Confidence intervals for a population mean | Year 12

01 / LEARN

Understand the idea.

Estimate plus and minus uncertainty

A mean interval starts at the observed sample mean and extends a margin in both directions. With known population spread, the margin is $E = z^{*} \frac{σ}{\sqrt{n}}$ . Choose $z^{*}$ from the desired central normal coverage; 95% uses approximately $1.96$ .

Independent observations from a normal population give an exact normal mean model when population spread is known. Otherwise, justify a large-sample normal approximation. Write the model qualification with the interval, retain intermediate digits, and round only the endpoints.

\bar{x} \pm \frac{z^{*} σ}{\sqrt{n}}

Read each component

The centre estimates the population mean; the margin measures sampling uncertainty under the model. The full width is $2 E$ , not $E$ . Check the sampling frame and measurement process because a numerical margin does not account for systematic bias.

If population spread is unknown, a sufficiently large independent sample may justify replacing $σ$ with $s$ in a normal approximation. The resulting interval $\bar{x} \pm z^{*} \frac{s}{\sqrt{n}}$ uses estimated spread and is approximate. A small unknown-spread sample needs a separately justified method.

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

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

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.

Worked solution

1
Check the stated mean model.
$\bar{x} = 102.4 g, σ = 8 g, n = 64$
A normal population, known standard deviation and independent random observations justify the normal mean interval.
2
Find the standard error.
$SE (\bar{X}) = \frac{8}{\sqrt{64}} = 1 g$
The interval estimates the population mean rather than individual packet masses.
3
Calculate the 95% margin.
$E = 1.96 \times 1 = 1.96 g$
The supplied confidence level determines the normal critical value.
4
Subtract and add the margin.
$\begin{aligned} L & = 102.4 - 1.96 = 100.44 \\ U & = 102.4 + 1.96 = 104.36 \end{aligned}$
The 95% mean interval is (100.44, 104.36) g, centred at the sample mean.

Answer

(100.44, 104.36) g.

EXAMPLE 02

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.

Worked solution

1
Calculate the known-spread standard error.
$SE (\bar{X}) = \frac{10}{\sqrt{100}} = 1 \min$
The 100-trip average varies less than individual trip times.
2
Use the given 90% critical value.
$E = 1.645 \times 1 = 1.645 \min$
The required coverage is 90%, so the margin uses 1.645.
3
Calculate the bounds.
$\begin{aligned} L & = 32.6 - 1.645 = 30.955 \\ U & = 32.6 + 1.645 = 34.245 \end{aligned}$
Adding and subtracting the same margin gives the symmetric interval.
4
Qualify the result.
$μ is estimated by (30.955, 34.245) \min$
This is an approximate 90% interval using a justified large-sample normal model; normality of individual trip times was not stated.

Answer

Approximately (30.955, 34.245) minutes.

EXAMPLE 03

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.

Worked solution

1
Estimate the standard error.
$\hat{SE} (\bar{X}) = \frac{s}{\sqrt{n}} = \frac{9}{\sqrt{100}} = 0.9 \min$
Population spread is unknown, so use the sample standard deviation under the requested large-sample approximation.
2
Calculate the approximate 95% margin.
$E = 1.96 \times 0.9 = 1.764 \min$
The normal critical value multiplies the estimated standard error.
3
Find the interval endpoints.
$\begin{aligned} L & = 32.5 - 1.764 = 30.736 \\ U & = 32.5 + 1.764 = 34.264 \end{aligned}$
The sample mean is the centre of the interval.
4
State the approximation.
$(30.736, 34.264) \min$
This approximate 95% interval relies on independent random sampling, estimated spread and an adequate large-sample normal approximation.

Answer

Approximately (30.736, 34.264) minutes.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

(48.04, 51.96).

The standard error is $\frac{6}{\sqrt{36}} = 1$ , so the 95% margin is $1.96$ . The endpoints are $50 - 1.96 = 48.04$ and $50 + 1.96 = 51.96$ , under the stated normal-interval assumptions.

Keep practising

6 worksheet questions · 13 marks

Open worksheet 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 $= 2.576 \times 2.5 = 6.44$ .
$(243.56, 256.44)$ .
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 $\frac{12}{\sqrt{144}} = 1$ .
Use the approximate margin $1.96 \times 1 = 1.96$ .
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 ↗

In this lesson

Understand the idea.

Estimate plus and minus uncertainty

Read each component

See the method in action.

Packet masses

Check the stated mean model.

Find the standard error.

Calculate the 95% margin.

Subtract and add the margin.

A 90% interval

Calculate the known-spread standard error.

Use the given 90% critical value.

Calculate the bounds.

Qualify the result.

A large sample with unknown population spread

Estimate the standard error.

Calculate the approximate 95% margin.

Find the interval endpoints.

State the approximation.

Try it yourself.

Check your understanding

Keep practising