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WAMath MethodsSyllabus dot point

How does the sample proportion behave across repeated random samples, and why is it approximately normal?

Describe the sampling distribution of the sample proportion, including its mean, standard error and approximate normality for large samples

WACE Year 12 Mathematics Methods Unit 4 the sample proportion: random sampling, the sampling distribution of p-hat, its mean p, the standard error, and approximate normality for large samples, with worked SCSA-style examples.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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What this dot point is asking
Random sampling and the sample proportion
Mean and standard error
The standard error and sample size
Why approximate normality

What this dot point is asking

SCSA Unit 4 introduces statistical inference by studying how the sample proportion behaves across repeated samples. This dot point asks you to describe its sampling distribution, the mean, the standard error and the approximate normality, which underpins the confidence interval that follows. It is examined in both sections.

Random sampling and the sample proportion

When a random sample of size $n$ is drawn, the number of successes $X$ follows a binomial distribution $X\sim B(n,p)$ , where $p$ is the unknown population proportion. The sample proportion is

\hat{p}=\frac{X}{n}.

Different samples give different values of

\hat{p}

, so

\hat{p}

is itself a random variable with its own distribution, called the sampling distribution.

Mean and standard error

Because $\hat{p}$ is $\dfrac{1}{n}$ times a binomial variable, its mean and standard deviation follow from the binomial mean $np$ and variance $np(1-p)$ .

Since $E(\hat{p})=p$ , the sample proportion is an unbiased estimator: on average it equals the true proportion. The standard error has $\sqrt{n}$ in the denominator, so larger samples give less variable, more precise estimates.

Mean and standard error of the sample proportion

A population has true proportion $p=0.4$ . For samples of size $n=100$ , find the mean and standard error of $\hat{p}$ .

Step 1: State the mean of $\hat{p}$

The sampling distribution of $\hat{p}$ is centred on the true population proportion. This is why $\hat{p}$ is called an unbiased estimator: on average it equals $p$ .

E(\hat{p}) = p = 0.4

Step 2: Compute the standard error

The standard error measures the typical spread of $\hat{p}$ values across samples. It uses the true $p$ (not $\hat{p}$ ) and shrinks as $n$ increases.

\mathrm{SD}(\hat{p}) = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4\times 0.6}{100}} = \sqrt{\frac{0.24}{100}} = \sqrt{0.0024} \approx 0.049

Final answer: $\hat{p}$ has mean $0.4$ and standard error approximately $0.049$ .

The standard error and sample size

The factor $\dfrac{1}{\sqrt{n}}$ in the standard error has an important consequence: precision improves only with the square root of the sample size, not with the size itself. To halve the standard error you must quadruple the sample, and to cut it to a third you need nine times as many observations. This diminishing return is why large national surveys reach a point where extra respondents add little precision, and it explains the sample sizes that appear in confidence-interval and margin-of-error questions. The standard error is largest when $p=0.5$ , because $p(1-p)$ peaks there, so a survey near a $50/50$ split is the hardest case to estimate precisely.

Why approximate normality

For large $n$ the binomial distribution of $X$ is well approximated by a normal distribution, so $\hat{p}=\dfrac{X}{n}$ is approximately normal too. A common rule of thumb is that both $np$ and $n(1-p)$ should be at least about $10$ . This normal approximation is what allows the confidence interval to use a $z$ -multiplier.

Exam-style practice questions

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

WACE 20226 marksCalculator-assumed. In a population, the true proportion supporting a proposal is

p=0.55

. A random sample of

n=200

is taken. (a) State the mean and standard error of

\hat{p}

. (b) Find

P(\hat{p}>0.6)

using the normal approximation.

Show worked answer →

A sampling-distribution probability.

(a) $E(\hat{p})=p=0.55$ ; $\mathrm{SD}(\hat{p})=\sqrt{\dfrac{0.55\times 0.45}{200}}=\sqrt{0.0012375}\approx 0.0352$ .

(b) Standardise: $z=\dfrac{0.6-0.55}{0.0352}\approx 1.42$ . So $P(\hat{p}>0.6)=P(Z>1.42)\approx 0.078$ .

Markers reward the mean, the standard error with the square root, and standardising with the normal approximation.

WACE 20244 marksCalculator-free. A sample proportion

\hat{p}

comes from samples of size

n

from a population with proportion

p

. (a) Explain why

\hat{p}

is an unbiased estimator of

p

. (b) Describe what happens to the standard error if the sample size is increased fourfold.

Show worked answer →

A conceptual question about the sampling distribution.

(a) The mean of the sampling distribution is $E(\hat{p})=p$ , so on average across repeated samples $\hat{p}$ equals the true proportion; it neither systematically overestimates nor underestimates $p$ , which is what unbiased means.

(b) The standard error is $\sqrt{\dfrac{p(1-p)}{n}}$ , proportional to $\dfrac{1}{\sqrt{n}}$ . Multiplying $n$ by $4$ multiplies $\sqrt{n}$ by $2$ , so the standard error is halved.

Markers reward identifying $E(\hat{p})=p$ for unbiasedness and the $\dfrac{1}{\sqrt{n}}$ relationship giving a halving.