The sample proportion $\hat{p}$ as a random variable, the sampling distribution of $\hat{p}$ for repeated samples of size $n$ from a population with true proportion $p$, and the normal approximation for large $n$

What this dot point is asking

VCAA wants you to treat the sample proportion $\hat{p}$ as a random variable, identify the mean and standard deviation of its sampling distribution, and apply the normal approximation to compute sample-proportion probabilities. The dot point is the statistical-inference precursor to confidence intervals.

What is a sample proportion

Suppose a population has a true proportion $p$ of "successes" (members with some characteristic: voters for party A, defective items, smokers, opinion-poll affirmatives). A random sample of $n$ items is drawn, and the number of successes in the sample is recorded as $X$.

The sample proportion is:

Because $X$ is random (depends on which $n$ items happen to be sampled), $\hat{p}$ is a random variable. It varies from sample to sample.

The sampling distribution of $\hat{p}$

Repeatedly drawing samples of size $n$ from the same population and computing $\hat{p}$ each time produces a distribution of $\hat{p}$-values: the sampling distribution of $\hat{p}$.

Two facts about this distribution:

Mean of $\hat{p}$

The expected value of the sample proportion equals the population proportion. The sample proportion is an unbiased estimator of $p$.

Standard deviation of $\hat{p}$

Two interpretations:

• The standard deviation falls as $\sqrt{n}$. Quadrupling the sample size halves the standard deviation.
• The standard deviation is largest when $p = 0.5$. At $p = 0$ or $p = 1$, SD($\hat{p}$) = 0 (no variability because every sample has the same proportion).

Conditions for the formula

The formula assumes:

• Independence. Each sample item is drawn independently. In practice, this requires either sampling with replacement, or sampling from a population large enough that each draw does not materially change the remaining proportion (typically, the population should be at least 10 times the sample size).
• Identical distribution. Each sampled item has the same probability $p$ of being a success.

These are the conditions of the binomial distribution: $X \sim \text{Bin}(n, p)$.

The normal approximation

For large $n$, the sampling distribution of $\hat{p}$ is approximately normal:

(Equivalently, $\hat{p}$ has mean $p$ and standard deviation $\sqrt{p(1-p)/n}$.)

When is "large $n$" large enough?

Standard conditions (VCAA cites both):

• $n p \geq 10$
• $n (1 - p) \geq 10$

Some texts use $5$ or $15$ as the threshold; VCAA accepts any reasonable convention. The conditions ensure the binomial is well-approximated by the normal.

Why the normal approximation works

$X \sim \text{Bin}(n, p)$ for large $n$ is approximately $N(n p, n p (1 - p))$ by the central limit theorem. Dividing by $n$ gives $\hat{p} = X / n$ approximately $N(p, p(1-p)/n)$.

Computing sample-proportion probabilities

To find $P(\hat{p} \leq c)$, $P(\hat{p} \geq c)$ or $P(a \leq \hat{p} \leq b)$:

1. Verify the normal approximation conditions ($n p \geq 10$ and $n (1 - p) \geq 10$).
2. State the approximate distribution: $\hat{p} \approx N(p, p(1-p)/n)$.
3. Standardise: $z = \frac{\hat{p} - p}{\sqrt{p(1-p)/n}}$.
4. Compute the probability using calculator (normCdf) or table.

Worked example

A factory produces 60 percent of items meeting specification. A sample of $n = 150$ items is taken. Find the probability that the sample proportion meeting spec is at least 0.55.

Mean: $E(\hat{p}) = 0.60$.

SD: $\sqrt{0.6 \times 0.4 / 150} = \sqrt{0.0016} = 0.04$.

Conditions: $n p = 90 \geq 10$, $n (1-p) = 60 \geq 10$. Normal approximation valid.

Standardise: $z = (0.55 - 0.60) / 0.04 = -1.25$.

$P(\hat{p} \geq 0.55) = P(Z \geq -1.25) \approx 0.8944$.

Sampling distribution shape and the central limit theorem

For small $n$, the sampling distribution of $\hat{p}$ is discrete (taking only values $0, 1/n, 2/n, \ldots, 1$) and may be skewed if $p$ is far from 0.5. As $n$ grows, the distribution becomes both more concentrated (smaller SD) and more bell-shaped (better normal approximation). This is the central limit theorem in action.

VCE Methods does not require formal statement of the CLT, but the underlying intuition is the reason the normal approximation works.

Examples in context

Example 1. Opinion poll variability. In a region $55\%$ favour a proposal ($p = 0.55$). A poll of $n = 400$ people gives a sample proportion $\hat p$ with mean $E(\hat p) = 0.55$ and standard deviation $\sqrt{\frac{0.55 \times 0.45}{400}} = \sqrt{0.000619} = 0.0249$. To find the chance the poll shows under $50\%$ support: standardise $z = \frac{0.50 - 0.55}{0.0249} = -2.01$, so $P(\hat p < 0.50) = P(Z < -2.01) \approx 0.022$.

Example 2. Choosing a sample size for precision. A manufacturer with a $20\%$ return rate ($p = 0.2$) wants $\text{SD}(\hat p) \le 0.025$. Solving $\sqrt{\frac{0.2 \times 0.8}{n}} \le 0.025$ gives $\frac{0.16}{n} \le 0.000625$, so $n \ge 256$. A sample of at least $256$ products achieves the target precision.

Try this

Q1. A population has $p = 0.3$. For a sample of $n = 100$, find the mean and standard deviation of $\hat{p}$. [2 marks]

• Cue. $E(\hat p) = 0.3$; $\text{SD}(\hat p) = \sqrt{\frac{0.3 \times 0.7}{100}} = \sqrt{0.0021} = 0.0458$.

Q2. For $p = 0.5$ and $n = 100$, find $P(\hat{p} > 0.6)$ using the normal approximation. [3 marks]

• Cue. $\text{SD} = \sqrt{\frac{0.25}{100}} = 0.05$; $z = \frac{0.6 - 0.5}{0.05} = 2$; $P(Z > 2) \approx 0.0228$.

Q3. Find the smallest $n$ so that $\text{SD}(\hat{p}) < 0.03$ when $p = 0.4$. [3 marks]

• Cue. $\frac{0.24}{n} < 0.0009 \Rightarrow n > 266.7$, so $n = 267$.