IB Maths AI HLHypothesis Testing for Population ParametersPaper 1 & 2~8 min read
Poisson Hypothesis Testing
Same machinery as the binomial test — one-tailed, discrete, inverse-function critical regions — but now the parameter is the mean number of occurrencesm in a fixed time period, modelled by X ∼ Po(m). The one extra trap that catches everyone: the claimed rate is usually quoted over a different window than the observation, so you must scale it to set m0 before doing anything else.
📘 What you need to know
Tests a mean ratem: has the mean number of occurrences per time period increased or decreased?
Always one-tailed — no two-tailed Poisson tests in this course.
Test statisticx = number of occurrences observed, modelled by X ∼ Po(m).
Scale the rate to the observation window first — e.g. 120/hour becomes m0 = 20 per 10 minutes.
Hypotheses: H0: m = m0; H1: m < m0 or m > m0.
Decision: p-value < significance level → reject; or test statistic in critical region → reject. Conclude in context.
What’s being tested?
You’re checking whether the underlying rate has shifted. A sample of one time period is observed, the count of occurrences x is the test statistic, and it’s modelled by a Poisson distribution with the claimed mean.
The model behind every Poisson testX ∼ Po(m)
wherex = number of occurrences in the time period
Poisson probabilities come straight from your GDC ✓
🧠 Scale the rate to the window — every time
State the time period for m explicitly. If the claim is “120 hits per hour” but you observe a 10-minute window, then m0 = 120 × 1060 = 20. The null hypothesis must use the mean for the observed period, not the headline rate.
The steps
🧭 Recipe — Poisson hypothesis test
Scale & hypotheses: find m0 for the observed window, then H0: m = m0, H1: m < m0 or m > m0. Define m and its time period.
Find the p-value (or critical region) from X ∼ Po(m0).
Decide: p-value < significance level → reject; or test statistic in critical region → reject.
Conclude in context, tentatively — the rate has / has not increased or decreased.
The p-value — “at least as extreme as x”H1: m < m0 → p-value = P(X ≤ x | m = m0)
H1: m > m0 → p-value = P(X ≥ x | m = m0)
Direction of the tail = direction of H₁ — match them!
Conclusion wording: reject → “sufficient evidence the mean number of occurrences has increased / decreased”. Accept → “insufficient evidence it has increased / decreased”. Always name the real-world quantity (hits, calls, faults) and its time period.
Critical value & critical region
Same discrete logic as the binomial. The critical value c is the boundary set by α%, found with the inverse Poisson — which again lands about one off, so you check the neighbour.
Critical region for a Poisson test at α%H1: m < m0 → region X ≤ c, c = largest integer with P(X ≤ c) ≤ α%
H1: m > m0 → region X ≥ c, c = smallest integer with P(X ≥ c) ≤ α%
Building the critical region (lower-tail example)
Discrete distribution → tail probability sits just under α%, never exactly on it. The neighbour-check pins the right integer.
🤔 Why ≤ α% and not = α% for the critical region?
Because the Poisson is discrete, you can’t fine-tune the tail to land exactly on α%. You take the most extreme integer that keeps the tail probability at or below α%. This is also why, for discrete tests, the true probability of a Type I error is ≤ α% rather than equal to it.
Worked examples
All five use the website data: the owner claims 120 hits per hour; a purchaser believes it’s fewer; a 10-minute window is observed with 11 hits, at a 5% significance level.
WE 1
Scale the rate & state hypotheses
The purchaser thinks the site gets fewer hits than claimed. Set up m0 for the observed period and state the hypotheses.
let m = mean number of hits in a 10-minute period120 per hour → 120 × (10/60) = 20 per 10 min“fewer hits” → testing for a decrease → one-tailed <.H₀: m = 20 H₁: m < 20
WE 2
Set up the model & critical-region condition
Write the distribution of the test statistic and the condition the critical value must satisfy at 5%.
let X ~ Po(m) = number of hits in a 10-minute periodH₁ is m < 20 → lower tail. c = largest integer with P(X ≤ c | m = 20) ≤ 0.05.region X ≤ c, want P(X ≤ c) ≤ 0.05
WE 3
Find the critical region
Use the inverse Poisson, then check the boundary to find the critical region.
test candidate values around the inverse-Poisson resultP(X ≤ 13 | m = 20) = 0.0661… > 0.05 → too big, reduceP(X ≤ 12 | m = 20) = 0.0390… ≤ 0.05 ✓critical region: X ≤ 12
WE 4
Perform the test & conclude
The site received 11 hits in the 10-minute window. Carry out the test at 5% and conclude in context.
compare the test statistic with the critical region11 < 12 → 11 is in the critical region X ≤ 12reject H₀sufficient evidence the website receives on average fewer hits than the owner claims.
WE 5
The p-value method (cross-check)
Confirm the same conclusion using a p-value instead of the critical region.
H₁: m < 20 → lower-tail p-value at x = 11p = P(X ≤ 11 | m = 20) = 0.0214…0.0214 < 0.05reject H₀same conclusion — fewer hits than claimed.
💡 Top tips
Scale the rate first and state the time period for m — most lost marks happen here.
“Fewer / decreased” → ≤ lower tail; “more / increased” → ≥ upper tail. Match the tail to H1.
Always one-tailed — never split into two tails for the Poisson.
Inverse Poisson is one off — test the boundary integer and its neighbour.
Both methods agree: critical region and p-value give the same decision; use one to check the other.
Same decision rule: p < α → reject, or test statistic inside the region → reject.
⚠ Common mistakes
Forgetting to scale — using m0 = 120 when the window is 10 minutes (should be 20).
Wrong tail — using P(X ≥ x) for a “fewer” test instead of P(X ≤ x).
Trusting inverse Poisson blindly — not checking the neighbouring integer for the critical value.
Attempting a two-tailed test — not required and easy to set up wrongly.
Comparing the wrong way — reject when p < α, not when p > α.
Vague conclusions — not naming the occurrence (hits) or its time period, and not staying tentative.
Next up — Hypothesis Testing for Correlation, where a t-test checks for linear correlation between two variables. The parameter is the population correlation coefficient ρ, the hypotheses centre on ρ = 0, and this one can be two-tailed when you’re testing for “any” correlation.
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