A Level formula reference and statistics notes
- Syllabus
- 9700–2028–2029
- Topic
- —
- Level
- A2
Learning objectives
—Formula reference—Hardy-Weinberg equationsUse p + q = 1 and p^2 + 2pq + q^2 = 1, where p and q are allele frequencies and p^2, 2pq and q^2 are genotype frequencies. Formulae are provided in the examination when needed.—Formula reference—Lincoln index and Simpson diversityUse the Lincoln index N = (n1 x n2) / m2 for mark-release-recapture estimates and Simpson’s index of diversity D = 1 - sum((n / N)^2) for biodiversity. Formulae are provided in the examination when needed.—Formula reference—Chi-squared, sample standard deviation, standard error and 95% CIUse chi-squared = sum((O - E)^2 / E), sample standard deviation, standard error and 95% confidence interval formulae when provided; interpret the result rather than memorising the formula symbols only.—Formula reference—t-test, Pearson, Spearman and degrees of freedomUse provided formulae for t-test, Pearson’s linear correlation and Spearman’s rank correlation; know how to calculate degrees of freedom for chi-squared and t-tests; judge significance using critical values.—Statistics notes—Choosing valid statistical methodsKnow when chi-squared, t-test, Pearson’s linear correlation and Spearman’s rank correlation are appropriate; connect data type, distribution, independence, sample size and correlation pattern to the choice of method.
Hardy-Weinberg links two allele and three genotype frequencies
The examination provides p + q = 1 and p² + 2pq + q² = 1. Here p and q are the frequencies of two alleles; p² and q² are homozygous genotype frequencies and 2pq is the heterozygous genotype frequency.
When a recessive phenotype frequency is given: 1. identify it as q²; 2. calculate q = √q²; 3. calculate p = 1 − q; 4. calculate p² and 2pq; 5. check that p + q = 1 and p² + 2pq + q² = 1. Multiply a frequency by population size only if the question asks for an expected number.
If 16% of a population shows the recessive phenotype, q² = 0.16, so q = 0.40 and p = 0.60. Expected genotype frequencies are p² = 0.36, 2pq = 0.48 and q² = 0.16; these sum to 1.00.
Do not assign q = 0.16 in this example. The equations model an equilibrium population under assumptions; a correct substitution does not show that the assumptions hold in the real population.
Lincoln estimates abundance; Simpson quantifies diversity
Lincoln index: N = (n₁ × n₂) ÷ m₂, where N is estimated population size, n₁ is the first captured and marked sample, n₂ is the total second sample, and m₂ is the marked number recaptured in the second sample.
Simpson's index in this syllabus: D = 1 − Σ(n/N)², where n is the number of individuals of each type and N is the total across all types. For each type calculate n/N, square it, sum the squared proportions, then subtract from 1. A larger D means greater diversity under this stated formula.
Lincoln: n₁ = 40, n₂ = 50 and m₂ = 10 gives N = 200. Simpson: type counts 50, 30 and 20 give N = 100 and D = 1 − (0.50² + 0.30² + 0.20²) = 1 − 0.38 = 0.62.
A small recapture m₂ inflates N and makes the estimate sensitive to one animal. Use the exact Simpson formula printed in the question/syllabus: alternative textbooks may name reciprocal or complement indices differently.
Keep discrepancy, spread and uncertainty formulae distinct
The examination provides: χ² = Σ[(O − E)²/E]; sample s = √[Σ(x − mean)²/(n − 1)]; SE = s/√n; and 95% CI ≈ mean ± (2 × SE). Use the symbols exactly as keyed in the supplied formula.
χ²: pair each observed O with its expected E, calculate every (O − E)²/E contribution, then sum.
Sample SD: find deviations from the mean, square and sum them, divide by n − 1, then square-root.
SE: divide sample SD by √n.
95% CI: calculate 2SE, subtract it from and add it to the mean to obtain lower and upper limits.
χ² measures total discrepancy between observed and expected category frequencies. SD measures spread of individual observations. SE measures precision of the estimated mean. The 95% CI expresses uncertainty around that mean. The numerical output must be followed by the relevant biological or inferential interpretation.
Do not lose the summation in χ² or SD, pair an O with the wrong E, or use n instead of n − 1 in the provided sample-SD formula. A 95% CI is not the range expected to contain 95% of individual observations.
Formula output needs degrees of freedom and a critical comparison
The provided t-test formula standardises the difference between two means using sample SDs and sizes. Pearson's r measures linear correlation from paired continuous data. Spearman's rₛ = 1 − [6ΣD²/(n³ − n)] measures monotonic correlation from paired ranks, where D is each rank difference.
Degrees of freedom are not provided as formulae: for chi-squared, v = c − 1; for the syllabus two-sample t-test, v = n₁ + n₂ − 2. Pearson and Spearman critical-value tables use sample size n rather than these df formulae. Confirm whether the table asks for df or n before reading it.
State H₀, calculate the statistic or coefficient with the supplied formula, identify df or n, select the stated probability threshold, then compare the magnitude with the correct critical value. If it crosses the critical threshold, reject H₀ and state the significant difference or correlation in the biological context.
The sign of r or rₛ gives direction; significance uses the coefficient's magnitude and the table. A strong or significant correlation does not prove causation, and no formula can compensate for choosing an invalid test.
Choose the valid method before using the formula
Statistical selection begins with the question: difference from expectation, difference between two means, or correlation between paired measurements. Then check data type, distribution, independence, pattern and sample size before calculating.
Chi-squared | observed versus expected nominal frequencies; syllabus questions use one row or one column of classes.
t-test | difference between two sample means; continuous data; populations normal; SDs approximately equal; valid even when each sample has fewer than 30 values.
Pearson | correlation; two continuous variables; normal population; scatter plot suggests linearity; at least 5 pairs, ideally 10+.
Spearman | correlation; ordinal/rankable or non-normal data; independent points; scatter suggests increasing/decreasing monotonic pattern; more than 5 pairs, ideally 10–30; random equal-chance selection.
Selection route: 1. name the biological question; 2. classify each variable; 3. inspect distribution and scatter pattern where relevant; 4. check independence, sample size and test-specific conditions; 5. state why the method fits; 6. state H₀; 7. calculate and interpret using the critical table.
Do not choose a test because its formula is supplied or familiar. Pearson is not a general test for any two numerical columns; a t-test is not valid for category counts; and statistical significance does not repair biased or non-independent sampling.