Loav · VCE Mathematical Methods · Units 3 and 4
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A confidence interval uses an observed sample proportion to estimate an unknown population proportion while acknowledging sampling variation. This guide develops the approximate normal interval, its assumptions, interpretation, margin of error and sample-size planning. It also explains why intervals change between samples and how simulation can reveal their long-run behaviour. You should understand a sample proportion and the normal distribution.
The parameter p is the fixed population proportion. The observed statistic p̂ = x/n changes when a new sample is taken. A single estimate gives a useful centre, but it does not show how much the estimate could vary under repeated sampling. The interval adds a model-based measure of uncertainty around that centre.
1. Build the approximate interval
For a large independent random sample, the sample proportion is approximately normal with mean p and standard deviation √[p(1 − p)/n]. In an actual estimation problem, p is unknown. Replace it in the spread formula with the observed sample proportion to obtain an estimated standard error.
p̂ = x/n
Estimated standard error = √[p̂(1 − p̂)/n]
Approximate confidence interval: p̂ ± z√[p̂(1 − p̂)/n]
The positive value z is selected so that the desired central probability lies between −z and z under the standard normal curve. For 95% confidence, each tail has area 0.025, so z is approximately 1.96. For 90% confidence z is approximately 1.645; for 99% confidence it is approximately 2.576.
The expression after ± is the margin of error m. The full interval width is 2m. Confusing the width with the margin doubles or halves the intended precision. Write the lower endpoint p̂ − m and upper endpoint p̂ + m separately, then label whether you are reporting proportions or percentages.
2. Conditions are part of the answer
The interval relies on an appropriate random sample and an independent-trial approximation. Sampling without replacement from a population very large relative to the sample can support that approximation. A voluntary website poll does not become representative merely because it has many responses. Non-response, coverage gaps and poorly defined questions can produce errors that this interval does not measure.
The sample should contain sufficiently many successes and failures for a normal approximation. A common course diagnostic is to check that np̂ and n(1 − p̂), the observed success and failure counts, both exceed five. This is a rule of thumb rather than a universal guarantee of accurate coverage. Very small samples or proportions near zero or one can make the displayed interval unreliable.
If p̂ equals zero or one, the formula produces a zero estimated standard error. That does not establish certainty about the population proportion; it signals a breakdown of this approximation. Similarly, endpoints outside [0,1] reveal that the simple formula is not behaving sensibly. Do not clip the endpoints and claim the original stated confidence properties remain verified.
3. Interpret confidence as a property of the procedure
Imagine repeatedly drawing samples in the same way and constructing an interval from each sample. If the assumptions and approximation work well, about 95% of the resulting 95% intervals will contain the fixed population proportion. The intervals vary; p does not.
For one completed interval, the unknown parameter either lies inside it or does not. In this repeated-sampling framework, it is inaccurate to describe p as randomly moving around with a 95% chance of falling inside those now-fixed endpoints. A suitable report is: “We are 95% confident that the population proportion is between these bounds,” supported by an explanation of the repeated-sampling meaning.
The interval does not say that 95% of individuals have values between its endpoints, nor that 95% of future sample proportions must lie inside this particular interval. It estimates a population proportion, not the range of individual binary responses. It also does not measure every possible error in how the sample was obtained.
Worked examples
Worked example 1: calculate and report a 95% interval
In an original random-sampling model, 232 of 400 sampled members of a large population report using a service. Construct an approximate 95% confidence interval for the population proportion p.
- The observed proportion is p̂ = 232/400 = 0.58. Successes and failures are 232 and 168, so the expected-count diagnostic using the sample estimate is comfortably satisfied.
- The estimated standard error is √[0.58(0.42)/400] ≈ 0.0246779.
- Using z = 1.96, the margin is approximately 0.0484.
- The interval is approximately (0.5316, 0.6284), or 53.16% to 62.84%.
Report: under the random-sampling and normal-approximation assumptions, we are 95% confident that between about 53.2% and 62.8% of the population uses the service. Across repeated samples, approximately 95% of intervals constructed this way would contain p.
The centre is the observed 58%, not a claim that the population proportion is exactly 58%. The margin is about 4.84 percentage points. It is not a relative error of 4.84% of 58%.
4. Confidence, precision and sample size
For the same observed sample and sample size, greater confidence requires a larger z and hence a wider interval. Covering the fixed parameter more often demands a less precise range. Holding z and the observed proportion fixed, a larger n reduces the margin in proportion to 1/√n.
m = z√[p̂(1 − p̂)/n]
For planning with an assumed proportion p0: n ≥ z2p0(1 − p0)/m2.
Conservative planning: use p0 = 0.5.
The planning expression comes from rearranging the margin formula; it is not a guarantee against sampling bias or model failure. If the population proportion is unknown, p(1 − p) ≤ 0.25 provides the largest spread and therefore a conservative choice. Round the required sample size upward because n must be an integer and rounding downward may violate the target.
Worked example 2: plan for a three-percentage-point margin
A planner wants a 95% interval with approximate margin no greater than 0.03 and has no reliable prior estimate of p. Use z = 1.96 and the conservative proportion 0.5.
- Require 1.96√(0.25/n) ≤ 0.03.
- Squaring and rearranging gives n ≥ 1.962(0.25)/0.032 ≈ 1067.1111.
- Round upward to n = 1068. The planned sample size must be at least 1068 under this approximation.
The target 0.03 is a margin, so the corresponding full width is 0.06. If the question instead specified total width 0.03, the margin would be 0.015 and the required sample would be much larger.
Worked example 3: compare two confidence levels
A sample of 400 has 200 successes, so p̂ = 0.5 and the estimated standard error is √(0.25/400) = 0.025. Compare approximate 95% and 99% intervals.
- At 95%, use z ≈ 1.96. The margin is about 0.0490, giving (0.4510,0.5490).
- At 99%, use z ≈ 2.575829. The margin is about 0.0644, giving (0.4356,0.5644).
- The approximate widths are 0.0980 and 0.1288. The 99% interval contains the 95% interval because both have the same centre but the former has a larger margin.
The sample itself did not become more informative when the confidence level changed. The reporting requirement changed. A narrower interval can be obtained with more data, lower confidence, or a different observed proportion, but those changes have different meanings.
5. Understand variation between intervals
Different random samples usually produce different x values and therefore different centres p̂. Their widths may also vary because p̂(1 − p̂) changes. For fixed n and z, intervals are widest when the observed proportion is near 0.5 and narrower near the boundaries. A narrow interval near zero is not automatically trustworthy if the normal approximation is poor.
In a simulation, choose a known p only so that coverage can be checked. Generate many independent binomial samples of size n, calculate p̂ for each, construct each interval, and record whether its lower endpoint is at most p and its upper endpoint is at least p. The fraction of intervals containing p estimates the procedure’s coverage in that simulation.
Do not count whether an interval contains its own p̂: every interval of this form is centred on p̂, so that check always succeeds and tells you nothing about coverage. Reset the sample count each repetition, retain the same true p during one experiment, and separate the sample size n from the number of simulated samples.
A run of 100 simulated intervals need not contain exactly 95 successful captures. The coverage count is itself random, and the interval formula is approximate. Increasing the number of simulations makes the estimated coverage more stable; increasing n changes the statistical procedure being investigated. Those are different experiments.
6. Read conclusions with the correct scope
An interval is useful evidence about a defined population at a defined time under a stated sampling process. It does not establish causation. It does not automatically generalise from one population to another. If a claimed proportion lies outside the interval, state that it lies outside the calculated range; do not draw a stronger conclusion than the calculated interval supports.
Keep full precision for p̂ and the square root until the last line. When reporting percentage endpoints, multiply each endpoint by 100. Use sensible rounding and state the confidence level beside the interval so the reader can interpret its width.
Practice and reasoned answers
1. A random sample has 238 successes in 500 trials. Find an approximate 95% confidence interval.
p̂ = 238/500 = 0.476. The estimated standard error is √[0.476(0.524)/500] ≈ 0.0223349. Using z ≈ 1.96 gives margin about 0.0438 and interval approximately (0.4322,0.5198). Report the random-sampling and large-sample assumptions with the interpretation.
2. There are 50 successes in 100 observations. Find an approximate 90% interval.
p̂ = 0.5 and estimated standard error = 0.05. With z ≈ 1.644854, the margin is about 0.08224 and the interval is (0.4178,0.5822). Using 1.96 would give a 95% interval rather than the requested 90% interval.
3. A report says a 95% interval means 95% of the population lies between its endpoints. Correct the interpretation.
The endpoints estimate a population proportion, not individual measurements. The 95% describes the long-run fraction of intervals from the same repeated-sampling procedure that contain the fixed population proportion. It does not describe where 95% of individual responses lie.
4. Use z = 1.96 to plan a conservative 95% margin of at most 0.04.
Use p0(1 − p0) = 0.25. Then n ≥ 1.962(0.25)/0.042 = 600.25, so the minimum planned integer is 601. Rounding to 600 would make the conservative margin slightly larger than the target.
5. A sample has 540 successes among 900 observations. Find its approximate 95% margin and full width.
p̂ = 0.6 and estimated standard error = √(0.24/900) ≈ 0.0163299. The margin is about 0.0320 and the full width is about 0.0640. The interval is approximately (0.5680,0.6320). Width is twice the distance from the centre to one endpoint.
6. Keeping confidence level and the planning proportion fixed, what change in n halves the margin of error?
Quadruple n. Since m is proportional to 1/√n, replacing n by 4n divides the margin by √4 = 2. Doubling n only divides it by √2. This is a planning comparison, so the assumed proportion is held fixed.
7. A sample has 280 successes among 400 observations. Find a 95% interval and say whether it contains 0.75.
p̂ = 0.7 and estimated standard error = √(0.21/400) ≈ 0.0229129. The interval is approximately (0.6551,0.7449). It does not contain 0.75. This statement describes the calculated interval; it does not establish that 0.75 is logically impossible as a population value.
Learning check
Construct an interval from a count, identify its centre and margin, and interpret confidence using repeated samples. Explain why larger samples do not fix bias and why higher confidence widens an interval. Finally, trace a simulation and identify the fixed true proportion, the varying sample proportions and the changing endpoints. These distinctions are the foundation of clear statistical reasoning.
Curriculum and study route
Loav Mathematical Methods Units 3 and 4. These module divisions are a teaching sequence. Use the current VCAA Mathematics study design, Mathematical Methods Units 3 and 4, Area of Study 4, and the VCAA Mathematical Methods FAQ for course requirements. Scope checked 15 October 2026. All worked scenarios, questions and explanations in this guide are original.
