Standard error | Year 12 | Ascent Library

01 / LEARN

The idea, step by step

What standard error measures

The standard error of a mean is the standard deviation of its sampling distribution. With independent observations and known population spread, it is $σ / \sqrt{n}$ . It describes variation between repeated means, not variation among individuals.

Standard error and standard deviation use the original unit. Variance uses squared units. A population may be very variable while its mean is estimated precisely from a sufficiently large independent sample.

Reverse a standard-error calculation

Recover $σ$ by multiplying the standard error by $\sqrt{n}$ . To reach a target standard error $s_{e}$ , require $n \geq {(σ / s_{e})}^{2}$ and round up.

A sample standard deviation can estimate an unknown population spread, but that adds uncertainty. The normal mean-interval examples here state when population spread is known; do not silently apply the known-spread formula to a small unknown-spread sample.

A common trap

Divide standard deviation by n.

Instead: Divide by the square root of n.

02 / WORKED EXAMPLES

See the thinking

Example 1

Two different spreads

Individual delivery mass has standard deviation 24 g. Find the standard error for n = 64.

$24 / \sqrt{64} = 24 / 8 = 3$ g.
Individual spread remains 24 g.

3 g.

Example 2

A precision target

For population standard deviation 9, find the smallest n giving standard error at most 0.8.

$n \geq {(9 / 0.8)}^{2} = 126.5625$ .
Round up; 126 would fail the stated inequality.

127.

Check your understanding

A mean from n = 100 has standard error 1.4 kg. Recover population standard deviation.

Show answer

14 kg.

$σ = 1.4 \sqrt{100} = 14$ kg.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find standard error for standard deviation 35 cm and n = 49.
2 marks
Population variance is 144. Find standard error when n = 36.
2 marks
Standard error is 0.6 and n = 81. Recover population standard deviation.
2 marks
For standard deviation 10, find the smallest n giving standard error at most 1.2.
2 marks
A report calls 8/√400 = 0.4 the population SD. Correct it.
2 marks
Compare standard errors for n = 25 and n = 100 from one population.
2 marks

Read the worked solutions

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

Find standard error for standard deviation 35 cm and n = 49.
2 marks
$35 / \sqrt{49} = 5$ .
5 cm.
Population variance is 144. Find standard error when n = 36.
2 marks
First $σ = \sqrt{144} = 12$ ; then $12 / \sqrt{36} = 2$ .
2 units.
Standard error is 0.6 and n = 81. Recover population standard deviation.
2 marks
$0.6 \sqrt{81} = 5.4$ .
5.4.
For standard deviation 10, find the smallest n giving standard error at most 1.2.
2 marks
$n \geq {(10 / 1.2)}^{2} = 69.444 \dots$ ; round up.
70.
A report calls 8/√400 = 0.4 the population SD. Correct it.
2 marks
The division by the square root of sample size converts individual spread to mean spread.
0.4 is the standard error of the mean; population SD is 8.
Compare standard errors for n = 25 and n = 100 from one population.
2 marks
The ratio is $\sqrt{25 / 100} = 1 / 2$ .
The second is half the first.

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Confidence intervals for a population mean

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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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