Standard error | Year 12 | Ascent Library

01 / LEARN

Understand the idea.

What standard error measures

Standard deviation describes spread among individuals; standard error describes spread among repeated estimates. For a mean of independent observations, $SE (\bar{X}) = \frac{σ}{\sqrt{n}}$ . A large individual spread can coexist with a precise estimated mean when enough independent observations are averaged.

Keep the units attached. Standard deviation and standard error have the same unit as the observations, whereas variance has squared units. If variance is supplied, take its square root before using a formula that requires standard deviation.

Reverse a standard-error calculation

To recover population spread, reverse the division: $σ = SE (\bar{X}) \sqrt{n}$ . To achieve standard error at most a positive target $s_{e}$ , solve $\frac{σ}{\sqrt{n}} \leq s_{e}$ , giving $n \geq {(\frac{σ}{s_{e}})}^{2}$ . Round the final bound upwards.

When $σ$ is unknown, $\frac{s}{\sqrt{n}}$ estimates the standard error using the sample standard deviation. That estimate adds uncertainty; it does not make a small-sample known-spread normal interval valid. Check the interval method and distribution assumptions separately.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Two different spreads

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

Worked solution

1
Identify the spread and sample size.
$σ = 24 g, n = 64$
The stated 24 g measures individual delivery masses.
2
Calculate the spread of sample means.
$SE (\bar{X}) = \frac{24}{\sqrt{64}} = \frac{24}{8} = 3 g$
For independent observations, averaging divides standard deviation by the square root of sample size.
3
Distinguish the two results.
$SD (X) = 24 g, SD (\bar{X}) = 3 g$
Taking a larger sample changes the variability of the average, not the population variability of individual masses.

Answer

3 g.

EXAMPLE 02

A precision target

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

Worked solution

1
Express the precision requirement.
$\frac{9}{\sqrt{n}} \leq 0.8$
The target applies to standard error, so divide population standard deviation by the square root of sample size.
2
Solve the positive inequality.
$\begin{aligned} \sqrt{n} & \geq \frac{9}{0.8} \\ n & \geq 126.5625 \end{aligned}$
All quantities are positive, so rearranging and squaring preserve the inequality.
3
Round upwards to an integer.
$n_{min} = 127$
An integer below the lower bound would not meet the requirement.
4
Check the neighbouring sample sizes.
$\begin{aligned} \frac{9}{\sqrt{126}} & \approx 0.8018 > 0.8 \\ \frac{9}{\sqrt{127}} & \approx 0.7986 \leq 0.8 \end{aligned}$
The adjacent integer fails while 127 succeeds, confirming minimality.

Answer

127.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

14 kg.

Reverse $1.4 = \frac{σ}{\sqrt{100}}$ : multiplying by $\sqrt{100} = 10$ gives $σ = 14 kg$ . The 1.4 kg describes repeated sample means, while 14 kg describes individuals.

Keep practising

6 worksheet questions · 12 marks

Open worksheet 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 $\frac{8}{\sqrt{400}} = 0.4$ 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
$\frac{35}{\sqrt{49}} = 5$ .
5 cm.
Population variance is 144. Find standard error when $n = 36$ .
2 marks
First $σ = \sqrt{144} = 12$ ; then $\frac{12}{\sqrt{36}} = 2$ .
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
$n \geq {(\frac{10}{1.2})}^{2} = 69.444 \dots$ ; round up.
$70$ .
A report calls $\frac{8}{\sqrt{400}} = 0.4$ 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 $\sqrt{\frac{25}{100}} = \frac{1}{2}$ .
The second is half the first.

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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.

What standard error measures

Reverse a standard-error calculation

See the method in action.

Two different spreads

Identify the spread and sample size.

Calculate the spread of sample means.

Distinguish the two results.

A precision target

Express the precision requirement.

Solve the positive inequality.

Round upwards to an integer.

Check the neighbouring sample sizes.

Try it yourself.

Check your understanding

Keep practising