C4.1.15 · Skills · Worked example
Working through the numbers
1. Observed values (from the grid)
6 sites had both species, 1 had only Species A, 1 had only Species B, and 2 had neither. Row totals: A present = 7, A absent = 3. Column totals: B present = 7, B absent = 3. Grand total = 10 sites.
2. Expected values (if the species were independent)
Expected = (row total × column total) ÷ grand total, for each cell: both present = (7×7)/10 = 4.9; A only = (7×3)/10 = 2.1; B only = (3×7)/10 = 2.1; neither = (3×3)/10 = 0.9.
3. Chi-squared statistic
χ² = Σ (O − E)² ÷ E for all four cells:
(6 − 4.9)² ÷ 4.9 = 0.25 + (1 − 2.1)² ÷ 2.1 = 0.58 + (1 − 2.1)² ÷ 2.1 = 0.58 + (2 − 0.9)² ÷ 0.9 = 1.34
χ² = 2.75 (degrees of freedom = (2−1)×(2−1) = 1)
4. Conclusion
The critical value at p = 0.05 with 1 degree of freedom is 3.84 (data booklet). Because 2.75 is less than 3.84, this result is not statistically significant — this particular sample does not provide strong evidence of association between the two species, even though the raw counts might look suggestive at a glance. A real study would need a larger sample before drawing a conclusion either way.