Statistical inference - TCE Mathematics Specialised (Tasmania)

Syllabus dot point

“Use the distribution of the sample mean and the central limit theorem to construct confidence intervals.”

TCESpecialist MathematicsUnit 49 min readUpdated 2 Jun 2026

Quick answer

The sample mean $\bar{X}$ has mean $\mu$ and standard deviation $\tfrac{\sigma}{\sqrt{n}}$ . The central limit theorem says $\bar{X}$ is approximately normal for large $n$ , which lets you build a confidence interval $\bar{x} \pm z\tfrac{\sigma}{\sqrt{n}}$ for the population mean.

What this dot point is asking

Statistical inference uses a sample to make justified statements about an entire population. The bridge is the sampling distribution of the sample mean.

Mean and standard deviation of the sample mean

If a sample of size $n$ is drawn from a population with mean $\mu$ and standard deviation $\sigma$ , then the sample mean $\bar{X}$ has

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

The quantity

\tfrac{\sigma}{\sqrt{n}}

is the standard error.

So $\bar{X}$ is an unbiased estimator of $\mu$ , and it becomes more precise as $n$ grows, since the standard error shrinks like $\tfrac{1}{\sqrt{n}}$ .

Central limit theorem

For a sufficiently large sample size, the distribution of the sample mean $\bar{X}$ is approximately normal,

\bar{X} \approx N\!\left(\mu, \, \frac{\sigma^2}{n}\right),

regardless of the shape of the original population distribution.

This is what makes inference possible even when the population itself is not normal. As a rule of thumb $n \geq 30$ is enough for the approximation to be good.

Confidence intervals

A confidence interval gives a range of plausible values for $\mu$ . Using the normal model for $\bar{X}$ , a confidence interval for the population mean is

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

where

z

is the critical value for the chosen confidence level. The common values are

z = 1.96

for 95 percent and

z = 2.576

for 99 percent confidence.

A 95 percent confidence interval

Worked example

A machine fills bottles whose contents have known standard deviation $\sigma = 4$ mL. A random sample of $n = 64$ bottles has mean $\bar{x} = 502$ mL. Construct a 95 percent confidence interval for the true mean fill $\mu$ .

The standard error is

\frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{64}} = \frac{4}{8} = 0.5 \text{ mL}.

For 95 percent confidence,

z = 1.96

. The margin of error is

1.96 \times 0.5 = 0.98 \text{ mL}.

The confidence interval is

502 \pm 0.98 = (501.02, \, 502.98) \text{ mL}.

We are 95 percent confident the true mean fill lies between about 501.0 mL and 503.0 mL.

The margin of error and choosing a sample size

The term added and subtracted, $E = z\dfrac{\sigma}{\sqrt{n}}$ , is the margin of error. It is half the width of the interval. Because it depends on $n$ only through $\dfrac{1}{\sqrt{n}}$ , you can solve for the sample size needed to meet a target precision.

Sample size for a target margin of error

To guarantee a margin of error of at most $E$ at a given confidence level (critical value $z$ ),

n \ge \left(\frac{z\sigma}{E}\right)^2.

Always round the result up to the next whole number, because a smaller

n

would give a larger margin than allowed.

Finding the required sample size

Worked example

A quality inspector knows the fill volumes of a drink have standard deviation $\sigma = 3$ mL. How large a sample is needed so that a 95% confidence interval for the mean has a margin of error no more than $0.8$ mL?

Use $z = 1.96$ , $\sigma = 3$ , $E = 0.8$ :

n \ge \left(\frac{1.96 \times 3}{0.8}\right)^2 = \left(\frac{5.88}{0.8}\right)^2 = 7.35^2 = 54.02.

Round up: $n = 55$ bottles. Taking only $54$ would push the margin just above $0.8$ mL, so the minimum acceptable sample size is $55$ .

Interpreting the interval correctly

The confidence level describes the long-run reliability of the method: if we repeated the sampling many times and built an interval each time, about 95 percent of those intervals would contain the true $\mu$ . It is not a probability statement about $\mu$ for one fixed interval.

To halve the width of an interval you must quadruple the sample size, because the width depends on $\tfrac{1}{\sqrt{n}}$ . This square-root law is why doubling precision is expensive: cutting the margin of error to a quarter needs sixteen times the data.

Reading a given interval backwards

Exam questions often hand you a completed interval and ask you to recover the sample mean, the standard error, or the confidence level. Two facts make this routine. The sample mean is the midpoint of the interval, $\bar{x} = \dfrac{L + U}{2}$ , and the margin of error is the half-width, $E = \dfrac{U - L}{2}$ . Once you have $E$ and $\dfrac{\sigma}{\sqrt{n}}$ , the implied critical value is $z = \dfrac{E}{\sigma/\sqrt{n}}$ , which tells you the confidence level ( $z = 1.96$ for 95%, $z = 2.576$ for 99%).

Confusing population spread with the standard error

A frequent error is using $\sigma$ instead of $\tfrac{\sigma}{\sqrt{n}}$ in the interval. The sample mean varies far less than an individual measurement, so you must divide by $\sqrt{n}$ . A related slip is the wrong interpretation: saying there is a 95 percent probability that $\mu$ lies in one computed interval. Once the interval is fixed, $\mu$ either is or is not inside it; the 95 percent refers to the procedure, not that single interval.

Exam-style questions

Questions in the style of TASC 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.

Practice6 marks

A population has standard deviation

\sigma = 12

. A random sample of

n = 100

values has mean

\bar{x} = 84.5

. (a) Construct a 95% confidence interval for the population mean

\mu

. (b) State the smallest sample size needed so that the margin of error for a 95% interval is at most

1.5

Show worked answer →

(a) The standard error is $\dfrac{\sigma}{\sqrt{n}} = \dfrac{12}{\sqrt{100}} = \dfrac{12}{10} = 1.2$ . For 95% confidence, $z = 1.96$ , so the margin of error is $1.96 \times 1.2 = 2.352$ . The interval is $84.5 \pm 2.352 = (82.15, 86.85)$ . (3 marks)

(b) We need $1.96\cdot\dfrac{12}{\sqrt{n}} \le 1.5$ . Rearranging, $\sqrt{n} \ge \dfrac{1.96 \times 12}{1.5} = 15.68$ , so $n \ge 15.68^2 = 245.86$ . Since $n$ must be a whole number and the margin must be at most $1.5$ , round up to $n = 246$ . Markers reward the standard error, the correct $z = 1.96$ , and rounding the sample size up. (3 marks)

Practice5 marks

The lifetimes of a brand of battery have standard deviation

\sigma = 8

hours. A sample of

64

batteries gives a 99% confidence interval of

(118.42, 123.58)

hours for the mean lifetime. (a) Find the sample mean

\bar{x}

. (b) Verify the interval is consistent with

z = 2.576

Show worked answer →

(a) The sample mean is the midpoint of the interval: $\bar{x} = \dfrac{118.42 + 123.58}{2} = \dfrac{242}{2} = 121$ hours. (2 marks)

(b) The standard error is $\dfrac{\sigma}{\sqrt{n}} = \dfrac{8}{\sqrt{64}} = \dfrac{8}{8} = 1$ . The half-width of the interval is $\dfrac{123.58 - 118.42}{2} = \dfrac{5.16}{2} = 2.58$ . The implied $z$ value is $\dfrac{\text{half-width}}{\text{standard error}} = \dfrac{2.58}{1} = 2.58 \approx 2.576$ , which matches the 99% critical value. Markers reward the midpoint as the mean and the consistency check of half-width against $z\cdot\dfrac{\sigma}{\sqrt{n}}$ . (3 marks)

Practise this

Sources & how we know this

Mathematics Specialised Level 4 (MTS415118) course document (2018 onwards) — TASC