The shape of the lesson
By the end of the lesson, learners can
- Line and belt transects: Press Sample along the transect.
- Spearman's rank correlation: Spearman's test is the one to reach for with field counts, which are rarely normally distributed.
- Pearson's linear correlation: Pearson's coefficient uses the actual values, not ranks, so it asks a stricter question: how close do the points lie to a straight line?
- Belt transect = quadrats along a line; gives abundance along a gradient.
- rs = 1 − 6ΣD2 ÷ (n3 − n); ties share mean ranks.
- Compare with the critical value at p = 0.05; reject the null if equal or greater.
- Correlation is not causation; Pearson needs normal, linear data.
One picture every idea returns to
A line transect is a tape laid across a habitat, recording the organisms that touch it at regular intervals; a belt transect places quadrats at intervals along the line, giving abundance as well as presence. Measuring an abiotic factor (light, soil moisture, pH) or a biotic factor at each point lets you test for a correlation with a species' abundance. Spearman's rank correlation, rs = 1 − 6ΣD2 ÷ (n3 − n), works on ranks and suits data that are not normally distributed. Pearson's correlation coefficient r measures how close the points are to a straight line and needs both variables to be normally distributed. Both run from −1 to +1; the value is compared with a critical value at p = 0.05 to accept or reject the null hypothesis of no correlation. A correlation never proves cause.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Walk the gradient, test the link.. When conditions change across a habitat, from shade to sun or from wet to dry, species change with them. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Line and belt transects | Press Sample along the transect. Project tab 5 of the lab (Biodiversity lab, transects and correlation) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Spearman's rank correlation | Spearman's test is the one to reach for with field counts, which are rarely normally distributed. | Pairs |
| 21 to 30 min | Teach and check: Pearson's linear correlation | Pearson's coefficient uses the actual values, not ranks, so it asks a stricter question: how close do the points lie to a straight line? | Frayer Model |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 17 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Pearson's linear correlation, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| A transect is systematic, not random sampling: it is used deliberately along a gradient. | For estimating the overall diversity of a uniform area, use random quadrats instead. |
| A significant correlation does not show that one variable causes the other. | Light might affect the plants, or both might depend on a third factor such as soil moisture or trampling. |
Transects and correlation: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Describe how a belt transect could be used to investigate how the abundance of a moss changes with distance from a stream.
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).
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).
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.
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.
State the difference between a line transect and a belt transect.
Answer key and marking notes
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
What extra information does a belt transect give compared with a line transect?
- Athe abundance of each species at intervals along the gradient
- Bthe total population of each species in the whole habitat
- Ca random sample that avoids any bias in choosing sites
- Dthe genetic variation within each species along the line
In which investigation is a transect more suitable than random quadrats?
- Ahow plant species change from the sea edge to the back of a sand dune system
- Bthe mean density of daisies on a uniform school field
- CSimpson's index of diversity for a whole meadow
- Dthe population size of beetles in a woodland
What is the null hypothesis when Spearman's rank correlation is used to test for a link between soil pH and the number of a moss?
- Athere is no correlation between soil pH and the number of moss plants
- Bthere is a positive correlation between soil pH and the number of moss plants
- Csoil pH causes the change in the number of moss plants
- Dthe number of moss plants is normally distributed
Five quadrats contain 4, 7, 7, 9 and 12 plants. What ranks are given to the two values of 7?
- A2.5 and 2.5
- B2 and 3
- C2 and 2
- D3 and 3
A belt transect gives rs = −0.91 for soil moisture against the number of a cactus species, and the critical value is 0.74. Which conclusion is correct?
- Athere is a significant negative correlation: cactus numbers fall as soil moisture rises
- Bthere is a significant positive correlation: cactus numbers rise with soil moisture
- Cthere is no significant correlation, because rs is negative
- Dsoil moisture is proven to kill the cactus
For 10 pairs of values, ΣD2 = 33. Calculate rs using rs = 1 − 6ΣD2 ÷ (n3 − n).
- A0.80
- B0.20
- C0.97
- D0.67
A student found rs = 0.79 for 8 pairs of values. Use the table to decide the correct conclusion.
- Asignificant at p = 0.05 but not at p = 0.01, so reject the null hypothesis at the 5% level
- Bsignificant at both p = 0.05 and p = 0.01, because 0.79 is above 0.738
- Cnot significant, because 0.79 is below the p = 0.01 value of 0.881
- Dnot significant, because rs must be above 0.900 for 8 pairs
Counts of a rare orchid in 20 quadrats along a transect are mostly 0 or 1, with a few quadrats of 15 to 20. The student wants to test for a correlation with light intensity. Which choice is best?
- ASpearman's rank, because the counts are clearly not normally distributed
- BPearson's, because there are more than 10 pairs of values
- CPearson's, because light intensity is measured on a continuous scale
- Dneither, because a correlation needs equal numbers in every quadrat
Quiz key
Quadrats placed along the line give density, frequency or percentage cover at each point, so abundance as well as presence. A transect does not estimate a whole population, it is systematic rather than random, and it says nothing about genetic variation.
A transect follows an environmental gradient, such as distance from the sea across dunes. A uniform field and a whole-meadow diversity index need random quadrats, and a mobile beetle population is estimated by mark-release-recapture.
The null hypothesis is that there is no correlation; the test decides whether it can be rejected. A positive correlation is an alternative hypothesis, causation cannot be tested by correlation, and normal distribution is an assumption of Pearson's test, not a hypothesis.
The 7s would occupy ranks 2 and 3, so each gets the mean, (2 + 3) ÷ 2 = 2.5, and the next value, 9, is ranked 4. Giving 2 and 3 treats them as different; 2 and 2 or 3 and 3 would upset the sum of the ranks.
0.91 is greater than 0.74, so the correlation is significant, and the negative sign means one variable falls as the other rises. A negative value can still be significant. Correlation does not prove cause.
6 × 33 = 198; 198 ÷ 990 = 0.20; rs = 1 − 0.20 = 0.80. 0.20 forgets to subtract from 1; 0.97 divides by n3 − n without multiplying by 6 (1 − 33 ÷ 990); 0.67 divides by n2 − 1 = 99 and leaves out the 6 (1 − 33 ÷ 99).
For n = 8, 0.79 is above 0.738 (p = 0.05) but below 0.881 (p = 0.01), so the correlation is significant at the 5% level only and the null hypothesis is rejected at that level. It is not significant at p = 0.01, so claiming both levels is wrong. Significance is judged at the chosen level, usually p = 0.05, so failing the stricter 0.01 test does not make it non-significant, and there is no fixed threshold of 0.900.
The counts are strongly skewed, so the normal-distribution condition for Pearson's test fails, and Spearman's rank test, which uses ranks, is appropriate. A larger sample does not fix skew, and one continuous variable is not enough: both must be normally distributed. Correlation tests do not need equal counts.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.