Level I · Quantitative

Learning Module 9
Tests of Independence

Key Outcomes Summary & Practice Problems

Learning Outcomes

What you must be able to do

Curriculum Year: 2026

LOS 1

Explain parametric and nonparametric tests of the hypothesis that the population correlation coefficient equals zero, and determine whether the hypothesis is rejected at a given level of significance.

LOS 2

Explain tests of independence based on contingency table data using the chi-square test statistic, and determine whether the hypothesis of independence is rejected.

1 · Parametric Test of Correlation: Pearson (t-test)

The Pearson correlation coefficient (also called bivariate or pairwise correlation) measures the strength of the linearrelationship between two continuous variables. When both variables are normally distributed, we test whether the population correlation ρ equals zero using a t-statistic.

PEARSON FORMULA

Sample correlation between X and Y:

r_XY = s_XY / (s_X × s_Y)

where s_XY = sample covariance, s_X and s_Y = standard deviations

The covariance drives the sign of r:
Positive s_XY → r > 0 (variables move together)
Negative s_XY → r < 0 (variables move oppositely)

T-TEST FOR Ρ = 0

Test statistic: t = r√(n − 2) / √(1 − r²)

Distribution: t with n − 2 degrees of freedom

Hypotheses (most common — test for any relationship):
Two-sided: H₀: ρ = 0 vs Hₐ: ρ ≠ 0
One-sided+: H₀: ρ ≤ 0 vs Hₐ: ρ > 0
One-sided−: H₀: ρ ≥ 0 vs Hₐ: ρ < 0

Decision: Reject H₀ if |t_calc| > t_critical (two-tailed)
or if t_calc > t_critical (one-tailed right)

Worked example: n=33, r=0.43051, df=31, critical t=2.45282 (1%)
t = 0.43051√31 / √(1−0.18534) = 2.656 > 2.453 → Reject H₀

Key insight — sample size and significance: The magnitude of r needed to reject H₀ decreases as n increases because (1) degrees of freedom increase → smaller critical values, and (2) the numerator r√(n−2) grows with n → larger test statistic. Example: r = 0.35 with n=12 gives t=1.182 (not significant at 5%), but the same r with n=32 gives t=2.046 (just significant). As n → ∞, even very small correlations become statistically significant — always consider economic significance alongside statistical significance.

Pearson is parametric: assumes both variables are normally distributed. If this assumption is violated, use the nonparametric Spearman rank correlation instead.
Degrees of freedom = n − 2: we lose 2 degrees of freedom because the test estimates two parameters (means of X and Y) from the data before computing the correlation.
Sign vs magnitude: the sign of r tells direction (positive = same direction, negative = opposite). The magnitude (closer to ±1) tells strength. |r| = 0.35 is the same strength whether positive or negative.
Correlation matrix: when testing multiple pairwise correlations simultaneously, calculate a separate t-statistic for each pair and compare each to the critical value. Reject H₀ individually for each pair that exceeds the critical value.

2 · Nonparametric Test of Correlation: Spearman Rank

When the population departs meaningfully from normality, or when data contain outliers, the Spearman rank correlation coefficient (r_S) is appropriate. It is computed on the ranks of the observations, not their actual values — making it robust to outliers and distributional violations.

📊 PEARSON CORRELATION (PARAMETRIC)

Uses actual observed values. Assumes bivariate normality. Sensitive to outliers. Measures linear relationship. Appropriate for continuous data from normal populations.

🔢 SPEARMAN RANK CORRELATION (NONPARAMETRIC)

Uses ranks of observations. No normality assumption. Robust to outliers. Measures monotonic relationship (including nonlinear). Appropriate when normality is violated, data are ordinal, or outliers are present.

SPEARMAN STEPS

1. Rank X observations from largest (rank 1) to smallest (rank n)
Repeat for Y. Ties: assign average of tied ranks (e.g., 3 and 4 tied → 3.5)

2. For each pair i: compute d_i = rank(X_i) − rank(Y_i) and d_i²

3. Spearman correlation:
r_S = 1 − [6 × Σd_i²] / [n(n² − 1)]

Example: n=35, Σd² = 2202
r_S = 1 − 6(2202)/[35(1225−1)] = 1 − 13212/42840 = 0.6916

TESTING R_S = 0

For large samples (n > 30), use same t-statistic as Pearson:

t = r_S × √(n − 2) / √(1 − r_S²) with df = n − 2

For small samples (n ≤ 30): requires specialized critical value tables

Example: r_S = 0.6916, n=35, df=33, critical t = ±2.0345
t = 0.6916√33 / √(1−0.4783) = 3.974/0.7224 = 5.500 → Reject H₀

H₀: r_S = 0 vs Hₐ: r_S ≠ 0 (most common — test for any monotonic relationship)

When to use Spearman vs. Pearson: use Spearman when (1) normality assumption is violated, (2) outliers are present and influential, (3) data are ordinal or ranked rather than continuous, or (4) you suspect a monotonic but non-linear relationship.
Monotonic vs linear: Pearson measures linear association only. Spearman captures monotonic association — any consistently increasing or decreasing relationship, whether linear or curved.
Ties in rankings: when two observations share a rank (tied values), assign each the average of the ranks they would have occupied. For example, if the 3rd and 4th largest values are tied, both get rank 3.5.
Spearman for financial data: financial returns, expense ratios, and other financial variables are often bounded and non-normally distributed, making Spearman the preferred correlation measure for many investment applications.

3 · Tests of Independence: Contingency Tables & Chi-Square

When data are categorical or discrete (e.g., investment type, credit rating, ESG category), correlation cannot be used. Instead, a contingency table (two-way table) organises the frequencies and a chi-square test tests whether the two classification dimensions are independent.

OBSERVED FREQUENCIES: 1,594 ETFS BY SIZE AND INVESTMENT TYPE

Investment Type	Small-Cap	Mid-Cap	Large-Cap	Total
Value	50	110	343	503
Growth	42	122	202	366
Blend	56	149	520	725
Total	148	381	1,065	1,594

EXPECTED FREQ

Under H₀ (independence), expected count for each cell:

E_ij = (Row total_i × Column total_j) / Overall total

Example — small-cap Value ETFs:
E = (503 × 148) / 1,594 = 74,444 / 1,594 = 46.703

If observed = expected in all cells → χ² = 0 (perfect independence).
Any deviation makes χ² > 0. Squared deviations → always positive.
Therefore: only one rejection region (right tail)

CHI-SQUARE STAT

Sum over all m cells of the scaled squared deviations:

χ² = Σ_m (O_ij − E_ij)² / E_ij

Degrees of freedom: df = (r − 1)(c − 1)
where r = number of row categories, c = number of column categories

ETF example: r=3 investment types, c=3 size groups
df = (3−1)(3−1) = 4; critical χ² at 5% = 9.4877
Calculated χ² = 32.080 > 9.488 → Reject H₀

Conclusion: ETF size and investment type are NOT independent

SIX-STEP PROCESS FOR CHI-SQUARE TEST OF INDEPENDENCE

H₀: classifications are independent (no relationship). Hₐ: classifications are not independent (relationship exists).

χ² = Σ(O_ij − E_ij)² / E_ij. Chi-square distributed with (r−1)(c−1) degrees of freedom.

Specify α (commonly 5%). This determines the critical value from the chi-square table.

Right-tailed only. Reject H₀ if χ²_calc > χ²_critical. One rejection region (no left tail).

Compute expected frequencies for all cells, then sum (O−E)²/E across all m = r×c cells.

Compare χ²_calc to critical value. State conclusion about independence in context.

STANDARDIZED RESIDUAL

Also called Pearson residual — measures how far each cell deviates:

Standardized residual = (O_ij − E_ij) / √E_ij

Positive → more observations than expected under independence
Negative → fewer observations than expected under independence

ETF example findings:
Medium-cap Growth: residual = +3.69 (more than expected)
Large-cap Growth: residual = −2.72 (fewer than expected)

Mosaic charts visualise these residuals with colour coding across all cells.

Chi-square is always right-tailed: because deviations are squared, χ² ≥ 0 always. Perfect independence → χ² = 0. Departures from independence → larger positive χ². There is no left-tail rejection region.
Degrees of freedom = (r−1)(c−1): NOT r×c. A 3×3 table has df=4, not 9. A 2×2 table has df=1. A 2×3 table has df=2. Common exam error is using r×c instead of (r−1)(c−1).
Independence means multiplicative: if two classifications are truly independent, the expected cell frequency equals (row total × column total) / grand total. Any meaningful deviation indicates a relationship.
Use for categorical data only: use the chi-square test of independence when data are categorical or discrete (investment type, ESG rating, sector classification). Use Pearson or Spearman correlation for continuous data.
Contrast with chi-square for variance: the chi-square distribution is used for two entirely different purposes in this curriculum — (1) testing a single population variance (LM8) and (2) testing independence in contingency tables (LM9). Do not confuse them.