Confidence intervals for a population proportion | Year 12

01 / LEARN

The idea, step by step

Estimate the unknown spread

Compute $\hat{p} = x / n$ , then estimated standard error $\sqrt{\hat{p} (1 - \hat{p}) / n}$ . An approximate interval is $\hat{p} \pm z^{*} \sqrt{\hat{p} (1 - \hat{p}) / n}$ . It is a normal or Wald interval.

Check random selection, approximate independence, and sufficiently many observed successes and failures. With small samples or estimates near 0 or 1, this symmetric method can be poor or give impossible endpoints. Clipping endpoints does not repair its coverage.

\hat{p} \pm z^{*} \sqrt{\hat{p} (1 - \hat{p}) / n}

State the population conclusion

Give confidence level, population, success definition and bounds. A margin of 0.04 means four percentage points, not a relative 4% change. Keep intermediate precision until the final result.

The margin describes sampling variation under the model. It does not include frame bias, leading questions or non-response bias. A very large voluntary poll may have a narrow but invalid interval for a broader population.

A common trap

Use the unknown p inside a calculated data interval.

Instead: Use the observed sample proportion to estimate standard error for this interval method.

02 / WORKED EXAMPLES

See the thinking

Example 1

A 95% interval

240 of 400 independent randomly sampled adults support a proposal. Find a 95% interval.

$\hat{p} = 0.6$ ; estimated SE $= \sqrt{0.24 / 400} = 0.0244949 \dots$ .
Margin = 1.96 × SE = 0.0480100… .
Both observed counts, 240 and 160, are large.

Approximately (0.5520, 0.6480).

Example 2

A different confidence level

180 of 300 observations are successes. Use 1.645 for a 90% interval.

The estimate is 0.6.
Estimated SE $= \sqrt{0.24 / 300} = 0.0282843 \dots$ .
Margin = 0.0465276… .

Approximately (0.5535, 0.6465).

Check your understanding

Find the approximate 95% margin for 100 successes among 200.

Show answer

0.0693, or 6.93 percentage points.

$1.96 \sqrt{0.25 / 200} = 0.069296 \dots$ .

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Find a 95% interval for 160 successes from 400 independent randomly sampled people.
2 marks
Find a 95% interval for 225 successes from 500 independent observations.
2 marks
An interval is 0.47 ± 0.035. Give endpoints and margin in percentage points.
2 marks
Why is the formula unsuitable for zero successes among 30?
2 marks
A random sample of app users is used to infer a proportion for all adults. What can its margin omit?
2 marks
Does a proportion interval (0.51,0.59) contain the binary outcomes of 95% of people?
2 marks

Read the worked solutions

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

Find a 95% interval for 160 successes from 400 independent randomly sampled people.
2 marks
Estimate 0.4; margin $1.96 \sqrt{0.24 / 400} = 0.0480100 \dots$ .
Approximately (0.3520, 0.4480).
Find a 95% interval for 225 successes from 500 independent observations.
2 marks
Estimate 0.45; margin $1.96 \sqrt{0.45 (0.55) / 500} = 0.0436072 \dots$ .
Approximately (0.4064, 0.4936).
An interval is 0.47 ± 0.035. Give endpoints and margin in percentage points.
2 marks
Subtract/add 0.035; multiply the margin by 100 for percentage points.
(0.435,0.505); 3.5 percentage points.
Why is the formula unsuitable for zero successes among 30?
2 marks
No observed success does not prove zero population probability. Another method is required.
It gives a misleading zero-width interval and the success-count condition fails.
A random sample of app users is used to infer a proportion for all adults. What can its margin omit?
2 marks
The sampling frame differs from the target population.
Coverage bias from excluding non-users.
Does a proportion interval (0.51,0.59) contain the binary outcomes of 95% of people?
2 marks
Individual outcomes are success/failure. Confidence refers to repeated coverage.
No; it estimates one aggregate population proportion.

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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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