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.

§-Syllabus dot point

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 4.88 min answerUpdated 2026-06-02

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 $0.28$ .

When is the sample large enough?

The normal approximation only works when the sample is large enough that the binomial count is not bunched against $0$ or $n$ . A standard rule of thumb requires both $np \ge 10$ and $n(1-p) \ge 10$ (some texts use $5$ ). For a true proportion near $0.5$ a sample of a few dozen already behaves normally, but for a rare event such as $p = 0.02$ you need a much larger $n$ before the bell shape emerges, because most samples would otherwise record zero successes. Checking this condition is a quick way to justify, in an exam answer, that you may use $z$ -values at all.

Comparing two sample sizes

Problem

A true proportion is $p = 0.4$ . Compare the standard error of $\hat{p}$ for $n = 100$ and for $n = 400$ .

Standard error for n = 100

\sqrt{\frac{0.4\times 0.6}{100}} = \sqrt{0.0024} = 0.04899.

Standard error for n = 400

\sqrt{\frac{0.4\times 0.6}{400}} = \sqrt{0.0006} = 0.02449.

Interpretation

Quadrupling the sample size from $100$ to $400$ halves the standard error, because the spread scales as $\dfrac{1}{\sqrt{n}}$ and $\sqrt{4} = 2$ . To halve the margin of error again you would need to quadruple the sample once more, to $1600$ . This diminishing return is why very precise polls are expensive.

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.

TCE 20233 marksCalculator-assumed. A reporter surveys

10

people and finds

6

prefer product A, giving a

95\%

confidence interval of

(29.6\%, 90.4\%)

. The editor asks for

1000

people instead, a

100

-times increase. Assuming the larger survey still finds

60\%

prefer product A, what is the new

95\%

confidence interval?

Show worked answer →

The standard error of the sample proportion is $\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}$ . With $\hat{p} = 0.6$ and the new $n = 1000$ :

\sqrt{\frac{0.6\times 0.4}{1000}} = \sqrt{0.00024} = 0.015492.

The $95\%$ margin of error is $1.96\times 0.015492 = 0.0304$ (about $3.04\%$ ).

So the new interval is $0.6 \pm 0.0304 = (0.5696, 0.6304)$ , i.e. about $(56.96\%, 63.04\%)$ .

The key idea: multiplying the sample size by $100$ divides the standard error (and the margin) by $\sqrt{100} = 10$ , shrinking the old margin from about $0.304$ to about $0.0304$ . Larger samples give narrower intervals.

TCE 20243 marksCalculator-free. In a large population the true proportion supporting a policy is

p = 0.25

. Random samples of size

n = 75

are taken. (a) State the mean of the sampling distribution of

\hat{p}

. (b) Find the standard deviation of

\hat{p}

, to four decimal places. (c) Within what range would about

95\%

of sample proportions lie?

Show worked answer →

(a) The sampling distribution is unbiased, so $E(\hat{p}) = p = 0.25$ .

(b) The standard error is

\sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.25\times 0.75}{75}} = \sqrt{0.0025} = 0.05 = 0.0500.

(c) About $95\%$ of values lie within two standard deviations of the mean: $0.25 \pm 2(0.05) = 0.25 \pm 0.10$ , i.e. from $0.15$ to $0.35$ . One mark each for the mean, the standard error, and the $\pm 2\sigma$ range.