Lesson 2: Inference Procedures and Their ConditionsAP Deep Dive

Inference questions are won or lost before any computation: in the choice of procedure and the verification of its conditions. This lesson is a clinic on both — how to recognise paired data disguised as two samples, how confidence intervals and tests differ in their standard errors, how to *plan* a sample size, and how power actually works with real numbers.

Paired versus two-sample: the recognition skill

Worked example 1. Nine cars each drive a test route twice — once on standard fuel, once on a premium blend — and the differences (premium − standard, mpg) have mean 2.4 mpg and standard deviation 1.2 mpg. Test whether premium improves economy.

This is one sample of differences: a paired t procedure. t = 2.4/(1.2/√9) = 2.4/0.4 = 6.0 with df = 8, giving p < 0.001 — convincing evidence of improvement. Now see the trap: analysed wrongly as two independent samples of nine, the car-to-car variability (some cars are simply thirstier) inflates the standard error and the effect drowns. Pairing *removes* the between-car variation by differencing first. Recognition cues: the same subjects measured twice, matched pairs, twins, before/after — in every case, the data arrive as pairs and the analysis must be on the differences. The conditions are checked on the differences too: it is the distribution of the *differences* that must be roughly normal (or n large), not the two original distributions.

Planning a sample size

Worked example 2. How large a sample guarantees a margin of error of at most 0.03 for a proportion at 95% confidence? Solve z*√(p(1−p)/n) ≤ 0.03 for n: n ≥ (1.96/0.03)² × p(1−p). With no prior estimate, use the conservative p = 0.5 (maximising p(1−p)): n ≥ 4268 × 0.25 = 1067.1, so n = 1068 (always round *up* — rounding down silently breaks the guarantee). If a pilot study suggests p ≈ 0.3, then p(1−p) = 0.21 and n = 897 suffices. Two examinable ideas: the margin of error shrinks like 1/√n (halving it costs 4× the sample), and the conservative choice costs sample size but buys certainty. For means the parallel formula is n ≥ (z*σ/ME)², with σ from prior data.

Power, computed once for real

Power is the probability a test rejects H₀ when a specific alternative is true. Worked example 3. Test H₀: p = 0.50 versus Hₐ: p > 0.50 at α = 0.05 with n = 100. Under H₀, p̂ has SD √(0.25/100) = 0.05, so we reject when p̂ > 0.50 + 1.645(0.05) = 0.582. Now suppose the truth is p = 0.60: p̂ is centred at 0.60 with SD √(0.6 × 0.4/100) = 0.049. Power = P(p̂ > 0.582 | p = 0.60) = P(z > (0.582 − 0.60)/0.049) = P(z > −0.37) ≈ 0.64. The Type II error probability is β ≈ 0.36 — with 100 observations, a real 10-point advantage escapes detection more than a third of the time. Power rises when the true value moves farther from H₀, when n grows, or when α is relaxed; it falls as α is tightened. Any "how could the researchers increase power?" question is answered from that list — and "increase the sample size" is the answer that costs money rather than error rate.

Two subtle asymmetries

Pooled versus unpooled standard errors. In a two-proportion *test*, H₀ asserts p₁ = p₂, so the standard error uses the pooled p̂ — the null's own assumption builds the yardstick. In a two-proportion *interval*, no equality is assumed, so the standard error uses each p̂ separately. Using the pooled SE in an interval (or unpooled in the test) is a recognised top-band error. Tests and intervals are dual: a two-sided test at level α rejects exactly the null values outside the (1 − α) confidence interval — if a 95% interval for a difference excludes 0, the two-sided test at α = 0.05 rejects. Questions that give you one and ask about the other are testing this duality, not new computation.

Conditions, stated with evidence

Name the procedure, then verify each condition from the problem's text — "Random: the seeds were randomly assigned to treatments" is a verification; "Random: met" is not. The recurring sets: proportions need success–failure counts (using p₀ for tests, p̂ for intervals) of at least 10 in each group; means need a graph check for skew/outliers or n ≥ 30 (and for paired data, the *differences* graphed); chi-square needs all *expected* counts ≥ 5; regression slope inference needs the LINE conditions — linear form, independent observations, roughly normal residuals with constant spread, random sampling/assignment. A condition that fails has consequences you should be able to name: non-random samples destroy generalisation; heavy skew with small n makes the stated p-value untrustworthy.

Common mistakes at the top band

• Analysing paired data as two independent samples (Work Example 1's trap), or pairing data that were never matched.
• Checking chi-square conditions on observed counts instead of expected counts.
• Writing "the probability that H₀ is true is 0.03" for p = 0.03. The p-value conditions on H₀ being true; it is never a probability *about* H₀.
• Rounding a required sample size down, or using p̂ from the data being planned (circularity) instead of a pilot or 0.5.
• Claiming a nonsignificant result proves no effect — with power 0.64 (Example 3), absence of evidence is weak evidence of absence.

AP Exam Experience

1. A researcher measures the blood pressure of 30 patients before and after a treatment programme and wants to test whether the mean change differs from zero. The appropriate procedure is a

• (A) two-sample t test
• (B) one-sample t test on the differences (paired t)
• (C) two-proportion z test
• (D) chi-square test of independence

Answer: (B). The same patients measured twice produce paired data; the analysis runs on the 30 differences. Using (A) discards the pairing and lets patient-to-patient variability swamp the treatment effect.

2. To halve the margin of error of a confidence interval (same confidence level), the sample size must be

• (A) doubled
• (B) multiplied by √2
• (C) multiplied by 4
• (D) multiplied by 2, then increased by the critical value

Answer: (C). Margin of error scales as 1/√n. Halving it requires √n to double, i.e. n to quadruple. This inverse-square relationship governs the cost of all precision in statistics.

3. A test of H₀: p = 0.5 against p > 0.5 has power 0.64 when the true p = 0.6. Which change would increase the power?

• (A) Decreasing α from 0.05 to 0.01
• (B) Using a sample of 400 instead of 100
• (C) Testing against the alternative p < 0.5 instead
• (D) Requiring stronger evidence before rejecting

Answer: (B). Larger n shrinks the standard error, separating the null and alternative distributions so the rejection region captures more of the alternative. (A) and (D) are the same change described twice — a stricter α moves the threshold up and *reduces* power; (C) points the test away from the truth entirely.

Free-response question. A seed company claims its treated seeds germinate at a higher rate. In a controlled trial, 224 of 400 treated seeds and 192 of 400 untreated seeds germinate. The seeds were randomly assigned to treatment or control.

(a) State appropriate hypotheses for testing the company's claim. (b) Verify the conditions for a two-proportion z test, citing evidence from the problem. (c) Calculate the test statistic and p-value. (d) State the conclusion in context, and describe the scope of conclusions the design supports.

Scoring guideline.

• (a) 1 point. H₀: p_treated = p_control (or p_t − p_c = 0) versus Hₐ: p_treated > p_control, with the parameters defined in context (true germination proportions for treated and untreated seeds of this type). A two-sided alternative loses the point — the claim being tested is directional, and the hypotheses must match it.
• (b) 2 points. 1 point for randomisation/independence: seeds were randomly assigned, and each group's seeds are independent of the other's (accept reference to the 10% condition if framed correctly). 1 point for the success–failure condition with *numbers*: successes and failures are 224, 176, 192, 208 — all ≥ 10, so the sampling distribution of p̂_t − p̂_c is approximately normal. Writing "conditions are met" without the counts earns neither point.
• (c) 2 points. 1 point for the pooled proportion and standard error: p̂_pooled = 416/800 = 0.52; SE = √(0.52 × 0.48 × (1/400 + 1/400)) = 0.0353. Pooling is required because the test's null asserts a common proportion — using the unpooled SE forfeits this point. 1 point for z = (0.56 − 0.48)/0.0353 = 2.26 with p-value ≈ 0.012 (one-sided; accept 0.011–0.013).
• (d) 2 points. 1 point for the linked conclusion: since p (0.012) < α = 0.05, reject H₀; there is convincing evidence that the treated seeds have a higher true germination proportion than untreated seeds. 1 point for scope: because treatment was randomly assigned, a causal conclusion is supported (the treatment caused the increase, for seeds like these); generalisation extends to the population of seeds the sample represents, not to all seed species. Causal language without invoking random assignment earns no scope point.

Choosing α: errors have prices, and the prices differ

Every test trades two errors: Type I (rejecting a true H₀, probability α) and Type II (failing to reject a false H₀, probability β). Which error is worse is a *subject-matter* question, and FRQs ask it directly. Screening blood donations: a Type II error (missing a contaminated unit) endangers a patient, while a Type I error (discarding a clean unit) costs a donation — so the screening is tuned for power, tolerating false alarms. Approving a new drug: a Type I error (declaring an ineffective drug effective) exposes the public to harm and false hope, so regulators demand small α and strong evidence. Notice the practical lever: α is chosen in advance (0.05 by convention, 0.01 when Type I is grave), but β is controlled mainly through sample size — a study that cannot afford Type II errors must buy observations. A related discipline: the direction of Hₐ (one-sided versus two-sided) must be fixed by the research question *before* seeing data. Choosing a one-sided alternative after noticing the sample difference doubles the effective Type I error rate while wearing the costume of α = 0.05, and rubrics for the germination FRQ in this lesson deduct for exactly that mismatch between claim and alternative.