Confidence intervals for a population mean | Year 12

01 / LEARN

The idea, step by step

Estimate plus and minus uncertainty

For independent observations with known population standard deviation, use $\bar{x} \pm z^{*} σ / \sqrt{n}$ . The critical value leaves the chosen central normal probability between $- z^{*}$ and $z^{*}$ . A 95% interval uses approximately 1.96, not 0.95.

The interval is exact for a normal population under the known-spread model. Otherwise, it is approximate and needs a justified sample-mean normal model. Keep intermediate digits and round the final endpoints.

\bar{x} \pm z^{*} σ / \sqrt{n}

Read each component

The sample mean is the centre. The margin of error is critical value times standard error; full width is twice the margin. A confidence interval does not compensate for an incomplete frame or biased questionnaire.

If population standard deviation is unknown but a sufficiently large independent sample supports a normal approximation, use sample standard deviation s to estimate standard error as $s / \sqrt{n}$ . The resulting normal interval is approximate, not exact. A small sample with unknown spread must not silently be handled this way; another method may be needed.

Rounded two-sided normal critical values
Confidence	Critical value
90%	1.645
95%	1.96
99%	2.576

A common trap

Multiply standard error by the confidence proportion.

Instead: Use the corresponding normal critical value.

02 / WORKED EXAMPLES

See the thinking

Example 1

Packet masses

A normal population has known SD 8 g. A random independent sample of 64 has mean 102.4 g. Find a 95% mean interval.

Standard error $= 8 / 8 = 1$ g.
Margin = 1.96 g.
Subtract and add the margin to 102.4.

(100.44, 104.36) g.

Example 2

A 90% interval

A large random sample of 100 trips has mean 32.6 minutes. Use known SD 10 minutes and critical value 1.645.

Standard error $= 10 / \sqrt{100} = 1$ .
Margin = 1.645 minutes.
Trip populations may be non-normal, so state the large-sample approximation.

Approximately (30.955, 34.245) minutes.

Example 3

A large sample with unknown population spread

A random independent sample of 100 delivery times has mean 32.5 minutes and sample standard deviation 9 minutes. Population spread is unknown. Use a large-sample normal approximation to find a 95% mean interval.

Estimate the standard error using $s / \sqrt{n} = 9 / 10 = 0.9$ minutes.
The approximate margin is $1.96 (0.9) = 1.764$ minutes.
Subtract/add the margin to 32.5. State that this uses an estimated spread and a justified large-sample approximation.

Approximately (30.736, 34.264) minutes.

Check your understanding

For mean 50, known SD 6 and n = 36, find a 95% interval.

Show answer

(48.04, 51.96).

Standard error = 1; margin = 1.96.

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A normal population has known SD 12. A sample of 36 has mean 80. Find a 95% interval.
2 marks
For mean 18.5, known SD 4 and n = 64, use 1.645 for a 90% interval.
2 marks
An interval is (12.2, 15.8). Recover its centre and margin.
2 marks
A known-spread normal interval uses n = 6 from a strongly skewed population. What is the concern?
2 marks
A mean estimate is 250 with standard error 2.5. Use 2.576 for a 99% interval.
2 marks
A large independent random sample has n = 144, mean 48.2 and sample standard deviation 12. Population spread is unknown. Use an approximate 95% normal interval for the mean.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A normal population has known SD 12. A sample of 36 has mean 80. Find a 95% interval.
2 marks
Standard error = 2; margin = 3.92.
(76.08, 83.92).
For mean 18.5, known SD 4 and n = 64, use 1.645 for a 90% interval.
2 marks
Standard error = 0.5; margin = 0.8225.
(17.6775, 19.3225).
An interval is (12.2, 15.8). Recover its centre and margin.
2 marks
Take the midpoint and half-width.
Centre 14; margin 1.8.
A known-spread normal interval uses n = 6 from a strongly skewed population. What is the concern?
2 marks
Known spread alone does not justify normality with a small skewed sample.
The mean distribution need not be approximately normal.
A mean estimate is 250 with standard error 2.5. Use 2.576 for a 99% interval.
2 marks
Margin = 2.576 × 2.5 = 6.44.
(243.56, 256.44).
A large independent random sample has n = 144, mean 48.2 and sample standard deviation 12. Population spread is unknown. Use an approximate 95% normal interval for the mean.
3 marks
Estimate standard error as 12/√144 = 1.
Use the approximate margin 1.96 × 1 = 1.96.
The interval uses sample spread and a large-sample normal approximation, not an exact known-spread model.
Approximately (46.24, 50.16).

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: sampling distributions and confidence intervals for means and proportions. Mean tasks distinguish known population spread from sufficiently large-sample estimated-spread normal intervals. Repeated-interval computational simulation, approximation conditions, design and interpretation are included.

Official curriculum source ↗

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