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A2 · Practice questions · Transects and correlation

Rank, compare, conclude.

Six original Cambridge-style questions on transects and correlation, with a full Spearman's rank and a Pearson's calculation.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Gradient, rank, critical value.

01
Analysis
[4 marks]

Describe how a belt transect could be used to investigate how the abundance of a moss changes with distance from a stream.

  • Lay a tape measure from the stream edge at right angles across the gradient, away from the water. ✓
  • Place a frame quadrat at regular intervals along the tape (for example every 1 m) and record the percentage cover of moss in each. ✓
  • At each quadrat measure an abiotic factor such as soil moisture (probe) or light intensity. ✓
  • Repeat with several parallel transects and calculate a mean for each distance. ✓
02
Calculation
[4 marks]

Seven quadrats give these pairs (light / lux, number of plants): (120, 2), (300, 5), (450, 4), (600, 8), (900, 7), (1100, 11), (1500, 12). Calculate Spearman's rank correlation coefficient using rs = 1 − 6ΣD2 ÷ (n3 − n), and state whether the correlation is significant (critical value for n = 7 at p = 0.05 = 0.786).

  • Ranks for light: 1, 2, 3, 4, 5, 6, 7. Ranks for plants: 1, 3, 2, 5, 4, 6, 7. ✓
  • D = 0, −1, 1, −1, 1, 0, 0; ΣD2 = 4. ✓
  • rs = 1 − (6 × 4) ÷ (343 − 7) = 1 − 24 ÷ 336 = 0.93. ✓
  • 0.93 is greater than 0.786, so reject the null hypothesis: there is a significant positive correlation between light intensity and plant number. ✓
03
Calculation
[3 marks]

Five pairs of values are x: 1, 2, 3, 4, 5 and y: 2, 4, 5, 4, 5. Σxy = 66, sx = 1.581 and sy = 1.225. Calculate Pearson's correlation coefficient, r = (Σxy − n x̄ ȳ) ÷ ((n − 1) sx sy).

  • x̄ = 3 and ȳ = 4, so n x̄ ȳ = 5 × 3 × 4 = 60. ✓
  • Numerator = 66 − 60 = 6; denominator = 4 × 1.581 × 1.225 = 7.75. ✓
  • r = 6 ÷ 7.75 = 0.77, a fairly strong positive linear correlation. ✓
04
Explain
[2 marks]

Explain why a student should use Spearman's rank correlation rather than Pearson's correlation to analyse counts of limpets against height on a shore.

  • Counts of organisms in quadrats are usually not normally distributed (many zeros or a skewed spread), and Pearson's test requires normal distributions. ✓
  • The relationship may not be linear; Spearman's test only needs a consistent rise or fall (monotonic relationship) because it uses ranks. ✓
05
Explain
[2 marks]

A significant positive correlation is found between soil nitrate concentration and the number of nettles. Explain why this does not show that nitrate causes nettles to grow.

  • Correlation shows only that the two variables change together. ✓
  • Another factor could cause both (for example animal dung adds nitrate and disturbs the soil), or the nettles could affect the nitrate; an experiment controlling other variables is needed to test cause. ✓
06
Recall
[2 marks]

State the difference between a line transect and a belt transect.

  • Line transect: records the species touching a tape at regular intervals, giving presence or absence along the line. ✓
  • Belt transect: quadrats are placed at intervals along the tape, giving abundance (density, frequency or percentage cover) at each point. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.