Skip to main content

Sample proportions and confidence intervals: QCE Maths Methods Unit 4

Syllabus dot point

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

9 min read

Quick answer

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

Jump to a section
  1. What this dot point is asking
  2. Sample proportion
  3. Sampling distribution of p^\hat{p}
  4. Normal approximation
  5. Confidence intervals
  6. Interpretation
  7. Sample size design
  8. Trade-offs
  9. When the normal approximation fails
  10. Exam-style questions

What this dot point is asking

QCAA wants you to treat the sample proportion p^\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 pp of "successes" and a random sample of nn is drawn with XX successes, the sample proportion is:

p^=Xn\hat{p} = \frac{X}{n}

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

Sampling distribution of p^\hat{p}

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

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

Two takeaways:

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

Normal approximation

For large nn:

p^≈N(p,p(1−p)n)\hat{p} \approx N\left(p, \frac{p(1-p)}{n}\right)

Conditions for the approximation (QCAA convention):

  • np≥10n p \geq 10
  • n(1−p)≥10n(1-p) \geq 10

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

Confidence intervals

A confidence interval for a population proportion combines:

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

The formula:

p^±z∗p^(1−p^)n\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

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

Standard z∗z^* values

Level z∗z^*
90% 1.6449 (round to 1.645)
95% 1.9600 (round to 1.96)
99% 2.5758 (round to 2.58)

Interpretation

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

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

This is NOT:

  • "There is a C%C \% probability that pp is in this interval." (Once the interval is constructed, pp either is or is not in it.)
  • "C%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 EE at C%C \% confidence:

n≥(z∗)2p^(1−p^)E2n \geq \frac{(z^*)^2 \hat{p}(1-\hat{p})}{E^2}

If p^\hat{p} is unknown in advance, use p^=0.5\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=400n = 400 adults gives 120 who do.

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

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

For 95 percent: z∗=1.96z^* = 1.96. Margin: 1.96×0.0229≈0.04491.96 \times 0.0229 \approx 0.0449.

CI: (0.30−0.045,0.30+0.045)=(0.255,0.345)(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 nn.

Sample size economics. Doubling nn reduces SE by a factor of 2\sqrt{2}. Quadrupling nn halves SE. Diminishing returns above n≈1000n \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 (np≥10n p \geq 10 and n(1−p)≥10n(1-p) \geq 10) ensure validity. Outside these conditions, an exact (binomial-based) interval would be needed, beyond Methods scope.

Common errors
Wrong z∗z^* for the level
1.96 for 95 percent; 1.645 for 90 percent. Mixing them up gives wrong widths.
Probability misinterpretation
"The probability pp is in this interval" is wrong. Use long-run-procedure language.
Sample size not rounded up
n=384.1n = 384.1 becomes n=385n = 385, not 384.
pp vs p^\hat{p} in formulas
In the sampling distribution SD: use pp (when known). In the CI standard error: use p^\hat{p} (when pp unknown).
Forgetting worst-case p=0.5p = 0.5
When no prior estimate of pp exists for sample-size design, use p=0.5p = 0.5 to maximise p(1−p)=0.25p(1-p) = 0.25 for the most conservative nn.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks
A poll of 600 voters found 354 supported candidate X. (a) Compute a 95 percent confidence interval for the true proportion. (b) Interpret the interval. (c) Find the smallest sample size needed to achieve a 95 percent confidence interval of half-width 0.02, assuming the true proportion is around 0.5.
Show worked answer →

(a) Confidence interval.

p^=354/600=0.59\hat{p} = 354/600 = 0.59.

SE: 0.59×0.41/600=0.000403≈0.0201\sqrt{0.59 \times 0.41 / 600} = \sqrt{0.000403} \approx 0.0201.

z∗=1.96z^* = 1.96 for 95 percent.

Margin: 1.96×0.0201≈0.03941.96 \times 0.0201 \approx 0.0394.

CI: (0.59−0.0394,0.59+0.0394)=(0.5506,0.6294)(0.59 - 0.0394, 0.59 + 0.0394) = (0.5506, 0.6294), approximately (0.551,0.629)(0.551, 0.629).

(b) Interpretation. If many similar samples of 600 voters were drawn and 95 percent confidence intervals were constructed from each, approximately 95 percent of these intervals would contain the true population proportion of voters supporting candidate X. (Avoid: "there is a 95 percent probability the true proportion is in this interval", which is incorrect.)

(c) Sample size. Margin: z∗p(1−p)/n≤0.02z^* \sqrt{p(1-p)/n} \leq 0.02. With worst-case p=0.5p = 0.5:

1.960.25/n≤0.021.96 \sqrt{0.25 / n} \leq 0.02

0.25/n≤0.02/1.96≈0.0102\sqrt{0.25 / n} \leq 0.02 / 1.96 \approx 0.0102

0.25/n≤0.0001040.25 / n \leq 0.000104

n≥0.25/0.000104≈2401n \geq 0.25 / 0.000104 \approx 2401.

Smallest n=2401n = 2401.

Markers reward correct SE formula, correct z∗z^* for 95 percent, the correct interpretation language, and rounding up to integer for sample size.

Practise this

Sources & how we know this

ExamExplained