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Confidence intervals for a mean (SACE Stage 2 Mathematical Methods)

Syllabus dot point

“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.”

8 min read

Quick answer

A confidence interval for the population mean is xˉ±z∗σn\bar{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}, where z∗z^* is the critical value for the confidence level (1.961.96 for 95%95\%). It gives a range of plausible values for μ\mu; "95%95\% confident" means 95%95\% of such intervals, over repeated sampling, would contain the true mean.

Jump to a section
  1. What this dot point is asking
  2. The formula
  3. Critical z-values
  4. Interpreting the interval correctly
  5. Using an interval to test a claim
  6. What changes the width
  7. Common errors
  8. Why it matters
  9. Exam-style questions

What this dot point is asking

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

The formula

Confidence interval for a mean

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

where xˉ\bar{x} is the sample mean, σ\sigma the population standard deviation, nn the sample size, and z∗z^* the critical z-value for the chosen confidence level.

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

Critical z-values

The critical value z∗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∗z^*
90%90\% 1.6451.645
95%95\% 1.9601.960
99%99\% 2.5762.576

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

Constructing a 95% confidence interval

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

σn=8064=808=10.\frac{\sigma}{\sqrt{n}}=\frac{80}{\sqrt{64}}=\frac{80}{8}=10.

Step 2: compute the margin of error

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

E=1.96×10=19.6.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±19.6⇒(1180.4, 1219.6).1200\pm 19.6 \quad\Rightarrow\quad (1180.4,\ 1219.6).

Final answer: we are 95%95\% confident the true mean lifetime lies between approximately 11801180 and 12201220 hours.

Interpreting the interval correctly

The interpretation is precise and frequently examined:

What "95% confident" means

A 95%95\% confidence level refers to the method: if you repeated the sampling many times and built an interval each time, about 95%95\% of those intervals would capture the true population mean μ\mu. It does not mean there is a 95%95\% probability that μ\mu lies in this particular interval - μ\mu is a fixed number, either in the interval or not.

A 99% interval is wider

Using the same sample of globes (xˉ=1200\bar{x}=1200, σ/n=10\sigma/\sqrt{n}=10), this example constructs a 99%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%99\% confidence level requires z∗=2.576z^*=2.576 instead of 1.961.96. 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=2.576×10=25.76.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±25.76⇒(1174.2, 1225.8).1200\pm 25.76 \Rightarrow (1174.2,\ 1225.8).

Raising the confidence level from 95%95\% to 99%99\% widened the interval from ±19.6\pm 19.6 to ±25.8\pm 25.8. More confidence always costs precision: to be more certain of capturing μ\mu, we must accept a less precise estimate.

Final answer: the 99%99\% confidence interval is (1174.2, 1225.8)(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 μ\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 21.921.9 litres lies below the 95%95\% interval (22.1,24.9)(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∗σn2z^*\dfrac{\sigma}{\sqrt{n}}, so three things move it. A higher confidence level raises z∗z^* and widens it; a larger sample size raises n\sqrt{n} and narrows it; and a more variable population (larger σ\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 nn 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

Misinterpreting confidence and using σ without √n

The classic mark-loser is saying "there is a 95%95\% chance μ\mu is in this interval" - confidence is about the long-run method, not a probability for one fixed interval. Numerically, the other frequent slip is forgetting the n\sqrt{n}, using z∗σz^*\sigma instead of z∗σ/nz^*\sigma/\sqrt{n}, which inflates the interval enormously.

Compute the standard error first

Work out σ/n\sigma/\sqrt{n} as a separate number before multiplying by z∗z^*. This keeps the calculation organised, makes the margin of error easy to state, and lets you reuse the standard error if the question also asks for a different confidence level.

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 questions

Questions in the style of SACE Board 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.

SACE 2022-style2 marks
Calculator-assumed. Wool fibre diameter is normally distributed with standard deviation 2.52.5 microns. A producer randomly chooses 80 fibres and finds a sample mean of 17.417.4 microns. Calculate the 99%99\% confidence interval for the population mean.
Show worked answer →

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

Standard error =2.580=2.58.944≈0.2795= \dfrac{2.5}{\sqrt{80}} = \dfrac{2.5}{8.944} \approx 0.2795.

Margin of error =2.576×0.2795≈0.720= 2.576 \times 0.2795 \approx 0.720.

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

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

SACE 2023-style2 marks
Calculator-assumed. After a new feeding plan, a random sample of 20 cows has a mean daily milk production of 23.523.5 litres. Assuming the standard deviation is still 3.263.26 litres, calculate a 95%95\% confidence interval for the mean daily production on the new plan.
Show worked answer →

Use xˉ±z∗σn\bar{x} \pm z^*\dfrac{\sigma}{\sqrt{n}} with xˉ=23.5\bar{x} = 23.5, σ=3.26\sigma = 3.26, n=20n = 20, z∗=1.96z^* = 1.96.

Standard error =3.2620=3.264.472≈0.729= \dfrac{3.26}{\sqrt{20}} = \dfrac{3.26}{4.472} \approx 0.729.

Margin of error =1.96×0.729≈1.43= 1.96 \times 0.729 \approx 1.43.

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

Marks: one for the standard error and z∗=1.96z^* = 1.96, one for the interval. Since the old mean 21.921.9 lies below the interval, this suggests the plan raised production.

SACE 2021-style3 marks
Calculator-assumed. A sample of n=100n = 100 batteries has mean life xˉ=42\bar{x} = 42 hours, with population standard deviation σ=6\sigma = 6 hours. (a) Construct a 90%90\% confidence interval for the mean life. (b) State what happens to the interval width if the confidence level is raised to 95%95\%.
Show worked answer →

(a) Standard error =6100=0.6= \dfrac{6}{\sqrt{100}} = 0.6. With z∗=1.645z^* = 1.645, the margin is 1.645×0.6=0.9871.645 \times 0.6 = 0.987, so the interval is 42±0.987=(41.0, 43.0)42 \pm 0.987 = (41.0,\ 43.0) hours. (2 marks)

(b) Raising confidence to 95%95\% increases z∗z^* to 1.961.96, which widens the interval (less precision for more confidence). (1 mark)

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

Practise this

Sources & how we know this

ExamExplained