A-Level Biology: Statistics and Data Analysis

A significant share of the marks in A level biology depend on mathematical skills, and the statistical tests are where many otherwise strong answers lose them: the wrong test chosen, degrees of freedom miscounted, or a conclusion that 'proves the hypothesis'. This material trains the whole chain from raw numbers to a correctly worded conclusion.

It starts with choosing a test. Chi-squared compares observed with expected frequencies in categories, such as the offspring of a genetic cross or woodlice choosing between chambers. An unpaired t-test compares the means of two sets of continuous, roughly normally distributed measurements. Spearman's rank correlation tests for an association between two variables measured on the same samples. Each scenario in the set gives you the kind of data and asks which test fits.

The calculations are done in full with original data: a standard deviation using n − 1, a chi-squared value for a 3 : 1 ratio, a Spearman's rank coefficient from ranked data with Σd², and in the written questions a t-value from two means and standard deviations. Degrees of freedom are practised separately (categories − 1 for chi-squared, n₁ + n₂ − 2 for an unpaired t-test), and every critical value you need is given in the question, as it would be in an exam.

The last step is the conclusion. You state the null hypothesis, compare the calculated value with the critical value at p = 0.05, and decide whether to reject it — then say what that means in biological terms, including the probability that the result arose by chance. Supporting ideas are tested too: what overlapping standard deviations suggest, percentage change, and why a strong correlation does not prove that one variable causes the other.

The material offers a multiple-choice quiz with worked explanations, a flashcard deck of the tests, formulae and definitions, a printable written-work sheet of eight open questions (full calculations and conclusions) answered by hand and checked, and an oral exam in which an examiner asks one question at a time about choosing and interpreting tests and gives feedback at the end.

The content is based on the mathematical requirements in the DfE A level biology subject content, as applied in current exam-board specifications. Formulae are written in the usual A level form; check your own board's formula list. It is practice material and does not replace your own board's specification.

Choose between chi-squared, an unpaired t-test and Spearman's rank correlation for a given data set
Calculate standard deviation, chi-squared, t and Spearman's rank coefficient from raw or summary data
Work out degrees of freedom and compare a calculated value with a critical value at p = 0.05
Write a null hypothesis and a correctly worded conclusion
Interpret standard deviations and error bars
Explain why correlation does not prove causation

Practice material written by Zestly, based on the mathematical requirements of the DfE A level biology subject content (statistical tests, standard deviation, significance).

Sample question

Scatter graph of the number of plants per m² against light intensity in lux at eight sites: (120, 5), (210, 8), (340, 9), (450, 10), (560, 14), (610, 12), (780, 17), (900, 16).

A student measured light intensity and the number of plants per m² at eight sites in a wood. Light intensity (lux): 120, 340, 560, 210, 780, 450, 900, 610. Plants per m² at the same sites, in the same order: 5, 9, 14, 8, 17, 10, 16, 12. Using $r_s = 1 - \frac{6\sum d^2}{n(n^2 - 1)}$ and a critical value of 0.738 for n = 8 at p = 0.05, what is Spearman's rank correlation coefficient and the correct conclusion?

$r_s = 0.95$; there is a significant positive correlation, so the null hypothesis is rejected
$r_s = 0.95$; the correlation is not significant, so the null hypothesis is not rejected
$r_s = 0.05$; the correlation is not significant, so the null hypothesis is not rejected
$r_s = 0.71$; there is a significant positive correlation, so the null hypothesis is rejected

See the answer

$r_s = 0.95$; there is a significant positive correlation, so the null hypothesis is rejected

Ranking light intensity gives 1, 3, 5, 2, 7, 4, 8, 6 and ranking plant density gives 1, 3, 6, 2, 8, 4, 7, 5. The differences are 0, 0, −1, 0, −1, 0, 1, 1, so Σd² = 4. $r_s = 1 - \frac{6 \times 4}{8(64 - 1)} = 1 - \frac{24}{504} = 0.95$. As 0.95 is greater than the critical value 0.738, there is less than a 5% probability that a correlation this strong arose by chance: the null hypothesis (no correlation) is rejected. The correlation is positive but does not by itself prove that light causes the higher density.

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