A2 · A2 3 Practical Skills in A2 Biology
Use chi-squared when you have counted categorical data and want to know whether observed numbers differ significantly from expected ones. Calculate χ² = Σ (O − E)² / E, work out degrees of freedom as (number of categories − 1), and compare your value with the critical value at p = 0.05.
A dihybrid cross gives 160 offspring, expected in a 9:3:3:1 ratio: 90, 30, 30, 10. Observed: 100, 26, 28, 6. The contributions are (100−90)²/90 = 1.11, (26−30)²/30 = 0.53, (28−30)²/30 = 0.13, (6−10)²/10 = 1.60, giving χ² = 3.37.
With 3 degrees of freedom the critical value at p = 0.05 is 7.82. Since 3.37 is less than 7.82, the difference is not significant: we accept the null hypothesis and the results are consistent with a 9:3:3:1 ratio, with any deviation due to chance.
| Question | Test |
|---|---|
| Do counts fit an expected ratio or association? | Chi-squared |
| Is there a difference between two means? | t-test |
| Is there a relationship between two variables? | Correlation (Spearman's rank or Pearson's) |
Write your answer, then see it marked against every mark scheme point before the model answer is revealed.
A2 2 / A2 3 statistics question — recurring in every A2 2 series · 5 marks
These questions are written in the style of the CCEA GCE Biology (2016) papers and are not reproduced from any live paper. CCEA owns the copyright in its question papers and mark schemes — always download the official paper and mark scheme from CCEA and use these worked answers alongside them. BioCCEA is an independent study tool and is not endorsed by CCEA.
Two A2-standard questions in CCEA paper style, each with the full mark scheme, a full-mark answer written the way you should write it, and the point where most candidates drop marks.
A2 2 / A2 3 statistics question — recurring in every A2 2 series · 5 marks
Expected numbers for 260 offspring in a 9:3:3:1 ratio are 146.25, 48.75, 48.75 and 16.25. Using χ² = Σ (O − E)² / E: (152 − 146.25)²/146.25 = 0.23; (39 − 48.75)²/48.75 = 1.95; (53 − 48.75)²/48.75 = 0.37; (16 − 16.25)²/16.25 = 0.004. χ² = 2.55. Degrees of freedom = 4 − 1 = 3, and the critical value at p = 0.05 is 7.82. Because 2.55 is less than 7.82, the difference between observed and expected results is not significant; any difference is due to chance, so the null hypothesis is accepted and the results do fit a 9:3:3:1 ratio.
Examiner insight: The final marking point requires three things in one sentence: the comparison with the critical value, the words 'not significant', and 'due to chance'. Answers that stop at the χ² value routinely score 3 out of 5.
A2 3 practical skills — choosing a statistical test · 3 marks
The null hypothesis is that there is no significant difference between the observed results and the expected results, and that any difference is due to chance. A chi-squared test is appropriate because genetic cross data are categoric frequency data — counts of individuals falling into discrete phenotype classes. A t-test compares the means of two sets of continuous measurements, so it cannot be applied to counts of phenotypes.
Examiner insight: Say 'frequencies' or 'counts in categories', not 'numbers'. The comparison with the t-test must reference continuous data and means to earn the third mark.
These questions are written in the style of the CCEA GCE Biology (2016) papers and are not reproduced from any live paper. CCEA owns the copyright in its question papers and mark schemes — always download the official paper and mark scheme from CCEA and use these worked answers alongside them. BioCCEA is an independent study tool and is not endorsed by CCEA.
Download the official CCEA paper and mark schemeGenerate 10 spec-worded questions on chi squared test a level biology, or photograph your written answer and see it marked line by line against the mark scheme.
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