Apply the sampling distribution of the sample proportion $\hat{p}$ (mean $p$, standard deviation $\sqrt{p(1-p)/n}$) and construct approximate confidence intervals $\hat{p} \pm z^* \sqrt{\hat{p}(1-\hat{p})/n}$ for a population proportion

What this dot point is asking

QCAA wants you to treat the sample proportion $\hat{p}$ as a random variable, apply the normal approximation to its sampling distribution, construct confidence intervals for a population proportion, and interpret the interval correctly. The dot point bridges Unit 3 binomial probability with Unit 4 statistical inference, and is heavily examined in PSMT and EA.

Sample proportion

If a population has true proportion $p$ of "successes" and a random sample of $n$ is drawn with $X$ successes, the sample proportion is:

Because $X$ is random, $\hat{p}$ is a random variable: it varies from sample to sample.

Sampling distribution of IMATH_10

Mean. $E(\hat{p}) = p$. The sample proportion is an unbiased estimator of $p$.

Standard deviation. $\text{SD}(\hat{p}) = \sqrt{\frac{p(1-p)}{n}}$.

Two takeaways:

• SD falls as $\sqrt{n}$: quadruple $n$ to halve SD.
• SD is maximised at $p = 0.5$; minimised at $p = 0$ or $p = 1$.

Normal approximation

For large $n$:

Conditions for the approximation (QCAA convention):

• IMATH_20
• IMATH_21

When these conditions hold, $\hat{p}$ is approximately normal with mean $p$ and SD $\sqrt{p(1-p)/n}$. Standardise via $Z = (\hat{p} - p)/\sqrt{p(1-p)/n}$ to compute probabilities.

Confidence intervals

A confidence interval for a population proportion combines:

• The point estimate $\hat{p}$ (centre).
• The standard error: $\sqrt{\hat{p}(1-\hat{p})/n}$ (using $\hat{p}$ in place of unknown $p$).
• The critical value $z^*$ for the confidence level.

The formula:

The margin of error is $\text{MoE} = z^* \times \text{SE}$.

Standard $z^*$ values

Interpretation

A $C \%$ confidence interval has the long-run interpretation:

Approximately $C \%$ of intervals constructed by this procedure across repeated samples would contain the true population proportion.

This is NOT:

• "There is a $C \%$ probability that $p$ is in this interval." (Once the interval is constructed, $p$ either is or is not in it.)
• "$C \%$ of the population have proportions in this interval." (The interval is about the parameter, not about individuals.)

The correct language refers to the long-run success rate of the procedure.

Sample size design

To achieve a margin of error at most $E$ at $C \%$ confidence:

If $\hat{p}$ is unknown in advance, use $\hat{p} = 0.5$ (worst case) for a conservative design.

Always round up to the next integer (cannot sample a fractional person).

Worked example

A study estimates the proportion of adults who exercise daily. A random sample of $n = 400$ adults gives 120 who do.

$\hat{p} = 120/400 = 0.30$.

SE: $\sqrt{0.30 \times 0.70 / 400} = \sqrt{0.000525} \approx 0.0229$.

For 95 percent: $z^* = 1.96$. Margin: $1.96 \times 0.0229 \approx 0.0449$.

CI: $(0.30 - 0.045, 0.30 + 0.045) = (0.255, 0.345)$.

Interpretation: approximately 95 percent of intervals constructed by this procedure would contain the true population proportion of adults who exercise daily.

Trade-offs

Confidence vs precision. Higher confidence (99 percent) requires a wider interval. Lower confidence (90 percent) gives a narrower interval. To improve both, increase $n$.

Sample size economics. Doubling $n$ reduces SE by a factor of $\sqrt{2}$. Quadrupling $n$ halves SE. Diminishing returns above $n \approx 1000$ for opinion polling.

When the normal approximation fails

For very small samples or proportions near 0 or 1, the approximation can be poor. QCAA conventions ($n p \geq 10$ and $n(1-p) \geq 10$) ensure validity. Outside these conditions, an exact (binomial-based) interval would be needed, beyond Methods scope.

Common errors

Wrong $z^*$ for the level. 1.96 for 95 percent; 1.645 for 90 percent. Mixing them up gives wrong widths.

Probability misinterpretation. "The probability $p$ is in this interval" is wrong. Use long-run-procedure language.

Sample size not rounded up. $n = 384.1$ becomes $n = 385$, not 384.

**$p$ vs $\hat{p}$ in formulas.** In the sampling distribution SD: use $p$ (when known). In the CI standard error: use $\hat{p}$ (when $p$ unknown).

Forgetting worst-case $p = 0.5$. When no prior estimate of $p$ exists for sample-size design, use $p = 0.5$ to maximise $p(1-p) = 0.25$ for the most conservative $n$.

In one sentence

The sample proportion $\hat{p} = X/n$ has mean $p$ and SD $\sqrt{p(1-p)/n}$, and for large $n$ (with $np \geq 10$ and $n(1-p) \geq 10$) is approximately normal; an approximate $C \%$ confidence interval for the population proportion is $\hat{p} \pm z^* \sqrt{\hat{p}(1-\hat{p})/n}$ with $z^* = 1.645, 1.96, 2.58$ for the 90, 95, 99 percent levels, interpreted as the long-run procedure containing the true $p$ in approximately $C \%$ of repeated samples.