subtract the margin of error from the sample mean to obtain the lower and upper bounds.

50.4±1.96=(48.44,  52.36).50.4 \pm 1.96 = (48.44,\; 52.36).

Step 4: Interpret the result

A confidence interval is a statement about the procedure, not a probability about this one interval. We are 95% confident the population mean lies between 48.4448.44 and 52.3652.36. To halve the margin of error we would need to quadruple the sample size, because the standard error involves n\sqrt{n}.

Final answer: The 95% confidence interval for μ\mu is (48.44,  52.36)(48.44,\; 52.36).

Effect of sample size and confidence level

A wider confidence level (say 99%99\% instead of 95%95\%) gives a larger zz and a wider interval: more confidence costs precision. A larger sample reduces the standard error and narrows the interval. Because the standard error has n\sqrt{n} in the denominator, increasing the sample size by a factor of 44 halves the width.

Common traps
Dividing by nn instead of n\sqrt{n}
The standard error is σn\dfrac{\sigma}{\sqrt{n}}, not σn\dfrac{\sigma}{n}. Forgetting the square root is the single most common error.
Misinterpreting the interval
A 95%95\% interval does not mean μ\mu has a 95%95\% chance of being inside this particular interval; the procedure captures μ\mu in 95%95\% of repeated samples.
Using the wrong zz
Match zz to the confidence level: 1.961.96 for 95%95\%, not 9595 or 1.651.65.
Confusing σ\sigma with sd(Xˉ)\text{sd}(\bar{X})
The population spread σ\sigma and the standard error σn\frac{\sigma}{\sqrt{n}} are different quantities; the interval uses the standard error.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2024-style8 marks
Calculator-assumed. A machine fills bottles. A random sample of n=49n = 49 bottles has mean volume xˉ=498\bar{x} = 498 mL, and the population standard deviation is σ=7\sigma = 7 mL. (a) Construct a 90%90\% confidence interval for the mean fill volume. (b) The target mean is 500500 mL. Comment on whether the interval is consistent with the machine meeting target. (c) What sample size gives a 90%90\% margin of error of 11 mL?
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A full inference question with interpretation.

(a) Standard error =749=77=1= \dfrac{7}{\sqrt{49}} = \dfrac{7}{7} = 1. For 90%90\%, z=1.645z = 1.645, margin =1.645= 1.645. Interval: 498±1.645=(496.36, 499.64)498 \pm 1.645 = (496.36,\ 499.64) mL.

(b) The target 500500 mL lies above the upper limit 499.64499.64, so 500500 is not in the interval. The data suggest the machine is filling slightly below target.

(c) Require 1.645×7n≤11.645 \times \dfrac{7}{\sqrt{n}} \le 1, so n≥1.645×7=11.515\sqrt{n} \ge 1.645 \times 7 = 11.515, hence n≥132.6n \ge 132.6. Round up to n=133n = 133.

Markers reward the standard error, the 90%90\% critical value, the interval, the comparison with target, and rounding nn up.

WACE 2022-style4 marks
Calculator-free. Explain why the central limit theorem is needed to justify a confidence interval for a mean when the population distribution is unknown, and state the role of the standard error.
Show worked answer →

A conceptual question on the theory.

A confidence interval xˉ±zσn\bar{x} \pm z\dfrac{\sigma}{\sqrt{n}} uses normal-distribution critical values such as z=1.96z = 1.96. These are only valid if Xˉ\bar{X} is normally distributed. When the population shape is unknown, the central limit theorem provides the justification: for a large sample, Xˉ\bar{X} is approximately normal regardless of the population's shape, so the zz-values apply.

The standard error σn\dfrac{\sigma}{\sqrt{n}} is the standard deviation of Xˉ\bar{X}; it sets the width of the interval and shrinks as nn grows, making the estimate more precise. Markers reward linking the normal critical values to the CLT and identifying the standard error as the spread of the sample mean.

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Sources & how we know this

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