TASMath MethodsSyllabus dot point

How do we estimate a population proportion from a sample and express the uncertainty?

Use the distribution of sample proportions to construct and interpret confidence intervals for a population proportion.

Sample proportions vary from sample to sample with an approximately normal distribution; a confidence interval uses this to give a plausible range for the unknown population proportion at a stated level of confidence.

Generated by Claude Opus 4.79 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

When you cannot survey a whole population, you estimate a proportion from a sample. The sample proportion $\hat{p} = \dfrac{\text{number of successes}}{n}$ is a point estimate of the true population proportion $p$ . Because a different sample would give a slightly different $\hat{p}$ , we describe the variability and report an interval.

Constructing a confidence interval

A confidence interval takes the point estimate and adds and subtracts a margin of error.

Since $p$ is unknown, we use $\hat{p}$ inside the standard error, which is why the formula contains $\hat{p}(1-\hat{p})$ rather than $p(1-p)$ .

The trade-offs

Higher confidence uses a larger $z$ , which widens the interval. A larger sample size $n$ shrinks the standard error and narrows the interval. So precision (a narrow interval) and confidence pull in opposite directions, and you buy precision by increasing $n$ .

Worked example: a 95% confidence interval

In a survey of $400$ voters, $220$ said they support a proposal. Construct a 95% confidence interval for the true proportion of supporters.

Point estimate:

\hat{p} = \frac{220}{400} = 0.55.

Standard error:

\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.55 \times 0.45}{400}} = \sqrt{\frac{0.2475}{400}} = \sqrt{0.00061875} \approx 0.02487.

For 95% confidence, $z \approx 1.96$ , so the margin of error is

E = 1.96 \times 0.02487 \approx 0.04875.

The confidence interval is

0.55 \pm 0.0488, \quad \text{that is} \quad (0.501,\ 0.599).

We are 95% confident that between about $50.1\%$ and $59.9\%$ of all voters support the proposal.

Summary

Compute $\hat{p}$ , then the standard error $\sqrt{\hat{p}(1-\hat{p})/n}$ , then multiply by the $z$ for your confidence level to get the margin of error, and add and subtract it from $\hat{p}$ . Remember that increasing confidence widens the interval while increasing the sample size narrows it, and always interpret the interval as a statement about the long-run reliability of the method.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 TASC4 marks200 random people are surveyed and 70 say they prefer honey to jam on their toast. A confidence interval for the true proportion who prefer honey is (0.2945, 0.4055). Find the level of confidence that generates this interval. Show full working.

Show worked answer →

The sample proportion is p-hat = 70/200 = 0.35, which is the centre of the interval.

The margin of error is half the interval width: 0.4055 - 0.35 = 0.0555 (equivalently 0.35 - 0.2945).

The standard deviation of the sample proportion is sqrt(p-hat (1 - p-hat)/n) = sqrt(0.35 times 0.65 / 200) = sqrt(0.0011375) = 0.033727.

The z-value used is margin / standard deviation = 0.0555 / 0.033727 = 1.645.

A z-value of 1.645 corresponds to a 90% confidence level (it leaves 5% in each tail). So the interval was generated at 90% confidence. Marks reward finding p-hat, the margin, the standard deviation, and identifying z = 1.645 as 90%.

2024 TASC2 marksA survey of 1000 people found that 50% own a puffer jacket made by a company. Show that the 95% confidence interval for this finding is approximately (46.90%, 53.10%), i.e. a margin of error of 3.10%, to 2 decimal places.

Show worked answer →

Here p-hat = 0.5 and n = 1000. For 95% confidence the z-value is 1.96.

Margin of error = 1.96 times sqrt(p-hat (1 - p-hat)/n)
= 1.96 times sqrt(0.5 times 0.5 / 1000)
= 1.96 times sqrt(0.00025)
= 1.96 times 0.015811 = 0.03099, i.e. about 3.10%.

Confidence interval = 0.5 plus or minus 0.0310 = (0.4690, 0.5310) = (46.90%, 53.10%). Showing the substitution into the margin formula and the final interval earns both marks.

2023 TASC2 marksA marketing company samples 500 people, with 53% in favour of Party A. They wish to publish 53% support with a 95% confidence interval. Calculate the 95% confidence interval for the true support for Party A.

Show worked answer →

With p-hat = 0.53, n = 500 and z = 1.96 for 95% confidence:

Margin of error = 1.96 times sqrt(0.53 times 0.47 / 500)
= 1.96 times sqrt(0.0004982)
= 1.96 times 0.022321 = 0.04375 (about 4.38%).

Confidence interval = 0.53 plus or minus 0.0438 = (0.4862, 0.5738), i.e. about (48.6%, 57.4%).

Note that the interval extends below 50%, which is why the follow-up part of this question explores how a higher confidence level would be needed if they wanted the lower bound to sit at exactly 50%.