Spearman's rank correlation; limitations of Pearson/Spearman practice questions.

Five students are ranked 1 to 5 on a maths test and on a physics test. The differences in ranks (d = maths rank − physics rank) are 0, 1, −1, 2, −2. Calculate Spearman's rank correlation coefficient rₛ.

• 0.50
• 0.75
• 0.90
• 1.00

Answer: 0.50

rₛ = 1 − 6Σd²/(n(n²−1)). Σd² = 0+1+1+4+4 = 10, so rₛ = 1 − 60/120 = 0.50.

A researcher measures the association between two variables whose scatter plot shows a clear increasing curve that is NOT a straight line. Which correlation measure best captures this monotonic but non-linear relationship?

• Pearson's product-moment correlation r
• Spearman's rank correlation rₛ
• The coefficient of determination R²
• The gradient of the least-squares line

Answer: Spearman's rank correlation rₛ

Spearman's rₛ measures the strength of any monotonic relationship, so it is appropriate when the trend increases consistently but is not linear.

A data set contains one extreme outlier caused by a recording error that cannot be removed. Which statement best explains why Spearman's rₛ may be preferred over Pearson's r here?

• Spearman's rₛ requires the data to be normally distributed
• Spearman's rₛ uses ranks, so it is less sensitive to the outlier
• Spearman's rₛ always gives a larger correlation than Pearson's r
• Spearman's rₛ can only be used for linear relationships

Answer: Spearman's rₛ uses ranks, so it is less sensitive to the outlier

Because Spearman uses ranks rather than raw values, a single extreme value affects it far less than it affects Pearson's r.

The heights and weights of 10 people give Pearson's r = 0.94 and Spearman's rₛ = 0.91. What does the closeness of these two values suggest about the relationship?

• The relationship is strong and approximately linear
• The relationship is non-monotonic
• There must be a calculation error since they should be equal
• The data contain no correlation

Answer: The relationship is strong and approximately linear

When r and rₛ are both high and close, the association is strong and roughly linear, so ranks and raw values tell the same story.

Four countries are ranked by GDP (1 = highest) and by life-expectancy rank. Ranks are: A (GDP 1, life 2), B (GDP 2, life 1), C (GDP 3, life 4), D (GDP 4, life 3). Calculate rₛ.

• 0.20
• 0.40
• 0.60
• 0.80

Answer: 0.60

d = −1, 1, −1, 1 so Σd² = 4. rₛ = 1 − 6·4/(4(16−1)) = 1 − 24/60 = 1 − 0.4 = 0.60.

Which of the following is a genuine limitation of BOTH Pearson's r and Spearman's rₛ?

• Neither can be negative
• Neither shows that one variable causes the other
• Both require the data to be ranked before use
• Both can only be applied to more than 30 data points

Answer: Neither shows that one variable causes the other

A high correlation, whether measured by r or rₛ, never proves causation; a lurking variable or coincidence may explain the association.