IB Maths AI HL Hypothesis Testing for Population Parameters Paper 1 & 2 ~8 min read

Binomial Hypothesis Testing

Now the parameter being tested is a proportion p, not a mean — has the share of a population with some characteristic gone up or down? The test statistic is just a count (number of “successes”), modelled by X ∼ B(n, p). Two things make this page different from the mean tests: these are always one-tailed, and you’ll meet the critical region found with the inverse binomial function.

📘 What you need to know

What’s being tested?

Everything keys off a single binomial model: a sample of n members, each either having the characteristic (“success”) or not, with population proportion p. The count of successes x is your test statistic.

The model behind every binomial test X ∼ B(n, p)   where  x = number of successes in the sample Binomial probabilities come straight from your GDC ✓

🤔 Why are these always one-tailed?

The syllabus only asks you to test whether a proportion has gone in one stated direction — “more effective”, “fewer than claimed”. A two-tailed binomial test would need you to split the probability awkwardly across two discrete tails, so it’s simply left out of the course. If you ever see “increased” or “decreased”, you’re one-tailed by design.

The steps

🧭 Recipe — binomial hypothesis test

  1. Hypotheses: H0: p = p0, then H1: p < p0 or p > p0. Define p as the population proportion.
  2. Find the p-value (or the critical region) from X ∼ B(n, p0).
  3. Decide: p-value < significance level → reject; or test statistic in critical region → reject.
  4. Conclude in context, tentatively — the proportion has / has not increased or decreased.
The p-value — “at least as extreme as x” H1: p < p0 → p-value = P(Xx | p = p0)
H1: p > p0 → p-value = P(Xx | p = p0) Direction of the tail = direction of H₁ — match them!
Conclusion wording: reject → “sufficient evidence the proportion has increased / decreased”. Accept → “insufficient evidence the proportion has increased / decreased”. Always tie it back to the real-world characteristic in the question.

Critical value & critical region

Instead of a p-value, an exam may ask for the critical region: the set of “extreme” counts that would make you reject H0. The critical value c is the boundary, set by the significance level α%.

Critical region for a binomial test at α% H1: p < p0 → region Xc,  c = largest integer with P(Xc) ≤ α%
H1: p > p0 → region Xc,  c = smallest integer with P(Xc) ≤ α%
Finding the critical value with inverse binomial
Inverse binomial gives a candidate value (usually ONE off the real critical value) Check the boundary probability against α% does P(tail) stay ≤ α%? nudge c by 1 if not P(X ≤ c) ≤ α% ✓ keep c P(X ≤ c+1) > α% (confirms boundary) P(X ≥ c) ≤ α% ✓ keep c P(X ≥ c−1) > α% (confirms boundary) Test statistic in the region → reject H₀
Discrete distribution → the tail probability lands just under α%, never exactly on it. The boundary check is what nails the right integer.

🧠 “Inverse binomial is one off — so check the neighbour”

The inverse binomial rarely lands on the exact critical value. After it gives you a candidate, test P at that value and at the one next door. Keep the integer that pushes the tail probability just under α% — checking the neighbour confirms you haven’t overshot.

Worked examples

All five use Dr Sabir’s treatment data: the existing treatment is effective in 85% of cases; she tests a new treatment on 60 patients and it works for 57, at a 1% significance level.

WE 1

State the hypotheses

Dr Sabir claims her new treatment is more effective than the existing 85%. State H0 and H1.

let p = proportion of the population for which the new treatment is effective “more effective” → testing for an increase → one-tailed >. H₀: p = 0.85  H₁: p > 0.85
WE 2

Set up the model & p-value expression

Write the distribution of the test statistic and the probability you need for the test.

let X ~ B(60, p) = number of the 60 patients the treatment works for H₁ is p > 0.85, so the p-value is the upper tail at the observed x = 57. p-value = P(X ≥ 57 | p = 0.85)
WE 3

Calculate the p-value

Use the GDC to find the p-value for the test.

evaluate the upper-tail binomial probability P(X ≥ 57 | p = 0.85) = 1 − P(X ≤ 56) = 0.01483… p = 0.0148 (3sf)
WE 4

Decide & conclude in context

Perform the test at the 1% significance level and state the conclusion in context.

compare p with the significance level 0.0148 > 0.01 accept H₀ insufficient evidence to suggest the new treatment is more effective than the existing treatment.
WE 5

The critical-region method (cross-check)

Find the critical region for the same test at 1%, and confirm the conclusion using x = 57.

H₁: p > 0.85 → region X ≥ c, want smallest c with P(X ≥ c) ≤ 0.01 P(X ≥ 57) = 0.01483… > 0.01  → 57 too small P(X ≥ 58) = 0.00388… ≤ 0.01  ✓ critical region: X ≥ 58 57 is NOT in the region → accept H₀, same conclusion as the p-value method.

💡 Top tips

⚠ Common mistakes

Next up — Poisson Hypothesis Testing, the same one-tailed, discrete, inverse-function approach but for the mean number of occurrences in a time period. Watch the units: you’ll often scale the claimed rate to match the observation window before setting m0.

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