CFA Level I Exam · Estimation and Hypothesis Testing

Parametric vs Nonparametric Tests for CFA Level I

Updated 7 October 2026 · Fact-checked

Parametric tests assume the data follow a specific distribution, usually normal, and test parameters such as the mean or variance. Nonparametric tests make fewer assumptions. Use them when data are ranked, have outliers, are not normal, or when the hypothesis is not about a parameter. Examples: Spearman rank correlation, sign test, chi-square independence test.

Understand Parametric vs Nonparametric Tests

A parametric test tests a statement about a population parameter, such as a mean, variance or correlation. It also relies on assumptions about the distribution, often that the population is normal. The t-test, chi-square test of a variance and F-test are parametric.

A nonparametric test makes few or no assumptions about the population distribution. Either it does not test a parameter at all, or it uses ranks or counts instead of the raw values. You give up some power in return for fewer assumptions. If the parametric assumptions truly hold, the parametric test is more powerful.

Use a nonparametric test in four situations:
- The data are ranks, not measurements (for example, analyst rankings of funds).
- The data are far from normal or have extreme outliers, and the sample is small.
- The hypothesis is not about a parameter, such as whether two characteristics are independent.
- Key assumptions of the parametric test are violated.

Three tests matter for the exam. The Spearman rank correlation test checks whether two variables have a monotonic relationship, using ranks. The sign test checks a median or paired differences by counting the positive and negative signs. The chi-square test of independence uses a contingency table of counts to test whether two categorical variables are independent.

For large samples, the central limit theorem often lets you use parametric tests on means even when data are not normal. Nonparametric tests matter most when samples are small or the data type forces them.

Key formulas to remember

Spearman rank correlation: rs = 1 − [6 Σdi²] ÷ [n(n² − 1)]; di is the difference between the two ranks of observation i. n is the number of observations. Tied values get the average of their ranks (this formula is exact only without ties).
Test statistic for Spearman correlation: t = rs × √(n − 2) ÷ √(1 − rs²), with n − 2 degrees of freedom; Same t-statistic form as the parametric correlation test. Reject H0 of zero correlation if |t| exceeds the critical value.
Chi-square test of independence: χ² = Σ (Oij − Eij)² ÷ Eij; Oij is the observed count in a cell, Eij the expected count if the variables are independent.
Expected frequency in a cell: Eij = (row i total × column j total) ÷ grand total; Computed under the null hypothesis that the two variables are independent.
Degrees of freedom for independence test: df = (r − 1)(c − 1); r is the number of rows and c the number of columns. The test is one-sided (right tail).
Sign test idea: Count + and − signs; under H0 each sign has probability 0.5; The number of positive signs follows a binomial distribution with p = 0.5. Zero differences are dropped from n.

How to solve Parametric vs Nonparametric Tests questions

Use this sequence for any question that asks which test to use or asks you to run one of these tests.

1Read what the data are: raw measurements, ranks, or category counts.
2Identify the claim: about a parameter (mean, variance, correlation) or about a relationship, independence or a median.
3Check assumptions: normality, sample size, outliers. If they fail or the data are ranks, choose a nonparametric test.
4Match the test: ranked pairs or monotonic link means Spearman; paired differences or a median means sign test; two categorical variables in a table means chi-square independence.
5State H0 and Ha. For independence, H0 is that the variables are independent. For Spearman, H0 is zero rank correlation.
6Compute the statistic (for chi-square, find expected counts first) and the degrees of freedom.
7Compare with the critical value, then state the decision and its meaning in plain words.

Quickest way: Three-clue test selector

When to use it: Use this when the question only asks which test is appropriate, or when you must eliminate options fast.

Categories in a table: chi-square independence.
Ranks or a non-normal, outlier-heavy sample for a correlation: Spearman.
Median or direction of paired changes: sign test.
Normal data, a mean or variance hypothesis: parametric test (t, chi-square variance, F).
Remember nonparametric means less power, so it is never chosen just because it sounds safer.

Common mistakes in Parametric vs Nonparametric Tests

Choosing a nonparametric test whenever the sample is small.
Students link small samples with weak assumptions.
Fix: Small samples from a normal population still suit a t-test. Choose nonparametric only if normality fails, data are ranks, or the claim is not about a parameter.
Claiming nonparametric tests are always better because they need fewer assumptions.
Fewer assumptions sounds like more robustness.
Fix: Remember the trade-off: when parametric assumptions hold, the parametric test has more power.
Using row totals only for expected counts in the chi-square test.
Students forget the formula uses both margins.
Fix: Use E = row total × column total ÷ grand total for every cell.
Using df = n − 1 for the independence test.
Mixing it up with the t-test.
Fix: Use (r − 1)(c − 1). A 3 × 2 table has 2 degrees of freedom.
Forgetting to rank both variables separately before computing d.
Students rank one series or use raw values.
Fix: Rank each variable on its own from smallest to largest, then subtract rank from rank.
Treating a significant chi-square result as proof of causation.
Significance feels like a strong conclusion.
Fix: It only shows the variables are probably not independent, not that one causes the other.

Worked examples

Example 1

An analyst ranks six funds by 3-year return and by expense ratio (1 = highest in each). Rank pairs (return, expense): (1,2), (2,1), (3,4), (4,3), (5,6), (6,5). Compute the Spearman rank correlation.

Show the solution

Find d for each pair: −1, 1, −1, 1, −1, 1.
Square each: 1 each, so Σd² = 6.
n = 6, so n(n² − 1) = 6 × 35 = 210.
rs = 1 − (6 × 6) ÷ 210 = 1 − 36 ÷ 210.
36 ÷ 210 = 0.1714, so rs = 0.8286.

Answer: rs ≈ 0.83, a strong positive rank correlation.

Example 2

A survey of 200 investors classifies them by risk tolerance (low, high) and by preferred asset (bonds, equities). Observed counts: low-bonds 60, low-equities 40, high-bonds 40, high-equities 60. Test independence at the 5% level. The critical chi-square value with 1 degree of freedom is 3.841. Which conclusion is correct?

Show the solution

Row totals: low 100, high 100. Column totals: bonds 100, equities 100. Grand total 200.
Expected count for each cell = 100 × 100 ÷ 200 = 50.
Each cell's (O − E)² ÷ E = 10² ÷ 50 = 2.
Sum over four cells: χ² = 8.
df = (2 − 1)(2 − 1) = 1. Since 8 > 3.841, reject H0.

Answer: Reject independence: risk tolerance and asset preference are related at the 5% level (χ² = 8).

Exam tips

Questions are usually conceptual: pick the test from the data type. Look for words like ranks, median, categories or outliers.
Expect three options, so eliminate any that apply a parametric test to ranked or categorical data.
For chi-square, compute expected counts from the margins first. Check that observed and expected totals match.
Remember the independence test is right-tailed and df = (r − 1)(c − 1).
There is no penalty for wrong answers, so answer every question. If you are unsure, eliminate what you can and guess.

Practice questions from Estimation and Hypothesis Testing

Parametric vs Nonparametric Tests: frequently asked questions

When should I use a nonparametric test instead of a parametric one?

Use it when data are ranks, when the distribution is clearly non-normal or has outliers in a small sample, or when the claim is not about a parameter. Otherwise the parametric test is more powerful.

What does the Spearman rank correlation test measure?

It measures the strength of a monotonic relationship between two variables using their ranks. It is less affected by outliers and does not need normal data. You test it with a t-statistic using n − 2 degrees of freedom.

What does the chi-square test of independence test?

It tests whether two categorical variables are independent, using a contingency table of counts. You compare observed counts with the counts expected under independence. A large statistic leads you to reject independence.

What is the sign test used for?

It tests a median or whether paired differences are centred on zero. You count positive and negative signs, and under H0 each is equally likely. It uses only direction, not size, so it has low power.