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C4.1.15—Chi-squared test

Chi-squared test explains how ecological evidence links organisms, resources and interactions to population size, distribution or community structure in a habitat.

Syllabus
First assessment 2025
Objective
C4.1.15
Level
HL

Exam analysis

Chance of appearing4%of analysed past papers
Latest appearanceMay 2022
Most common paperPaper2
Typical marks1

Common command terms

  • Identify
  • Outline
  • Determine
  • Explain
  • Calculate
  • State

Scoring notes

Common mistake
Using non-random sampling when the estimate requires representative quadrat positions.

Recent exam appearances

May 2022Paper2 ["HL"] · TZ23(d)[ 1 ]C4.1.15—Chi-squared test
May 2022Paper2 ["HL"] · TZ23(c)[ 1 ]C4.1.15—Chi-squared test
May 2022Paper2 ["HL"] · TZ23(b)[ 1 ]C4.1.15—Chi-squared test
November 2020Paper1 ["HL"] · TZ014[ 1 ]C4.1.15—Chi-squared test
May 2018Paper3 ["HL"] · TZ21(b)[ 3 ]C4.1.15—Chi-squared test

Coverage 2017–2022 · Updated 16 Jul 2026

Expected Counts Describe No Association

For two species scored present or absent in random quadrats, the null hypothesis H₀ states that their distributions are independent: there is no association between presence of one and presence of the other.

200 quadrats Bell present Bell absent Row total
ling present 89 45 134
ling absent 31 35 66
column total 120 80 200

E=row total×column totalgrand totalE=\frac{\text{row total}\times\text{column total}}{\text{grand total}}

Expected both present = (134 × 120) ÷ 200 = 80.4. Keep expected values unrounded during the calculation so rounding does not distort χ².

Chi-Squared Measures Departure from Independence

χ2=∑(O−E)2E\chi^2=\sum\frac{(O-E)^2}{E}

Cell O E (O−E)²/E
both present 89 80.4 0.920
ling present, bell absent 45 53.6 1.380
ling absent, bell present 31 39.6 1.868
both absent 35 26.4 2.802
sum 6.97

For a 2 × 2 table, df = (2−1)(2−1) = 1. At p = 0.05 the critical value is 3.84. Because 6.97 > 3.84, reject H₀: the species show a statistically significant association.

Association is not proof of competition or causation. The species may respond to the same abiotic factor or interact in another way; biological evidence must identify the mechanism.

Explain Community Change with Signs, Mechanism and Evidence

Question What a strong explanation identifies
Who interacts? populations and whether each gains or loses
What is transferred or limited? food, habitat, nutrients, protection, disease or a chemical
Where does change begin? individual pair, predator level, producer/resource level or introduced species
How does it propagate? competition, feedback, trophic cascade or niche restriction
What supports causation? manipulation, control, replication and a predicted response
What supports association? representative sampling, expected counts and a statistical decision

Use observations to find a pattern, statistics to judge whether it is unlikely under a null model, and experiments or field manipulations to test the proposed mechanism.

Community structure emerges from many population interactions. Naming a relationship is only the start; explanation requires its mechanism, direction of effect, feedback through the network and the limits of the evidence.

Chi-squared test

[ 3 ]

Outline how chi-squared can be used to test for an association between the distributions of the two species.

Concept essentials

  • Chi-squared test must be linked to the correct ecological unit, method or species interaction.
  • The evidence for chi-squared test depends on measurable abundance, distribution, survival or resource patterns.
  • Chi-squared test is clearest when cause, ecological process and population outcome are kept distinct.
  • Field or graph evidence for chi-squared test needs biological interpretation, not values alone.