Effect of sample size and variability | Year 12

01 / LEARN

Understand the idea.

Three controls on interval width

For a known-spread mean interval, width is $2 z^{*} \frac{σ}{\sqrt{n}}$ . More confidence increases $z^{*}$ ; more population variability increases $σ$ ; more independent observations increase the denominator. Compare one factor at a time while stating what stays fixed.

When confidence and spread are fixed, the margin ratio is $\frac{E_{new}}{E_{old}} = \sqrt{\frac{n_{old}}{n_{new}}}$ . A new sample may have a different mean. For a proportion interval, the estimated spread can change too, so its realised margin need not follow this ratio exactly.

width = \frac{2 z^{*} σ}{\sqrt{n}}

Plan precision without losing validity

To divide a margin by a positive factor $r$ , multiply sample size by $r^{2}$ . Reducing a margin from 3 to 1 therefore needs nine times the observations, with confidence and population spread fixed. Precision gains slow as sample size increases.

Plan validity alongside precision. Extra convenience-sample responses cannot restore excluded groups or remove leading wording. A wider, well-designed study can be more informative about the target population than an extremely narrow interval based on a biased selection process.

Check coverage with a computational simulation

To investigate coverage, draw 2,000 independent samples from $N (100, 20^{2})$ . For each mean, build its known-spread interval and record whether it contains the fixed value 100. Compare sample sizes 25 and 100, using both 95% and 99% confidence.

The supplied simulation uses NumPy default_rng, seed 20261106, separate samples for each size and the same means across confidence levels at each size. Compare widths and capture fractions: quadrupling sample size halves width; 99% intervals are wider. Finite capture counts fluctuate around nominal coverage.

One simulation: 2,000 sample intervals per row; normal population mean 100, SD 20
Sample size	Confidence	Captures	Capture fraction	Mean width
25	95%	1903	0.9515	15.680
25	99%	1978	0.9890	20.608
100	95%	1902	0.9510	7.840
100	99%	1988	0.9940	10.304

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Half the margin

A margin is 4 at n = 50. Find n for margin 2 with spread/confidence fixed.

Worked solution

1
Form the required margin ratio.
$\frac{E_{new}}{E_{old}} = \frac{2}{4} = \frac{1}{2}$
The new margin must be half the old margin.
2
Use the square-root sample-size relation.
$\begin{aligned} \sqrt{\frac{50}{n_{new}}} & = \frac{1}{2} \\ \frac{50}{n_{new}} & = \frac{1}{4} \end{aligned}$
Spread and confidence are fixed, leaving only the inverse square-root dependence on sample size.
3
Solve for the sample size.
$n_{new} = 4 \times 50 = 200$
Four times the observations halve the margin.

Answer

200.

EXAMPLE 02

Competing changes

Population SD doubles and n quadruples. Compare margins at the same confidence.

Worked solution

1
Express both changes.
$σ_{new} = 2 σ, n_{new} = 4 n$
The question changes variability and sample size simultaneously, while confidence is fixed.
2
Substitute into the margin formula.
$E_{new} = z^{*} \frac{2 σ}{\sqrt{4 n}} = z^{*} \frac{2 σ}{2 \sqrt{n}}$
The square root of the fourfold sample size supplies a factor of two in the denominator.
3
Compare with the original margin.
$E_{new} = z^{*} \frac{σ}{\sqrt{n}} = E_{old}$
The doubling of population spread exactly cancels the sample-size reduction in standard error, giving equal margins.

Answer

Equal margins.

03 / YOUR TURN

Try it yourself.

Check your understanding

Margin is 6 at n = 100. Find it at n = 900 with other quantities fixed.

Show answer & reasoning

The sample size rises by a factor of $\frac{900}{100} = 9$ , so the margin is divided by $\sqrt{9} = 3$ . Equivalently, $E_{new} = 6 \sqrt{\frac{100}{900}} = 2$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Margin is 5 at $n = 36$ . Find it at $n = 144$ .
2 marks
A sample of 80 has a given margin. How many observations give one-third that margin?
2 marks
SD changes from 8 to 12 with confidence and $n$ fixed. Compare widths.
2 marks
Compare 99% and 95% widths using critical values 2.576 and 1.96.
2 marks
Both $n$ and population SD are multiplied by four. Compare margins.
2 marks

Use the supplied repeated-interval simulation table. Compare interval width when

n

rises from 25 to 100 at fixed confidence. Does the observed capture fraction have to equal the advertised confidence exactly?

3 marks

One simulation: 2,000 sample intervals per row; normal population mean 100, SD 20
Sample size	Confidence	Captures	Capture fraction	Mean width
$25$	95%	$1903$	$0.9515$	$15.680$
$25$	99%	$1978$	$0.9890$	$20.608$
$100$	95%	$1902$	$0.9510$	$7.840$
$100$	99%	$1988$	$0.9940$	$10.304$

Read the worked solutions

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

Margin is 5 at $n = 36$ . Find it at $n = 144$ .
2 marks
Four times the sample halves the margin.
$2.5$ .
A sample of 80 has a given margin. How many observations give one-third that margin?
2 marks
Multiply by $3^{2} = 9$ .
$720$ .
SD changes from 8 to 12 with confidence and $n$ fixed. Compare widths.
2 marks
$\frac{12}{8} = 1.5$ .
The new width is 1.5 times the old width.
Compare 99% and 95% widths using critical values 2.576 and 1.96.
2 marks
$\frac{2.576}{1.96} = 1.314285 \dots$ .
Ratio approximately 1.3143.
Both $n$ and population SD are multiplied by four. Compare margins.
2 marks
Spread factor 4 divided by square-root sample factor 2 gives 2.
The new margin is twice the old.

Use the supplied repeated-interval simulation table. Compare interval width when

n

rises from 25 to 100 at fixed confidence. Does the observed capture fraction have to equal the advertised confidence exactly?

3 marks

One simulation: 2,000 sample intervals per row; normal population mean 100, SD 20
Sample size	Confidence	Captures	Capture fraction	Mean width
$25$	95%	$1903$	$0.9515$	$15.680$
$25$	99%	$1978$	$0.9890$	$20.608$
$100$	95%	$1902$	$0.9510$	$7.840$
$100$	99%	$1988$	$0.9940$	$10.304$

Quadrupling sample size halves standard error and width under the known-spread model.

Each row contains a finite number of simulated intervals; capture fraction is an estimate of long-run coverage.

The normal population, fixed sample size and stated independent draws make the simulation method reproducible.

Width halves; no, empirical coverage varies randomly.

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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.

Three controls on interval width

Plan precision without losing validity

Check coverage with a computational simulation

See the method in action.

Half the margin

Form the required margin ratio.

Use the square-root sample-size relation.

Solve for the sample size.

Competing changes

Express both changes.

Substitute into the margin formula.

Compare with the original margin.

Try it yourself.

Check your understanding

Keep practising