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Further Mathematics

Poisson & Geometric Hypothesis Testing

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  Edexcel A Level Further Maths · Further Statistics 1

Poisson & Geometric
Hypothesis Testing

The Big Idea: When you suspect a Poisson rate (λ) or a Geometric success probability (p) has changed, you assume it hasn't changed (H₀), then ask: "if H₀ were true, how likely was I to see a result this extreme — or even more extreme?" If that probability is smaller than your significance level, the result was too surprising to be a coincidence, so you reject H₀.

Summary — What This Chapter Covers

  • Hypothesis testing is a formal way of checking whether an observed value is consistent with a claimed parameter (λ for Poisson, p for Geometric), or whether it's suspiciously extreme.
  • Both tests follow the exact same 5-step skeleton: hypotheses → rescale if needed → find the tail probability → compare to α → conclude in context.
  • A critical region is a pre-calculated set of "surprising" values of X. If your observed X falls inside it, you automatically reject H₀ — no probability calculation needed at that point.
  • Because Poisson and Geometric are discrete distributions, you can almost never hit α exactly. The probability you actually land on is called the actual significance level.
  • For Geometric tests, the inequality direction flips between H₁ and the critical region — this is the single most common source of exam errors.

Topic 1 — Poisson Hypothesis Testing

Setting Up the Test

Imagine you run a small shop, and you know from years of data that you get an average of 3 customer complaints per week. That's your Poisson mean, λ = 3. One week, after training your staff differently, you get 7 complaints. Is that just a bad-luck week, or has something actually changed?

This is exactly the question Poisson hypothesis testing answers. If X ~ Po(λ), we test the mean λ using:

Hypotheses
H₀: λ = (claimed value)     H₁: λ ≠ … or λ < … or λ > …
H₀ is always "nothing has changed." H₁ is what you're trying to find evidence for. Choose ≠ for "has it changed at all", < for "has it decreased", > for "has it increased" — this comes from how the question is phrased, not from the data.

You'll always be given an observed value, x, and a significance level, α (commonly 5% = 0.05, sometimes 1% or 10%). The whole test then boils down to one question:

"Assuming H₀ is true, what's the probability of getting a result this extreme, or more extreme?"

If that probability is small (less than α, or α/2 for a two-tailed test), your observed result was too unlikely to be explained by chance alone — so you reject H₀. If the probability isn't that small, the result is still plausible under H₀, so you don't reject H₀.

Watch the Time/Space Interval — Rescaling λ

Critical Detail
A Poisson mean is only valid over the interval it was measured on. If λ = 8 per day but your observed value x is over a 12-hour period, you cannot compare them directly — you must rescale λ to match x's interval first. Here, 8 likes per 24 hours becomes 4 likes per 12 hours, so you actually test X ~ Po(4), not Po(8).
Worked Example

Mr Viajo believes his travel blog receives an average of 8 likes per day (24 hours). After a new advertising campaign, he tests at the 5% significance level whether this has changed. Over a randomly chosen 12-hour period, the blog receives 7 likes.

(i) Hypotheses. Let λ be the average number of likes per day.
H₀: λ = 8    H₁: λ ≠ 8  (two-tailed, since we're testing for "a change" in either direction)
(ii) Critical regions. First rescale: 8 likes per 24 hrs = 4 likes per 12 hrs. So for the 12-hour window, X ~ Po(4). Since this is two-tailed at 5%, we split into 2.5% in each tail.

Lower tail — want P(X ≤ c) as close to 2.5% as possible without exceeding it:
P(X = 0) = 0.0183 < 2.5% ✓    P(X ≤ 1) = 0.0916 > 2.5% ✗
→ Critical region (lower): X = 0

Upper tail — want P(X ≥ c) as close to 2.5% as possible without exceeding it:
P(X ≥ 9) = 1 − P(X ≤ 8) = 1 − 0.9786 = 0.0214 < 2.5% ✓
P(X ≥ 8) = 1 − P(X ≤ 7) = 1 − 0.9489 = 0.0511 > 2.5% ✗
→ Critical region (upper): X ≥ 9

Full critical region: X = 0 or X ≥ 9

(iii) Actual significance level. Add the two actual tail probabilities: 0.0183 + 0.0214 = 0.0397 (3.97%) — close to but under the target 5%.
(iv) Conclusion. The observed value is X = 7. Since 7 is not in the critical region (X = 0 or X ≥ 9), there is insufficient evidence to reject H₀. The mean number of likes per 12-hour period has not changed from 4 (equivalently, from 8 per day).

Finding the Critical Region

A critical region is just the set of "extreme enough to reject H₀" values, worked out in advance — before you even see the observed data. Once you have it, testing is instant: is x inside the region or not?

One-tailed, H₁: λ < …
Find the largest c such that P(X ≤ c) ≤ α
Test integers going up. The critical region is X ≤ c. Checking c + 1 should give a probability that exceeds α — this confirms you've found the right boundary.
One-tailed, H₁: λ > …
Find the smallest c such that P(X ≥ c) ≤ α
Same idea, but watch out: tables usually give P(X ≤ ...), so P(X ≥ c) = 1 − P(X ≤ c − 1). It's very easy to accidentally find c − 1 instead of c here — always double check which one your calculation actually represents.
Two-tailed, H₁: λ ≠ …
Critical region is X ≤ c₁ or X ≥ c₂
Split α into α/2 in each tail. Find c₁ so P(X ≤ c₁) is as close to α/2 as possible without exceeding it, and c₂ so P(X ≥ c₂) is as close to α/2 as possible without exceeding it.

Actual Significance Level

Because Poisson only takes whole-number values (0, 1, 2, 3, ...), you can never land exactly on α — there's always a "gap" between your critical value and the target. Whatever probability you actually end up with in the tail(s) is called the actual significance level.

What it means
The actual significance level is the true probability of wrongly rejecting H₀ when H₀ is in fact correct (a Type I error). It's normally slightly less than α — but some exam questions instead ask for the critical region that's "as close as possible to α", which might mean going slightly over.

Example: if P(X ≤ 6) = 0.0298 and P(X ≤ 7) = 0.0510, with α = 0.05 — normally X ≤ 6 is chosen (doesn't exceed α). But if the question asks for the region "as close as possible to α", you'd pick X ≤ 7, giving an actual significance level of 0.0510.

Practice Question 1

A call centre receives an average of 5 complaints per hour. Following a new training scheme, the manager tests at the 5% significance level whether the mean has decreased. In a randomly chosen hour, only 1 complaint is received. (a) State the hypotheses. (b) Find the critical region. (c) State the conclusion.

Practice Question 2

A website usually gets 6 sign-ups per day. After a redesign, 11 sign-ups occur in one day. Test at the 5% level whether the mean has increased, using the critical region method.

Topic 2 — Geometric Hypothesis Testing

Setting Up the Test

Picture flipping what's supposed to be a fair coin (p = ½) and counting how many flips it takes to get your first tail. If it takes 5 flips, is the coin still fair, or is it biased? Geometric hypothesis testing answers exactly this kind of question — but instead of counting how many events happen in an interval (Poisson), we're counting how many trials it takes to get the first success.

Hypotheses
H₀: p = (claimed value)     H₁: p ≠ … or p < … or p > …
If X ~ Geo(p), you're testing the probability of success, p, per trial. The observed value x is the number of trials up to and including the first success.
Expected number of trials
E(X) = 1/p
This is your intuition-check. A fair coin (p = ½) should take on average 1/(½) = 2 flips to see the first tail. If your observed x is much bigger than 1/p, that suggests p is actually smaller than claimed (harder to succeed, so more attempts needed). If x is much smaller than 1/p, that suggests p is actually bigger than claimed.

The Flip Rule — Most Important Idea in This Topic

The #1 Exam Trap
For Geometric distributions, a lower success probability p means it takes more trials, on average, to succeed. A higher p means it takes fewer trials. This creates an inequality flip between H₁ and the "extreme values":
H₁ says……extreme (surprising) values of X are…
p < claimed valueX ≥ x (took unusually many trials)
p > claimed valueX ≤ x (took unusually few trials)
This is the opposite direction to what students instinctively expect (and opposite to how Poisson works), so it's the single biggest source of dropped marks in this topic. Always pause and ask: "does a lower p mean more or fewer trials?" before writing your inequality.
Two-tailed: which side is extreme?
Compare x with E(X) = 1/p
If x < 1/p → extreme values are X ≤ x.   If x > 1/p → extreme values are X ≥ x. Then check whether that one-sided probability is less than α/2.

Finding the Critical Region

Just like Poisson, you can pre-calculate the critical region — but the useful formulas here are slightly different, and based on the geometric tail probabilities:

Key formulas
P(X ≥ c) = (1 − p)^(c − 1)     P(X ≤ c) = 1 − (1 − p)^c
These let you solve directly for c using logs, instead of testing integers one by one (though testing integers still works and is often safer).
If H₁: p < …
Critical region is X ≥ c, where P(X ≥ c) = (1 − p)^(c − 1) is as close to α as possible without exceeding it.
Note the inequality is opposite to the direction in H₁ — this is the flip rule again.
If H₁: p > …
Critical region is X ≤ c, where P(X ≤ c) = 1 − (1 − p)^c is as close to α as possible without exceeding it.
Using logs safely
When solving inequalities like (1−p)^(c−1) ≤ α by taking logs of both sides, remember log(1 − p) is negative (since 0 < 1 − p < 1). Dividing both sides of an inequality by a negative number flips the inequality sign — a very easy step to forget under exam pressure.
Worked Example

Palamedes builds a spinner numbered 1–40 and claims P(landing on '1') = 1/40 exactly. Odysseus is suspicious and runs a two-tailed test at the 10% level (5% in each tail), counting how many spins it takes to land on '1' for the first time.

(a) Hypotheses. H₀: p = 1/40    H₁: p ≠ 1/40
(b) Critical regions. Let X ≤ c₁ or X ≥ c₂ be the critical region, with P(X ≤ c₁) = 0.05 and P(X ≥ c₂) = 0.05.

Lower tail: solve 1 − (39/40)^c₁ = 0.05 → (39/40)^c₁ = 0.95 → c₁ = log(0.95)/log(39/40) = 2.026
Testing c₁ = 2: P(X ≤ 2) = 0.04938 (under 0.05, good)    c₁ = 3: P(X ≤ 3) = 0.0731 (over)
c₁ = 2 is the critical value.

Upper tail: solve (39/40)^(c₂ − 1) = 0.05 → c₂ − 1 = log(0.05)/log(39/40) = 118.325 → c₂ = 119.325
Testing c₂ = 119: P(X ≥ 119) = 0.05041 (over, but closer)    c₂ = 120: P(X ≥ 120) = 0.04915 (under)
c₂ = 119 is the critical value (chosen since the question asks for "as close as possible", even though it slightly exceeds 5%).

Critical region: X ≤ 2 or X ≥ 119

(c) Actual significance level. P(X ≤ 2) + P(X ≥ 119) = 0.049375 + 0.050413 = 0.0998 (to 3 s.f.) — very close to the target 10%.
(d) Conclusion. The spinner lands on '1' on the very first spin, so X = 1. Since 1 is inside the critical region (X ≤ 2), there is sufficient evidence to reject H₀ — Palamedes' claim that p = 1/40 is not supported.
Practice Question 3

A gardener claims that 1 in 5 seeds of a certain flower germinate (p = 0.2). A friend is suspicious that the true germination probability is lower. They plant seeds one at a time and record that it takes 9 seeds to get the first one to germinate. Test at the 5% significance level.

Practice Question 4

A basketball player claims her free-throw success probability is p = 0.6. A coach suspects it's actually higher. In a session, she scores on her very first attempt. Using the critical region approach at the 5% level, is there sufficient evidence to support the coach's suspicion? (Hint: find the critical region X ≤ c first.)

What to Memorise

ConceptPoisson X ~ Po(λ)Geometric X ~ Geo(p)
Parameter testedMean rate, λSuccess probability, p
Observed value x meansCount of events in an intervalTrial number of first success
Sanity checkRescale λ to match x's intervalCompare x with E(X) = 1/p
H₁: parameter < claim → extreme XX ≤ x (fewer events)X ≥ x (more trials needed) — flipped!
H₁: parameter > claim → extreme XX ≥ x (more events)X ≤ x (fewer trials needed) — flipped!
Critical region formulaUse cumulative Poisson tables/calculatorP(X≥c)=(1−p)^(c−1), P(X≤c)=1−(1−p)^c
Two-tailed splitα/2 in each tailα/2 in each tail
The 5-Step Skeleton (memorise this order)
1. Write H₀ and H₁ (with correct inequality direction) and state α.
2. Rescale if needed (Poisson only) or compute E(X) = 1/p (Geometric only).
3. Find P(X as extreme as observed, or more extreme) under H₀.
4. Compare to α (or α/2 for two-tailed). Smaller → reject H₀.
5. Write the conclusion in context, in words, referring back to the original scenario.

Concepts Checklist

Exam Tips & Common Mistakes

  • Forgetting to rescale λ. If the question gives a rate over one interval (e.g. per day) but the observed data is over a different interval (e.g. per 12 hours), you must rescale λ before doing anything else. Marking schemes award a method mark just for this step.
  • Mixing up P(X ≥ c) and P(X ≤ c − 1). Cumulative tables/calculators usually only give P(X ≤ k) directly. Remember P(X ≥ c) = 1 − P(X ≤ c − 1). Students very often accidentally compute 1 − P(X ≤ c) instead, which is actually P(X ≥ c + 1) — off by one.
  • The Geometric "flip rule" gets ignored. Students instinctively write X ≤ x for H₁: p < ... (copying the Poisson pattern), when actually a lower p needs more attempts, so the extreme region is X ≥ x. Always physically reason it through: "lower success chance → more tries needed → higher X is surprising."
  • Exceeding α without realising it. When finding critical values, you must choose the value that does not exceed α (or α/2), unless the question specifically says "as close as possible even if it means exceeding α" — read the question wording carefully every time.
  • Not stating a conclusion in context. "Reject H₀" alone loses marks. You must translate it back into the scenario: e.g. "there is sufficient evidence that the mean number of accidents per hour has increased."
  • Forgetting to split α for two-tailed tests. Two-tailed tests always use α/2 in each tail, not α in each tail — a very common slip that inflates the critical region incorrectly.
  • Sign errors when using logs. Since 0 < 1 − p < 1, log(1 − p) is negative. Dividing an inequality by a negative number flips its direction — easy to forget mid-calculation.
What Examiners Actually Look For
Clear, labelled hypotheses (with a defined parameter in words) · correct rescaling shown as a step · the specific probability calculation shown (not just the final number) · explicit comparison against α or α/2 · a conclusion written in full sentences, in context, referencing the original scenario — not just "reject H₀".
Also in the full note
  • Exam Tips & Common Mistakes
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