Spearman’s Rank Correlation | 斯皮尔曼等级相关

📚 Spearman’s Rank Correlation | 斯皮尔曼等级相关

Spearman’s rank correlation is a non-parametric statistical test widely used in A-Level Biology to investigate whether there is a monotonic relationship between two variables that are measured on at least an ordinal scale. It does not require data to follow a normal distribution and is especially useful when sample sizes are small or when the relationship is not linear. Understanding how to calculate, interpret, and report the Spearman’s rank correlation coefficient (often denoted as rₛ) is an essential skill for Cambridge International A-Level Biology students, particularly within the context of ecological studies, behaviour observations, and other investigations where raw data are converted to ranks.

斯皮尔曼等级相关是一种非参数统计检验，在 A-Level 生物学中广泛用于考察至少以顺序尺度测量的两个变量之间是否存在单调关系。它不要求数据呈正态分布，当样本量较小或关系并非线性时尤其适用。理解如何计算、解释并报告斯皮尔曼等级相关系数（通常记作 rₛ）是剑桥国际 A-Level 生物学学生必须掌握的技能，这在生态研究、行为观察以及其他需要将原始数据转换为等级的研究中尤为重要。

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1. What is Spearman’s Rank Correlation? | 什么是斯皮尔曼等级相关？

Spearman’s rank correlation coefficient (rₛ, also called Spearman’s rho) is a measure of the strength and direction of association that exists between two variables measured on at least an ordinal scale. The test works by converting each set of data into ranks and then applying a formula that compares how closely the two rankings agree. Because it relies on ranks instead of raw values, it is robust to outliers and does not assume that the underlying data are normally distributed.

斯皮尔曼等级相关系数（rₛ，也称斯皮尔曼 rho）是衡量至少以顺序尺度测量的两个变量之间关联强度和方向的指标。该检验的原理是先将每组数据转换为秩次，然后应用一个比较两组秩次吻合程度的公式。因为它依赖秩次而非原始数值，所以对异常值不敏感，也不要求数据呈正态分布。

A value of +1 indicates a perfect positive monotonic association (as one variable increases, the other variable also increases in rank). A value of −1 indicates a perfect negative monotonic association (as one variable increases, the other decreases in rank). A value close to 0 suggests no monotonic relationship. It is important to remember that Spearman’s correlation assesses monotonic relationships, meaning that the variables tend to move in the same or opposite direction, but not necessarily at a constant rate.

数值 +1 表示完全正单调关联（当一个变量增大时，另一个变量的秩次也增大）。数值 −1 表示完全负单调关联（当一个变量增大时，另一个变量的秩次减小）。接近 0 的数值表明不存在单调关系。必须牢记，斯皮尔曼相关评估的是单调关系，即变量趋向于同向或反向变化，但变化速率不一定恒定。

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2. When to Use Spearman’s Rank Correlation in Biology | 生物学中何时使用斯皮尔曼等级相关？

In A-Level Biology investigations, Spearman’s rank is the test of choice when you wish to test for a correlation between two variables and one or more of the following conditions apply: the data are ordinal (e.g. abundance scores on a DAFOR scale, dominance ranks, behavioural intensity categories), the data are continuous but do not meet the assumption of bivariate normality required by Pearson’s correlation, or the sample size is small (typically n < 30). It is also suitable when you suspect the relationship is monotonic but not strictly linear, such as a logarithmic or sigmoidal trend.

在 A-Level 生物学探究中，当你希望检验两个变量之间的相关性且满足以下一个或多个条件时，斯皮尔曼等级相关是首选的检验方法：数据为顺序数据（例如 DAFOR 多度评分、优势等级、行为强度类别），数据为连续数据但不满足皮尔逊相关所要求的双变量正态性假设，或样本量较小（通常 n < 30）。当你怀疑关系为单调但并非严格线性（如对数或 S 形趋势）时，该检验同样适用。

Typical biological contexts include investigating the relationship between light intensity and the percentage cover of shade-tolerant plants, the association between soil pH and the number of earthworm burrows in a quadrat, or the rank correlation between the level of pollutants in a river and a biotic index based on invertebrate families. In each case, variables can be ranked and tested without assuming a linear model.

典型的生物学情境包括：探究光照强度与耐阴植物覆盖百分比之间的关系、土壤 pH 值与样方中蚯蚓洞穴数量之间的关联、河水污染物浓度与基于无脊椎动物科的生物指数之间的等级相关。在上述每种情形中，都可以对变量进行排序并开展检验，而不必假设线性模型。

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3. Null and Alternative Hypotheses | 零假设与备择假设

For a Spearman’s rank correlation, the hypotheses are stated in terms of the population rank correlation coefficient, often symbolised by ρₛ (rho). The null hypothesis (H₀) asserts that there is no monotonic relationship between the two variables in the population: ρₛ = 0. The alternative hypothesis (H₁) can be two-tailed (there is a monotonic relationship; ρₛ ≠ 0) or one-tailed (there is a positive monotonic relationship, ρₛ > 0, or a negative monotonic relationship, ρₛ < 0). In most A-Level Biology exam contexts, a two-tailed test is used because we are interested in detecting any association, whether positive or negative.

在斯皮尔曼等级相关中，假设是关于总体等级相关系数（通常记为 ρₛ）来陈述的。零假设（H₀）声称总体中两个变量之间不存在单调关系：ρₛ = 0。备择假设（H₁）可以是双尾的（存在单调关系；ρₛ ≠ 0），也可以是单尾的（存在正单调关系，ρₛ > 0；或存在负单调关系，ρₛ < 0）。在大多数 A-Level 生物考试情境中，使用双尾检验，因为我们关注的是检测是否存在任何关联，无论是正还是负。

Always state your hypotheses clearly at the start of an investigation. For example: H₀: There is no significant correlation between percentage cover of heather and soil moisture rank. H₁: There is a significant correlation between percentage cover of heather and soil moisture rank. The term ‘significant’ refers to statistical significance, which will be determined by a critical value comparison later.

务必在调查开始时清晰地陈述假设。例如：H₀：石楠覆盖百分比与土壤湿度秩次之间无显著相关。H₁：石楠覆盖百分比与土壤湿度秩次之间存在显著相关。’显著’一词指统计显著性，将由后续临界值比较来确定。

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4. Step-by-Step Calculation | 分步计算

The calculation of Spearman’s rank correlation coefficient involves the following steps. All working should be presented in a clear table to minimise errors and gain marks in examinations.

斯皮尔曼等级相关系数的计算包含以下步骤。所有演算过程应呈现在清晰的表格中，以减少失误并在考试中获得分数。

Step 1: List the paired measurements for your two variables, X and Y. Ensure you have n pairs of data, where n is the number of observations.

步骤 1：列出两个变量 X 和 Y 的配对测量值。确保有 n 对数据，n 为观测次数。

Step 2: Rank the X values from smallest to largest, assigning the rank 1 to the smallest value. If two or more X values are equal (tied), assign to each the average of the ranks they would have occupied. Repeat this independently for the Y values.

步骤 2：将 X 值从小到大排序，最小值赋秩为 1。如果有两个或多个 X 值相等（结），则给每个值赋予它们原本应占秩次的平均值。对 Y 值独立重复此操作。

Step 3: For each pair, calculate the difference d between the rank of X and the rank of Y: d = rank(X) − rank(Y).

步骤 3：对每一对数据，计算 X 的秩次与 Y 的秩次之差 d：d = rank(X) − rank(Y)。

Step 4: Square each difference to obtain d², then sum all the d² values to find ∑ d².

步骤 4：将每个差值平方得到 d²，然后对所有 d² 求和得到 ∑ d²。

Step 5: Apply the Spearman’s rank formula. If there are no tied ranks, use:

步骤 5：应用斯皮尔曼等级公式。如果没有相同秩次，使用如下公式：

rₛ = 1 − (6 ∑ d²) / [n (n² − 1)]

If there are tied ranks, a correction factor should be used, but for A-Level purposes, the standard formula is often acceptable unless the number of ties is large. Always check your syllabus requirements.

如果存在相同秩次，应使用校正因子；但在 A-Level 阶段，除非结的数量很多，通常使用标准公式即可。务必核对课程大纲要求。

Step 6: Interpret the calculated rₛ value. The coefficient will always lie between −1 and +1.

步骤 6：解释计算得到的 rₛ 值。该系数始终介于 −1 和 +1 之间。

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5. Worked Example: Species Richness and Altitude | 实例：物种丰富度与海拔

A student investigates whether plant species richness changes with altitude on a mountain slope. Ten sites at different altitudes are sampled, and the number of plant species per 10 m² quadrat is recorded. The student suspects the relationship is monotonic but not necessarily linear and decides to use Spearman’s rank correlation.

某学生探究植物物种丰富度是否随山坡海拔变化。在不同海拔的 10 个地点进行取样，记录每 10 m² 样方中的植物种数。该学生怀疑这种关系是单调但不一定是线性的，因此决定使用斯皮尔曼等级相关。

The data table with rankings (to be drawn by the student) might look like this:

Site	Altitude (m)	Species richness	Rank (Alt.)	Rank (Rich.)	d	d²
1	200	18	1	4	−3	9
2	350	22	2	7	−5	25
3	480	19	3	5.5	−2.5	6.25
4	560	19	4	5.5	−1.5	2.25
5	670	25	5	9	−4	16
6	800	21	6	8	−2	4
7	920	14	7	2	5	25
8	1050	16	8	3	5	25
9	1180	12	9	1	8	64
10	1250	10	10	1	9	81

Note: For species richness, sites 3 and 4 both have 19 species, so they share the tied ranks (5+6)/2 = 5.5. Sites 7 and 10 have unique values. Sum of d² = 9+25+6.25+2.25+16+4+25+25+64+81 = 257.5. With n = 10, we calculate rₛ = 1 − (6 × 257.5) / [10(100 − 1)] = 1 − (1545) / 990 = 1 − 1.5606 = −0.5606 (rounded to −0.561).

注意：对于物种丰富度，地点 3 和 4 都拥有 19 种，因此它们共享相同秩次 (5+6)/2 = 5.5。地点 7 和 10 具有独特数值。d² 总和为 9+25+6.25+2.25+16+4+25+25+64+81 = 257.5。n = 10，计算 rₛ = 1 − (6 × 257.5) / [10(100 − 1)] = 1 − (1545) / 990 = 1 − 1.5606 = −0.5606（四舍五入为 −0.561）。

The negative rₛ value suggests a negative correlation: as altitude rank increases, species richness rank tends to decrease. The magnitude (0.561) suggests a moderately strong association. Whether this is statistically significant will be assessed in a later section.

负的 rₛ 值表明负相关：随着海拔秩次升高，物种丰富度秩次趋于降低。数值大小 0.561 表明中等强度的关联。该关联是否具有统计显著性将在后续部分加以评估。

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6. Interpreting the Correlation Coefficient (rₛ) | 解释相关系数 rₛ

The Spearman coefficient rₛ ranges from −1 to +1. The sign indicates the direction of the monotonic relationship: positive rₛ means that high ranks of X tend to pair with high ranks of Y; negative rₛ means that high ranks of X tend to pair with low ranks of Y. The magnitude (absolute value) reflects the strength of the association. An rₛ value of 0.8 or above is generally regarded as a strong correlation, 0.5–0.8 as moderate, and below 0.5 as weak. However, statistical significance depends on sample size and critical values, not just on the absolute size of rₛ.

斯皮尔曼系数 rₛ 的取值范围是 −1 到 +1。符号表明了单调关系的方向：正 rₛ 意味着 X 的高秩次往往与 Y 的高秩次配对；负 rₛ 意味着 X 的高秩次往往与 Y 的低秩次配对。绝对值反映了关联的强度。通常认为 rₛ 绝对值在 0.8 以上为强相关，0.5–0.8 为中等相关，0.5 以下为弱相关。然而，统计显著性取决于样本量和临界值，而不仅仅取决于 rₛ 的绝对大小。

Do not interpret Spearman’s coefficient as a measure of linear slope. A high positive rₛ does not mean that plotting raw values gives a straight line; it means that the ranking order is highly consistent. Always state your interpretation in terms of ranks and monotonic association, not in terms of ’cause and effect’, because correlation does not imply causation.

请勿将斯皮尔曼系数解释为线性斜率的量度。高正 rₛ 值并不意味着原始数值作图会形成一条直线；它意味着秩次排序高度一致。请始终从秩次和单调关联的角度进行解释，而不要使用’因果关系’之类的表述，因为相关性并不意味着因果性。

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7. Testing for Significance | 显著性检验

To determine whether the observed rₛ is statistically significant, you must compare the absolute value of your calculated rₛ with a critical value from a Spearman’s rank correlation coefficient table. The critical value depends on the number of data pairs n and on the chosen significance level, usually α = 0.05 (5%). For a two-tailed test, you read the critical value at p = 0.05 for your n. If |rₛ| is greater than or equal to the critical value, you reject H₀ and conclude that there is a significant correlation. If |rₛ| is less than the critical value, you fail to reject H₀.

为判断观测到的 rₛ 是否具有统计显著性，必须将计算得到的 rₛ 的绝对值与斯皮尔曼等级相关系数表中的临界值进行比较。临界值取决于数据对的数量 n 以及所选显著性水平（通常 α = 0.05

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