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3.7.9Additional lessons

Chi-Squared Test: Skills — IB Biology SL Revision Notes & Topical Questions

5 min read 6 parts Lesson 90 of 142

📚 IB Biology SL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 6Association Between Species

Positive association
  • Positive association Species found together more often than expected by chance, as when they depend on each other.
Negative association
  • Negative association Species found together less often than expected, as when they compete.
Independent
  • Independent If distributions are unrelated, any apparent pattern is due to chance.
Quadrat data
  • Quadrat data Random quadrats record whether each species is present or absent, giving counts in a table.

Part 2 of 6Hypotheses

Hypotheses
Null hypothesis
  • Null hypothesis There is no significant association between the distributions of the two species.
Alternative hypothesis
  • Alternative hypothesis There is a significant association, either positive or negative.
Aim
  • Aim The test shows whether differences between observed and expected values are likely to be due to chance.
Significance
  • Significance A probability below 0.05 (5%) is the usual level for rejecting the null hypothesis.

Part 3 of 6Carrying out the Test

The test compares what was observed with what chance would predict.

1
Record observed values (O) in a 2 × 2 table of presence and absence for both species.
2
Calculate each expected value: E = (row total × column total) ÷ grand total.
3
Calculate χ² = Σ (O − E)² ÷ E, adding the four values.
4
Find degrees of freedom: (rows − 1) × (columns − 1), which is 1 for a 2 × 2 table.
5
Compare χ² with the critical value at p = 0.05, which is 3.84 for one degree of freedom.

Part 4 of 6Interpreting the Result

Greater than critical value
  • Greater than critical value Reject the null hypothesis and conclude that the association is significant.
Less than critical value
  • Less than critical value Do not reject the null hypothesis, so the apparent association may be due to chance.
Direction
  • Direction Compare observed and expected values to see whether the association is positive or negative.
Limitations
  • Limitations A significant association does not show the cause, and low expected values make the test unreliable.

Part 5 of 6Worked Example

Question

  • In 100 quadrats, both species were present in 40, only A in 10, only B in 15 and neither in 35.
  • Test for association at p = 0.05.

  1. TotalsA present 50, A absent 50, B present 55, B absent 45.
  2. Expected valuesboth present 27.5, only A 22.5, only B 27.5, neither 22.5.
  3. χ² = (12.5² ÷ 27.5) + (12.5² ÷ 22.5) + (12.5² ÷ 27.5) + (12.5² ÷ 22.5) = 25.25.
  4. Degrees of freedom = 1, so the critical value is 3.84, and 25.25 is greater than 3.84.

Answer Reject the null hypothesis: there is a significant positive association.

Part 6 of 6Exam Checklist & Tips

3.7.9 – Key Points for the Exam

  • Chi-squared test A statistical test comparing observed and expected frequencies.
  • Null hypothesis A statement that there is no significant difference or association.
  • Degrees of freedom (rows − 1) × (columns − 1) in a contingency table.
  • Critical value The value of χ² that must be reached for significance at a chosen probability.

Common Mistakes

  • Using observed values instead of expected values when calculating E.
  • Using the wrong degrees of freedom.
  • Saying that a significant result proves one species causes the distribution of the other.
  • Claiming to prove the null hypothesis, when it can only be accepted as not rejected.

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