§-Syllabus dot point

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How do we use one sample to estimate a plausible range for the population mean?

A confidence interval gives a range of plausible values for the population mean, built from the sample mean plus or minus a critical z-value times the standard error.

How to construct and correctly interpret a confidence interval for a population mean, the critical z-values for common confidence levels, and worked calculations.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The formula
Critical z-values
Interpreting the interval correctly
Using an interval to test a claim
What changes the width
Common errors
Why it matters

What this dot point is asking

A single sample mean $\bar{x}$ is a point estimate of $\mu$ , but it is almost never exactly right. A confidence interval gives an honest range of plausible values for $\mu$ , with a stated level of confidence.

The formula

The interval is centred on $\bar{x}$ and extends a margin of error $z^*\dfrac{\sigma}{\sqrt{n}}$ either side.

Critical z-values

The critical value $z^*$ comes from the standard normal: it is the z-score leaving the right proportion in the central region.

Common critical values

Confidence level	$z^*$
$90\%$	$1.645$
$95\%$	$1.960$
$99\%$	$2.576$

A higher confidence level needs a larger $z^*$ , giving a wider interval.

Constructing a 95% confidence interval

A sample of $n=64$ light globes has mean lifetime $\bar{x}=1200$ hours and the population standard deviation is $\sigma=80$ hours. This example constructs a $95\%$ confidence interval for the true mean lifetime.

Step 1: compute the standard error

The standard error measures how much the sample mean varies from sample to sample. Dividing the population standard deviation by the square root of the sample size:

\frac{\sigma}{\sqrt{n}}=\frac{80}{\sqrt{64}}=\frac{80}{8}=10.

Step 2: compute the margin of error

For a $95\%$ confidence level the critical value is $z^*=1.96$ . The margin of error is how far the interval extends each side of the sample mean:

E=1.96\times 10=19.6.

Step 3: write the confidence interval

The interval is centred on the sample mean and extends by the margin of error in each direction:

1200\pm 19.6 \quad\Rightarrow\quad (1180.4,\ 1219.6).

Final answer: we are $95\%$ confident the true mean lifetime lies between approximately $1180$ and $1220$ hours.

Interpreting the interval correctly

The interpretation is precise and frequently examined:

A 99% interval is wider

Using the same sample of globes ( $\bar{x}=1200$ , $\sigma/\sqrt{n}=10$ ), this example constructs a $99\%$ confidence interval and shows why higher confidence produces a wider interval.

Step 1: use the larger critical value and compute the new margin of error

A $99\%$ confidence level requires $z^*=2.576$ instead of $1.96$ . The larger critical value directly widens the margin of error, because we need a broader net to be more certain of capturing the true mean:

E=2.576\times 10=25.76.

Step 2: form the interval and compare

Centring the interval on the same sample mean and applying the new margin:

1200\pm 25.76 \Rightarrow (1174.2,\ 1225.8).

Raising the confidence level from

95\%

99\%

widened the interval from

\pm 19.6

\pm 25.8

. More confidence always costs precision: to be more certain of capturing

\mu

, we must accept a less precise estimate.

Final answer: the $99\%$ confidence interval is $(1174.2,\ 1225.8)$ hours.

Using an interval to test a claim

A confidence interval doubles as an informal test of a claimed mean. If a claimed value of $\mu$ falls inside the interval, the sample is consistent with that claim; if it falls outside, the sample provides evidence against it at the corresponding level. In the feeding-plan example, the old mean of $21.9$ litres lies below the $95\%$ interval $(22.1, 24.9)$ , which is evidence that the new plan genuinely raised production rather than the increase being due to sampling chance. SACE frequently asks you to construct an interval and then comment on whether a previous or claimed value is plausible in this way.

What changes the width

The width of the interval is $2z^*\dfrac{\sigma}{\sqrt{n}}$ , so three things move it. A higher confidence level raises $z^*$ and widens it; a larger sample size raises $\sqrt{n}$ and narrows it; and a more variable population (larger $\sigma$ ) widens it. Only the sample size is usually within the researcher's control, which is exactly why the next dot point focuses on choosing $n$ to hit a target width. Understanding these levers lets you explain, not just compute, why an interval is as wide as it is.

Common errors

Why it matters

Confidence intervals are the headline inference technique of Topic 6 and a guaranteed exam item, valued both for the calculation and for the precise interpretation. They build directly on the standard error and Central Limit Theorem, and lead into the next dot point on margin of error and choosing a sample size.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20222 marksCalculator-assumed. Wool fibre diameter is normally distributed with standard deviation

2.5

microns. A producer randomly chooses 80 fibres and finds a sample mean of

17.4

microns. Calculate the

99\%

confidence interval for the population mean.

Show worked answer →

Use $\bar{x} \pm z^*\dfrac{\sigma}{\sqrt{n}}$ with $\sigma = 2.5$ , $n = 80$ , $z^* = 2.576$ for $99\%$ .

Standard error $= \dfrac{2.5}{\sqrt{80}} = \dfrac{2.5}{8.944} \approx 0.2795$ .

Margin of error $= 2.576 \times 0.2795 \approx 0.720$ .

Interval $= 17.4 \pm 0.720 = (16.7,\ 18.1)$ microns.

Marks: one for the standard error and $z^* = 2.576$ , one for the interval. A common slip is dividing by $n$ instead of $\sqrt{n}$ .

SACE 20232 marksCalculator-assumed. After a new feeding plan, a random sample of 20 cows has a mean daily milk production of

23.5

litres. Assuming the standard deviation is still

3.26

litres, calculate a

95\%

confidence interval for the mean daily production on the new plan.

Show worked answer →

Use $\bar{x} \pm z^*\dfrac{\sigma}{\sqrt{n}}$ with $\bar{x} = 23.5$ , $\sigma = 3.26$ , $n = 20$ , $z^* = 1.96$ .

Standard error $= \dfrac{3.26}{\sqrt{20}} = \dfrac{3.26}{4.472} \approx 0.729$ .

Margin of error $= 1.96 \times 0.729 \approx 1.43$ .

Interval $= 23.5 \pm 1.43 = (22.1,\ 24.9)$ litres.

Marks: one for the standard error and $z^* = 1.96$ , one for the interval. Since the old mean $21.9$ lies below the interval, this suggests the plan raised production.

SACE 20213 marksCalculator-assumed. A sample of

n = 100

batteries has mean life

\bar{x} = 42

hours, with population standard deviation

\sigma = 6

hours. (a) Construct a

90\%

confidence interval for the mean life. (b) State what happens to the interval width if the confidence level is raised to

95\%

Show worked answer →

(a) Standard error $= \dfrac{6}{\sqrt{100}} = 0.6$ . With $z^* = 1.645$ , the margin is $1.645 \times 0.6 = 0.987$ , so the interval is $42 \pm 0.987 = (41.0,\ 43.0)$ hours. (2 marks)

(b) Raising confidence to $95\%$ increases $z^*$ to $1.96$ , which widens the interval (less precision for more confidence). (1 mark)

Marks reward the correct $z^*$ , the standard error, and recognising the confidence-width trade-off.