Confidence intervals for a population proportion | Year 12

01 / LEARN

Understand the idea.

Estimate the unknown spread

When $p$ is unknown, first calculate $\hat{p} = \frac{x}{n}$ and estimate its standard error with $\sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}$ . The normal, or Wald, interval places the margin $z^{*}$ times this estimated standard error on each side of $\hat{p}$ .

Check random selection, approximate independence, and sufficiently many observed successes and failures before using this approximation. Small counts or proportions near zero or one can give poor coverage or impossible bounds. Cutting bounds back to the valid range does not repair the method.

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

State the population conclusion

Report the confidence level, population, success definition and interval bounds together. A proportion margin of 0.04 means four percentage points; it is not a relative 4% increase or decrease. Preserve intermediate precision before rounding the final endpoints.

The interval measures sampling uncertainty under its assumptions. It does not allow for missing groups, selective replies or leading questions. For example, a large voluntary poll can produce a small estimated margin without providing a valid interval for all residents.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A 95% interval

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

Worked solution

1
Calculate the estimate and check counts.
$\begin{aligned} \hat{p} & = \frac{240}{400} = 0.6 \\ n - x & = 400 - 240 = 160 \end{aligned}$
The stated independent random sample has 240 successes and 160 failures, supporting the normal-interval approximation.
2
Estimate the standard error.
$\hat{SE} (\hat{p}) = \sqrt{\frac{0.6 (0.4)}{400}} \approx 0.0244949$
The unknown population proportion is replaced by its sample estimate for this Wald interval.
3
Calculate the 95% margin.
$E = 1.96 \sqrt{\frac{0.24}{400}} \approx 0.0480100$
The 95% normal critical value multiplies the unrounded estimated standard error.
4
Calculate and interpret the endpoints.
$\begin{aligned} L & = 0.6 - E \approx 0.5520 \\ U & = 0.6 + E \approx 0.6480 \end{aligned}$
We are approximately 95% confident that the population proportion of adults supporting the proposal lies between these bounds, under the sampling assumptions.

Answer

Approximately (0.5520, 0.6480).

EXAMPLE 02

A different confidence level

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

Worked solution

1
Calculate the sample proportion and counts.
$\begin{aligned} \hat{p} & = \frac{180}{300} = 0.6 \\ n - x & = 120 \end{aligned}$
The observed success and failure counts are both large; random selection and approximate independence are still required for population inference.
2
Estimate the standard error.
$\hat{SE} (\hat{p}) = \sqrt{\frac{0.6 (0.4)}{300}} \approx 0.0282843$
A smaller sample than in the preceding example gives a larger estimated standard error at the same proportion.
3
Use the supplied 90% multiplier.
$E = 1.645 \sqrt{\frac{0.24}{300}} \approx 0.0465276$
Confidence is 90%, so use 1.645 rather than the 95% critical value.
4
Form the approximate interval.
$\begin{aligned} L & = 0.6 - E \approx 0.5535 \\ U & = 0.6 + E \approx 0.6465 \end{aligned}$
The approximate 90% interval concerns the population success proportion under the model, not a guaranteed range for each future sample.

Answer

Approximately (0.5535, 0.6465).

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

0.0693, or 6.93 percentage points.

Here $\hat{p} = \frac{100}{200} = 0.5$ . The approximate 95% margin is $1.96 \sqrt{\frac{0.5 (0.5)}{200}} \approx 0.0693$ , equivalent to 6.93 percentage points. This is a half-width, so the full interval width is twice as large.

Keep practising

6 worksheet questions · 12 marks

Open worksheet 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 $0.47 \pm 0.035$ . 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 $1.96 \sqrt{\frac{0.24}{400}} = 0.0480100 \dots$ .
Approximately $(0.3520, 0.4480)$ .
Find a 95% interval for 225 successes from 500 independent observations.
2 marks
Estimate 0.45; margin $1.96 \sqrt{\frac{0.45 (0.55)}{500}} = 0.0436072 \dots$ .
Approximately $(0.4064, 0.4936)$ .
An interval is $0.47 \pm 0.035$ . 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 ↗

In this lesson

Understand the idea.

Estimate the unknown spread

State the population conclusion

See the method in action.

A 95% interval

Calculate the estimate and check counts.

Estimate the standard error.

Calculate the 95% margin.

Calculate and interpret the endpoints.

A different confidence level

Calculate the sample proportion and counts.

Estimate the standard error.

Use the supplied 90% multiplier.

Form the approximate interval.

Try it yourself.

Check your understanding

Keep practising