Approximate confidence intervals for a population proportion $p$ based on the sample proportion $\hat{p}$, including the standard 90, 95 and 99 percent intervals and their interpretation

What this dot point is asking

VCAA wants you to construct confidence intervals for a population proportion $p$ from sample data, choose the correct critical value $z^*$ for the requested confidence level, interpret the interval correctly (avoiding the common probability-misstatement), and solve sample-size design problems based on margin-of-error targets.

The confidence interval formula

Given a sample of size $n$ with sample proportion $\hat{p}$, an approximate $C \%$ confidence interval for the population proportion $p$ is:

The components are:

• Point estimate. $\hat{p}$, the centre of the interval.
• Standard error (SE). $\sqrt{\hat{p}(1 - \hat{p}) / n}$, the estimated standard deviation of $\hat{p}$.
• Critical value $z^*$. The number of standard errors corresponding to the confidence level.
• Margin of error (MoE). $z^* \times \text{SE}$, the half-width of the interval.

The interval is $(\hat{p} - \text{MoE}, \hat{p} + \text{MoE})$.

Note. $\hat{p}$ vs $p$ in the standard error

In the sampling-distribution formula, $\text{SD}(\hat{p}) = \sqrt{p(1-p)/n}$ uses the true (and usually unknown) $p$.

In the confidence-interval formula, the standard error uses $\hat{p}$ in place of $p$ because $p$ is unknown. This is the plug-in standard error. For Methods purposes the substitution is the standard convention.

Standard $z^*$ values

These come from the inverse-normal: $z^* = \text{invNorm}((1 + C/100) / 2)$ for confidence level $C$ percent.

The 95 percent value of 1.96 is the most commonly used and worth memorising.

Interpretation: what the confidence level means

A $C \%$ confidence interval has the following correct interpretation:

If the sampling procedure were repeated many times (each time computing a new $\hat{p}$ and a new confidence interval from a fresh sample of size $n$), approximately $C \%$ of the constructed intervals would contain the true population proportion $p$.

The interval describes a procedure that works $C \%$ of the time, not a probability statement about a single specific interval.

Common misstatements VCAA marks against

These are widespread but incorrect:

• "There is a 95 percent probability that the true proportion lies in this interval." (Wrong: once the interval is constructed, $p$ either is or is not in it; there is no probability.)
• "95 percent of the population have a proportion in this interval." (Wrong: the interval is about the parameter $p$, not about individuals.)
• "We are 95 percent sure that $\hat{p}$ is in this interval." (Wrong: $\hat{p}$ is the centre of the interval by construction.)

A correct interpretation refers to the long-run success rate of the procedure.

Worked construction

A sample of $n = 500$ patients includes 175 with a particular condition. Construct a 95 percent confidence interval for the true population prevalence $p$.

Sample proportion. $\hat{p} = 175 / 500 = 0.35$.

Standard error. $\sqrt{0.35 \times 0.65 / 500} = \sqrt{0.000455} \approx 0.02133$.

$z^*$ for 95 percent. $1.96$.

Margin of error. $1.96 \times 0.02133 \approx 0.0418$.

Confidence interval. $(0.35 - 0.0418, 0.35 + 0.0418) = (0.308, 0.392)$.

Interpretation: 95 percent of intervals constructed by this procedure would contain the true population prevalence.

Margin of error and sample size

The margin of error decreases as the sample size grows. Specifically:

To halve the margin of error, the sample size must quadruple (because of the $\sqrt{n}$ in the denominator).

To design a study with a specific margin of error, solve for $n$:

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

Worst-case sample size

If no prior estimate of $\hat{p}$ is available, use $\hat{p} = 0.5$ in the design formula, which maximises $\hat{p}(1 - \hat{p}) = 0.25$. This gives the largest required $n$ and so a conservative design.

For 95 percent confidence with $\text{MoE} = 0.03$:

$n \geq (1.96)^2 \times 0.25 / (0.03)^2 = 0.9604 / 0.0009 \approx 1067$.

The "1000 sample" convention in public opinion polling comes from approximately this calculation.

Confidence level and margin trade-off

Higher confidence level requires wider interval. For fixed $n$ and $\hat{p}$:

• 90 percent CI is narrower than 95 percent CI is narrower than 99 percent CI.
• The trade-off is between certainty (higher confidence) and precision (narrower interval).

To increase confidence without widening, you must increase $n$.

When the normal approximation is invalid

The formula assumes:

• $n \hat{p} \geq 10$ and $n (1 - \hat{p}) \geq 10$ (or 5, depending on convention).
• The sample is random and independent.

For very small samples or proportions very close to 0 or 1, the normal approximation breaks down and an exact binomial-based interval is needed. VCE Methods stays inside the regime where the normal approximation is valid; an out-of-regime question would typically flag the issue.

Calculator workflow (Paper 2)

Most CAS systems have a built-in confidence-interval function for proportions, often as zInterval_1Prop or similar. The inputs are usually $x$ (number of successes), $n$ (sample size), and confidence level. The output is the interval.

For Paper 2 questions, the calculator output is acceptable, but always show the formula or set-up: $\hat{p}$, SE, $z^*$ at minimum.

Examples in context

Example 1. Customer satisfaction survey. A business surveys $400$ customers and finds $260$ are satisfied, so $\hat{p} = \frac{260}{400} = 0.65$. The standard error is $\sqrt{\frac{0.65 \times 0.35}{400}} = \sqrt{0.000569} = 0.02385$. For a $95\%$ interval, $z^* = 1.96$, giving margin $1.96 \times 0.02385 = 0.0467$. The interval is $(0.603, 0.697)$: across repeated samples, about $95\%$ of such intervals capture the true satisfaction rate.

Example 2. Designing a poll. A pollster wants a $95\%$ margin of error no more than $0.025$ and has no prior estimate of $p$, so uses the worst case $\hat{p} = 0.5$. Then $n \ge \frac{(1.96)^2(0.25)}{(0.025)^2} = \frac{0.9604}{0.000625} = 1536.6$, so at least $1537$ people must be surveyed.

Try this

Q1. A sample of $250$ has $\hat{p} = 0.4$. Find the standard error. [2 marks]

• Cue. $\sqrt{\frac{0.4 \times 0.6}{250}} = \sqrt{0.00096} = 0.0310$.

Q2. From $600$ items, $90$ are faulty. Construct a $95\%$ confidence interval for the true fault rate. [3 marks]

• Cue. $\hat{p} = 0.15$; SE $= \sqrt{\frac{0.15 \times 0.85}{600}} = 0.01458$; margin $1.96 \times 0.01458 = 0.0286$; interval $(0.121, 0.179)$.

Q3. State why "there is a 95% probability the true proportion lies in (0.121, 0.179)" is an incorrect interpretation. [2 marks]

• Cue. Once computed, the interval either contains $p$ or not; the $95\%$ refers to the long-run success rate of the procedure across many samples, not this single interval.