What is simulating to see the variability?

A useful way to understand this is simulation. If you repeatedly generate samples from a population with a known p and plot all the resulting p values in a histogram, you see a roughly bell-shaped cluster centred on p. Larger samples produce a tighter, taller cluster, visually confirming that the standard error falls as n rises.

TASMath MethodsSyllabus dot point

How does a sample proportion vary from sample to sample, and what distribution does it follow?

Understand random sampling and describe the distribution of the sample proportion, including its mean and standard deviation.

Why random sampling matters, how the sample proportion behaves as a random variable, and the mean and standard deviation of its approximately normal sampling distribution for TCE Mathematics Methods Unit 4.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-30

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

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What this dot point is asking

Before you can build a confidence interval, you need to understand how a sample proportion behaves. This dot point is the bridge between the binomial work of Unit 3 and the interval estimation that follows.

Why random sampling matters

A sample is random when every member of the population has a known, nonzero chance of being chosen and the selection is free of bias. Random sampling is what justifies using the sample to make claims about the whole population. A convenience or self-selected sample can be systematically unrepresentative, so the methods of this topic simply do not apply to it.

The sample proportion as a random variable

Imagine taking many independent random samples of the same size $n$ and recording $\hat{p}$ each time. The collection of those values has its own distribution, called the sampling distribution of the sample proportion. Because the count of successes is binomial, dividing by $n$ rescales it, and the centre and spread follow directly.

Because the mean of $\hat{p}$ equals $p$ , the sample proportion is an unbiased estimator: on average across many samples it lands on the true value.

Approximate normality

Simulating to see the variability

A useful way to understand this is simulation. If you repeatedly generate samples from a population with a known $p$ and plot all the resulting $\hat{p}$ values in a histogram, you see a roughly bell-shaped cluster centred on $p$ . Larger samples produce a tighter, taller cluster, visually confirming that the standard error falls as $n$ rises.

Describing a sampling distribution

Problem

A factory produces components of which a true proportion $p = 0.2$ are defective. A quality inspector takes random samples of $n = 100$ components. Describe the sampling distribution of the sample proportion of defectives.

Solution

The mean of the sampling distribution is the true proportion:

E(\hat{p}) = p = 0.2.

The standard deviation (standard error) is

\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.2 \times 0.8}{100}} = \sqrt{\frac{0.16}{100}} = \sqrt{0.0016} = 0.04.

So $\hat{p}$ is approximately normal with mean $0.2$ and standard deviation $0.04$ . Most samples will give a defective proportion within about two standard deviations of $0.2$ , that is roughly between $0.12$ and $0.28$ .

Summary

A random sample lets the sample proportion stand in for the population proportion. Treat $\hat{p}$ as a random variable whose sampling distribution is centred at the true $p$ , with standard error $\sqrt{p(1-p)/n}$ , and is approximately normal for large $n$ . This unbiased, shrinking-spread behaviour is exactly what makes the confidence interval in the next dot point work.

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.

2023 TASC3 marksA reporter surveys 10 people and finds 6 prefer product A, giving a 95% confidence interval of (29.6%, 90.4%). The editor says this sample is too small and asks for 1000 people instead, a 100-times increase. Assuming the larger survey still finds 60% prefer product A, what will the 95% confidence interval be now?

Show worked answer →

The standard deviation of the sample proportion is sqrt(p-hat (1 - p-hat)/n). With p-hat = 0.6 and the new n = 1000:
sqrt(0.6 times 0.4 / 1000) = sqrt(0.00024) = 0.015492.

The 95% margin of error is 1.96 times this standard deviation:
1.96 times 0.015492 = 0.0304 (about 3.04%).

So the new interval is 0.6 plus or minus 0.0304 = (0.5696, 0.6304), i.e. about (56.96%, 63.04%).

Note the key idea the question is testing: multiplying the sample size by 100 divides the standard deviation (and hence the margin of error) by sqrt(100) = 10, shrinking the old margin of 0.304 down to about 0.0304. Larger samples give narrower intervals.