# Spearman's Rank Correlation ## What is Spearman's Rank Correlation? - Spearman's rank is a statistical test used to determine whether **correlation exists between two variables** - It is particularly useful when data **does not follow a normal distribution** - The test produces a correlation coefficient ($r_s$) that indicates the strength and direction of any relationship

Spearman's rank correlation coefficient ($r_s$) measures the strength of association between two ranked variables

## When to Use Spearman's Rank - Use when investigating whether two variables are related - Appropriate when data shows **non-normal distribution** - Useful for ecological data, behavioural studies, or any data that cannot be assumed to be normally distributed - Can detect both positive and negative correlations

Correlation does not prove causation - just because two variables are correlated doesn't mean one causes the other

## The Method ### Step 1: Create a scatter graph - Plot your data points with one variable on each axis - Visually inspect for possible linear correlation - This helps you form predictions about the relationship ### Step 2: State a null hypothesis - The null hypothesis assumes **no correlation** between the two variables - Example: "There is no correlation between temperature and enzyme activity" ### Step 3: Calculate Spearman's rank coefficient ($r_s$) Use the following formula: $$r_s = 1 - \frac{6 \times \sum D^2}{n^3 - n}$$ Where: - $r_s$ = Spearman's rank coefficient - $D$ = difference in rank between paired values - $n$ = number of paired observations (sample size)

The formula will always be provided in exams - focus on understanding when to use it and how to interpret results

### Step 4: Compare to critical values - Refer to a critical values table that relates $r_s$ to probability levels - Look up the critical value for your sample size ($n$) at the **0.05 probability level** - If your calculated $r_s$ is **greater than** the critical value, correlation is significant - If your calculated $r_s$ is **less than** the critical value, any correlation is not significant ## Interpreting Results ### If $r_s$ > critical value (at p = 0.05) - There is **significant correlation** between the variables - Less than 5% probability that the correlation occurred by chance - **Reject the null hypothesis** - The correlation is genuine ### If $r_s$ < critical value (at p = 0.05) - There is **no significant correlation** between the variables - Greater than 5% probability that any apparent correlation is due to chance - **Accept the null hypothesis** - Any observed correlation is not reliable ### Understanding the coefficient value | $r_s$ value | Interpretation | |-------------|----------------| | +1.0 | Perfect positive correlation | | +0.7 to +1.0 | Strong positive correlation | | +0.3 to +0.7 | Moderate positive correlation | | -0.3 to +0.3 | Weak or no correlation | | -0.3 to -0.7 | Moderate negative correlation | | -0.7 to -1.0 | Strong negative correlation | | -1.0 | Perfect negative correlation | ## Worked Example A researcher investigated whether there was a correlation between daily sunlight hours and the number of flowers produced by a plant species. They studied 12 individual plants over one growing season. ### Raw data collected | Plant | Sunlight hours per day | Number of flowers | |-------|------------------------|-------------------| | 1 | 8.5 | 45 | | 2 | 6.2 | 28 | | 3 | 7.8 | 38 | | 4 | 9.1 | 52 | | 5 | 5.5 | 22 | | 6 | 8.0 | 41 | | 7 | 6.8 | 31 | | 8 | 7.2 | 35 | | 9 | 8.8 | 48 | | 10 | 5.9 | 25 | | 11 | 7.5 | 37 | | 12 | 9.5 | 55 | ### Scatter graph of the data Following Step 1 of the method, we plot the data to visually inspect for correlation:

The scatter graph suggests a strong positive correlation - as sunlight hours increase, flower production tends to increase. Now we test this statistically. **Null hypothesis**: There is no correlation between daily sunlight hours and flower production. **Sample size**: $n = 12$ ### Step 1: Rank each variable Rank the data for each variable separately (rank 1 = smallest value) | Plant | Sunlight hours | Rank for sunlight | Flowers | Rank for flowers | |-------|----------------|-------------------|---------|------------------| | 1 | 8.5 | 10 | 45 | 9 | | 2 | 6.2 | 3 | 28 | 3 | | 3 | 7.8 | 8 | 38 | 7 | | 4 | 9.1 | 11 | 52 | 11 | | 5 | 5.5 | 1 | 22 | 1 | | 6 | 8.0 | 9 | 41 | 8 | | 7 | 6.8 | 5 | 31 | 4 | | 8 | 7.2 | 6 | 35 | 5 | | 9 | 8.8 | 10 | 48 | 10 | | 10 | 5.9 | 2 | 25 | 2 | | 11 | 7.5 | 7 | 37 | 6 | | 12 | 9.5 | 12 | 55 | 12 | ### Step 2: Calculate the difference in ranks ($D$) Find the difference between ranks for each plant, then square it | Plant | Rank sunlight | Rank flowers | Difference ($D$) | $D^2$ | |-------|---------------|--------------|------------------|-------| | 1 | 10 | 9 | 1 | 1 | | 2 | 3 | 3 | 0 | 0 | | 3 | 8 | 7 | 1 | 1 | | 4 | 11 | 11 | 0 | 0 | | 5 | 1 | 1 | 0 | 0 | | 6 | 9 | 8 | 1 | 1 | | 7 | 5 | 4 | 1 | 1 | | 8 | 6 | 5 | 1 | 1 | | 9 | 10 | 10 | 0 | 0 | | 10 | 2 | 2 | 0 | 0 | | 11 | 7 | 6 | 1 | 1 | | 12 | 12 | 12 | 0 | 0 | | **Total** | | | | $\sum D^2 = 6$ | ### Step 3: Apply the formula $$r_s = 1 - \frac{6 \times \sum D^2}{n^3 - n}$$ $$r_s = 1 - \frac{6 \times 6}{12^3 - 12}$$ $$r_s = 1 - \frac{36}{1728 - 12}$$ $$r_s = 1 - \frac{36}{1716}$$ $$r_s = 1 - 0.021$$ $$r_s = 0.979$$ ### Step 4: Compare to critical value For $n = 12$ at probability level 0.05, the critical value of $r_s$ is **0.591** | $n$ | Critical value of $r_s$ (p = 0.05) | |-----|-------------------------------------| | 10 | 0.648 | | 11 | 0.618 | | **12** | **0.591** | | 13 | 0.566 | | 14 | 0.545 | ### Conclusion The calculated value of $r_s = 0.979$ is **greater than** the critical value of 0.591 at $n = 12$ and $p = 0.05$. Therefore: - We can **reject the null hypothesis** - There **is a significant positive correlation** between sunlight hours and flower production - There is less than 5% probability this correlation occurred by chance - As sunlight hours increase, flower production tends to increase

This correlation does not prove that increased sunlight causes more flowers - other variables may be involved

## Key Points to Remember - Spearman's rank is used for **non-normally distributed data** - Always state a **null hypothesis** before testing - The test uses **ranked data** rather than raw values - Calculate $r_s$ using the provided formula - Compare $r_s$ to **critical values** at the 0.05 probability level - If $r_s$ > critical value, correlation is **significant** - **Correlation ≠ causation** - correlated variables may both be influenced by a third factor ## Common Exam Requirements You should be able to: - Explain when Spearman's rank correlation is appropriate - State a null hypothesis correctly - Rank data from smallest to largest - Calculate differences in ranks and square them - Substitute values into the formula (which will be provided) - Compare your result to critical values - State whether to accept or reject the null hypothesis - Interpret what the result means in biological context