📚 Case Study Practice: Plant Distribution and Soil pH | 案例分析实战演练：植物分布与土壤pH
This article provides a full worked case study tailored to AQA Year 13 Biology, focusing on data analysis, Spearman’s rank correlation, and evaluation skills. You will work through a realistic ecological investigation examining the relationship between the abundance of a plant species and soil pH in a traditional hay meadow. By the end, you should be confident in tackling similar statistical application questions in your A-level exam.
本文为AQA Year 13生物课程量身打造，呈现一个完整的案例分析，重点关注数据分析、Spearman秩相关计算与实验评估能力。你将逐步解构一个真实的生态调查——探究传统草甸中某种植物丰度与土壤pH之间的关系。阅读完毕后，你应该能够自信地应对A-level考试中类似的统计学应用题。
1. Introduction to the Case Study | 案例简介
A group of A-level students investigated the distribution of common bent grass (Agrostis capillaris) in a lowland hay meadow in Somerset. They wanted to find out whether the percentage cover of this grass was related to the pH of the soil.
一群A-level学生对萨默塞特一处低地干草草甸中普通翦股颖（Agrostis capillaris）的分布进行了调查。他们想弄清楚这种草的覆盖度百分比是否与土壤pH有关。
The meadow had been managed without artificial fertilisers for over twenty years, creating a mosaic of soil conditions. The students suspected that soil pH might influence nutrient availability and therefore affect plant growth.
这片草甸二十多年来从未施用人工化肥，形成了复杂的土壤条件镶嵌。学生们猜测土壤pH可能影响养分有效性，进而影响植物生长。
They used a systematic random sampling approach to collect paired data on grass cover and pH. Your task is to help them analyse these data, select an appropriate statistical test, and draw a biologically meaningful conclusion.
他们采用了系统随机取样法，收集了成对的草类盖度和pH数据。你的任务是帮助他们分析这些数据，选择合适的统计检验方法，并得出有生物学意义的结论。
2. Field Sampling Methods | 野外取样方法
Two 20-metre tape measures were laid perpendicularly to form a grid over a representative 20 m × 20 m area. Ten coordinates were generated using a random number table.
两条20米卷尺垂直摆放，在具有代表性的20 m × 20 m区域内形成一个网格。利用随机数表生成了10个坐标点。
At each coordinate, a 0.5 m × 0.5 m frame quadrat was placed. The percentage cover of Agrostis capillaris was estimated independently by two students and the mean recorded.
在每个坐标处放置一个0.5 m × 0.5 m的样方框。两名学生分别独立估算Agrostis capillaris的覆盖度百分比，并记录平均值。
A soil sample was taken from the centre of each quadrat using a trowel. In the laboratory, 10 g of soil was mixed with 25 cm³ of distilled water, and the pH was measured with a calibrated digital pH meter.
用泥铲在每个样方中央取一份土样。在实验室中，将10 g土壤与25 cm³蒸馏水混合，用已校准的数字pH计测定pH值。
All measurements were taken on the same morning in June to minimise variation in light and temperature.
所有测量均于六月的一个上午完成，以尽量减少光照和温度的变化。
3. Data Collected | 收集的数据
The raw data from the ten quadrats are presented in Table 1. Quadrat number, mean percentage cover and soil pH are shown.
表1列出了来自十个样方的原始数据，包括样方编号、平均覆盖度百分比和土壤pH值。
| Quadrat | Cover (%) | Soil pH |
|---|---|---|
| 1 | 5 | 5.2 |
| 2 | 12 | 5.8 |
| 3 | 8 | 5.4 |
| 4 | 20 | 6.3 |
| 5 | 15 | 5.9 |
| 6 | 3 | 5.0 |
| 7 | 25 | 6.5 |
| 8 | 18 | 5.5 |
| 9 | 10 | 6.1 |
| 10 | 22 | 6.4 |
Examine the data briefly: as percentage cover increases, the soil pH appears to rise as well. But is this pattern statistically significant? We need a formal test.
快速浏览数据：随着覆盖度百分比升高，土壤pH似乎也在上升。但这种模式是否具有统计显著性？我们需要一个正式的检验。
4. Formulating a Hypothesis | 提出假设
Every statistical test begins with a null hypothesis. The null hypothesis (H₀) always states that there is no association between the two variables. For this investigation:
每一次统计检验都从零假设开始。零假设（H₀）总是声称两个变量之间没有关联。对于本调查：
H₀: There is no correlation between the percentage cover of Agrostis capillaris and soil pH in the meadow.
H₀：草甸中Agrostis capillaris的覆盖度百分比与土壤pH之间不存在相关性。
H₁: There is a correlation between the percentage cover of Agrostis capillaris and soil pH (two-tailed, as we are not predicting the direction).
H₁：Agrostis capillaris的覆盖度百分比与土壤pH之间存在相关性（双尾检验，因为我们未预测方向）。
Note that H₁ could be one-tailed if we had prior evidence, but using a two-tailed test is safer when the direction is uncertain.
注意，如果之前有证据，H₁可以是单尾的，但在方向不确定时使用双尾检验更稳妥。
5. Choosing a Statistical Test | 选择统计检验
We are looking for a correlation between two continuous variables. The first choice might be the Pearson product‑moment correlation. However, Pearson’s test assumes the data are normally distributed and measured on an interval scale. Percentage cover estimates are often subjective and may not be normally distributed. Moreover, we have only ten data points.
我们正在寻找两个连续变量之间的相关性。首选可能是Pearson积差相关。然而，Pearson检验假设数据呈正态分布并且是在等距尺度上测量的。覆盖度百分比的估计往往带有主观性，可能不是正态分布。此外，我们只有十个数据点。
A Spearman’s rank correlation coefficient (rs) is non‑parametric and works by ranking the data. It does not require the data to be normally distributed and is less sensitive to outliers, making it ideal for small ecological samples.
Spearman秩相关系数（rs）是一种非参数检验，通过对数据排序来工作。它不要求数据呈正态分布，并且对异常值不敏感，因此非常适合小型生态学样本。
The test statistic rs ranges from −1 (perfect negative correlation) to +1 (perfect positive correlation). A value close to 0 suggests no correlation.
检验统计量 rs 的取值范围为 −1（完全负相关）到 +1（完全正相关）。接近 0 的值表示没有相关性。
Given the nature of our data, Spearman’s rank is the most appropriate choice.
鉴于我们数据的性质，Spearman秩相关是最合适的选择。
6. Step-by-Step Calculation of Spearman’s Rank | 逐步计算Spearman秩相关
Follow these steps to compute the correlation coefficient.
按照以下步骤计算相关系数。
Step 1: Rank the cover values from 1 (smallest) to 10 (largest). Rank the pH values separately in the same way.
步骤1：将覆盖度数值从1（最小）到10（最大）依次排序。同样地，对pH值单独排序。
Step 2: Calculate the difference (D) between the rank of cover and the rank of pH for each quadrat.
步骤2：计算每个样方覆盖度秩与pH秩之间的差（D）。
Step 3: Square each difference to get D², then sum these squared differences to obtain ΣD².
步骤3：将每个差值平方得到 D²，然后将这些平方差相加得到 ΣD²。
The formula for Spearman’s rank correlation coefficient is:
Spearman秩相关系数的公式为：
rs = 1 − (6 Σ D²) / (n (n² − 1))
where n is the number of paired observations. Table 2 shows the full calculation.
其中 n 是配对观测值的数量。表2展示了完整的计算过程。
| Quadrat | Cover (%) | Cover rank | pH | pH rank | D | D² |
|---|---|---|---|---|---|---|
| 1 | 5 | 2 | 5.2 | 2 | 0 | 0 |
| 2 | 12 | 5 | 5.8 | 5 | 0 | 0 |
| 3 | 8 | 3 | 5.4 | 3 | 0 | 0 |
| 4 | 20 | 8 | 6.3 | 8 | 0 | 0 |
| 5 | 15 | 6 | 5.9 | 6 | 0 | 0 |
| 6 | 3 | 1 | 5.0 | 1 | 0 | 0 |
| 7 | 25 | 更多咨询请联系16621398022（同微信）
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