What you need to know
- 5.14Data from habitatThe place where an organism lives. studies need appropriate processing.
- A.1.9Select and use chi-squaredA statistical test of whether observed numbers differ significantly from the numbers expected., Student’s t-test and a correlationA relationship in which two variables change together. A correlation does not prove that one causes the other. coefficient.
- A.1.10Range and standard deviationA measure of how spread out data values are around the mean. as measures of dispersion.
Critical values in questions are supplied teaching numbers, not an official formula sheet. Spearman’s rank is taught as the correlation coefficient used with ordinal ecological ranks. Do not invent mark allocations. Fictional data only.
By the end, you should be able to…
- Choose chi-squared, Spearman’s rankA statistical test of whether there is a significant correlation between two variables, using their ranks. or a t-test to match the data.
- Calculate χ² and Spearman’s Rs on small fictional sets.
- Interpret a comparison with a stated critical value without claiming proof.
Build your understanding
Match the test to the question
chi-squaredA statistical test of whether observed numbers differ significantly from the numbers expected. tests whether observed frequencies differ from those expected under a simple hypothesis, such as “speciesA group of organisms that can interbreed to produce fertile offspring. and a habitatThe place where an organism lives. feature are independent”. Use counts, not percentages, and independent category data. Student’s t-test compares two means, for example mean percentage coverThe percentage of the area of a quadrat covered by one species. in grazed versus ungrazed plots, when the data are roughly continuous. A correlationA relationship in which two variables change together. A correlation does not prove that one causes the other. coefficient tests whether two variables, such as soil moisture and cover, vary together.
In ecology the correlation is often Spearman’s rankA statistical test of whether there is a significant correlation between two variables, using their ranks., because you rank each variable and the relationship need not be a perfect straight line. The specification says “the correlation coefficient”; Spearman’s formula is the one this course uses for ranked field data. Pearson’s product-moment correlation is enrichment unless a question gives it.
Worked fictional numbers
Chi-squared: χ² = Σ (O − E)² / E. A 2 × 2 fictional count of a plant present or absent on wet versus dry soil might give χ² = 5.4. If the question states that the critical value at p = 0.05 with 1 degree of freedom is 3.84, 5.4 > 3.84, so you reject independence at that level. You have evidence of an association, not proof that moisture caused the pattern.
Spearman: rank each variable, find d = difference in ranks for each pair, then Rs = 1 − [6Σd² / (n(n² − 1))]. For a fictional n = 6 with Σd² = 14, Rs = 1 − (84 / 210) = 0.60. Compare with a stated critical value. A significant Rs is a correlation, not causationWhen a change in one variable directly causes a change in another. Showing it needs controlled experiments or a known mechanism. — a third factor, or sampling along a hidden gradient, can still be responsible.
Spread, not just the mean
Range is highest minus lowest. It uses only two values, so one outlier inflates it. standard deviationA measure of how spread out data values are around the mean. uses every value and is usually a fairer measure of spread around the mean. Overlapping spreads around two means warn you not to over-claim a difference even if the means look apart. The t-test is how you judge that formally when the question asks.
Never hunt for a test after seeing the numbers you like. Write the hypothesis and the test choice first. All values on this page are fictional teaching data.
NOT: a significant test does not prove that the abiotic factor caused the species pattern.
- Chi-squared example: 5.4 is greater than the stated critical value 3.84 at p = 0.05 and 1 degree of freedom
- Spearman example: n = 6 and Σd² = 14 gives Rs = 0.60
- Rs = 1 − [6Σd² / (n(n² − 1))]
- χ² = Σ (O − E)² / E, using counts
- Range uses only two values; standard deviation uses every value
Two traps this diagram blocks
- Do not calculate chi-squared from percentages.
- Do not describe Rs or χ² as proof of causation.
From the diagram alone
Question
A study ranks soil moisture and percentage cover at six sites. Name a suitable correlation method and state what a significant result would not show. [3]
Show a model answer
Spearman rank correlation is suitable because the two variables can be ranked. A significant Rs would show that they vary together. It would not show that moisture caused the cover.
| Question | Test | Data |
|---|---|---|
| Are these frequencies independent of a category? | Chi-squared | Counts in categories |
| Do two variables change together? | Spearman’s rank (correlation) | Paired ranks or measurements |
| Do two means differ? | Student’s t-test | Two samples of continuous data |
| How spread out is one sample? | Range or standard deviation | One set of repeats |
Make your reasoning visible
Counts into chi-squared
Do not put percentages or means into χ². Recalculate expected values from totals.
Significant is not causal
Always end with association or difference, then a biological caveat.
Quote the critical value you were given
Do not memorise a whole table as if it were specification content.
Key explanations
- Choose the test from the question, before looking for a pleasing number.
Avoid this misconception
The claim: A significant test proves the abiotic factor caused the speciesA group of organisms that can interbreed to produce fertile offspring. pattern.
The correction: It supports a non-random association or difference. Cause still needs a mechanism and, ideally, more than one line of evidence.
Check your understanding
Try each question before opening the suggested answer.
Practice question
Retrieval and explanation · suggested answer
Which test compares mean cover of a moss in two woods, and which tests whether a snail is associated with damp quadrats?
Show suggested answer
Means: Student’s t-test. Association of counts: chi-squared.
Practice question
Retrieval and explanation · suggested answer
Observed 18, expected 12. Calculate that cell’s contribution to χ².
Show suggested answer
(18 − 12)² / 12 = 36/12 = 3.0.
Practice question
Retrieval and explanation · suggested answer
Why might standard deviation be more useful than range for ten soil-moisture readings?
Show suggested answer
Range uses only the two extreme values, so one odd puddle dominates. Standard deviation uses all ten readings.
Take the idea further
Apply the same specification knowledge to an unfamiliar problem. Any extra assumptions are supplied in the question.
Stretch question
For n = 5, Σd² = 10, Rs = 1 − 60/(5×24) = 0.50. The stated critical value at p = 0.05 is 0.90. Interpret, and say what extra data would help.
Show a hint
Compare Rs with the critical value, then think about sample size.
Explore the reasoning
0.50 is below 0.90, so there is not significant evidence of a correlation at p = 0.05. n = 5 is small, so even a real weak relationship would be hard to detect. More paired stations, and a plotted scatter, would help; do not reverse the conclusion just because the Rs looks “quite positive”.
Recall this lesson
Lesson recall questions
Question 1 of 5: Describe, in order, how to place quadrats randomly in a uniform grassland so abundance estimates are unbiased.
Quadrat and transect study
Practice question
Retrieval and explanation · suggested answer
Describe, in order, how to place quadrats randomly in a uniform grassland so abundance estimates are unbiased.
Lesson: 5B 6: Investigating Abundance and Distribution of Organisms
Build the sequence from memory first. Then reveal the model order.
Tap the steps in the order you recall. Tap a placed step to send it back.
- 01Next step
- 02Waiting
- 03Waiting
Tap to place
Put your biology to work
Explore where these ideas appear in the world. Enrichment, not extra required recall.
p = 0.05 is a convention
5B · Statistics and EcologyMathematical skill A.1.9 in the Pearson specification lists chi-squared, the t-test and a correlation coefficient as tests you can be asked to select and use. The 0.05 significance…Specification 5.14Read & use the scienceClose article
Why this matters
Mathematical skill A.1.9 in the Pearson specification lists chi-squared, the t-test and a correlation coefficient as tests you can be asked to select and use. The 0.05 significance level used in school tables is a convention for deciding whether a result is surprising under the null hypothesis. It is not a law of nature and not a Pearson mark scheme. Treat every worked example here as teaching arithmetic. A result just above the threshold is not automatically unimportant, and one just below is not automatically convincing. Effect size, sampling method and repeated studies also shape what can reasonably be concluded.
Back to the specification
Pearson Edexcel IAL Biology · Issue 2, February 2021 · Unit 4, pp. 29–32. The application extends beyond required recall.
- 5.14 CORE PRACTICAL 11: carry out a study of the ecology of a habitat, such as using quadrats and transects to determine the distribution and abundance of organisms, and measuring abiotic factors appropriate to the habitat
Use the science
If χ² is just below the critical value, have you proved there is no association?
Explore a possible response
No. You have not found significant evidence at that level. A larger sample or a better design might still detect a real, smaller effect.
Explore the source
- Statement on statistical significance and replicabilityAmerican Statistical Association
A wildlife app is not a random sample
5B · Statistics and EcologyA phone app can gather thousands of wildlife sightings, but people tend to record animals near paths and towns or photograph conspicuous species. US Geological Survey researchers explain…Specification 5.14Read & use the scienceClose article
Why this matters
A phone app can gather thousands of wildlife sightings, but people tend to record animals near paths and towns or photograph conspicuous species. US Geological Survey researchers explain that citizen-science records can be powerful while also being non-random, with missed species and occasional false identifications. A large sample does not automatically remove bias. Core Practical 11 instead asks students to choose quadrats or transects to fit a habitat question and record effort consistently. A statistical test cannot repair a flawed sampling design after the fact. App records can complement planned field surveys when their observation process is understood.
Back to the specification
Pearson Edexcel IAL Biology · Issue 2, February 2021 · Unit 4, pp. 29–32. The application extends beyond required recall.
- 5.14 CORE PRACTICAL 11: carry out a study of the ecology of a habitat, such as using quadrats and transects to determine the distribution and abundance of organisms, and measuring abiotic factors appropriate to the habitat
Use the science
Why could 10,000 app sightings near footpaths misrepresent abundance across an entire reserve?
Explore a possible response
Effort is concentrated near footpaths, so unsampled areas may differ. High sample size does not make convenience locations representative.
Explore the source
- Occupancy models for citizen-science dataUS Geological Survey