Non-parametric tests

CIE A-Level Further Mathematics· 15 min read

1. When to use non-parametric tests★★☆☆☆⏱ 3 min

Non-parametric (distribution-free) tests do not require the underlying population to follow a specific known distribution. They are the appropriate choice when assumptions required for parametric tests are violated.

📘 Definition

Non-parametric test

A hypothesis test that makes no assumptions about the shape or parameters of the population distribution the sample is drawn from.

Example:

Used when testing the median of a heavily skewed population with small sample size.

Data is ordinal (ranked but not continuous numerical)
Population distribution is known to be non-normal
Sample size is too small to confirm normality
Testing a claim about a population median rather than mean

📐 Worked Example

A researcher wants to test if the median annual income in a town is greater than $50,000. Incomes are heavily right-skewed, sample size is 7. Should a non-parametric test be used?

1
Check assumptions for a parametric t-test for the mean:
2
The population distribution is heavily skewed, so normality assumption is violated. Small sample size means the Central Limit Theorem does not apply.
3
The claim is about the median, not the mean.
4
Conclusion: Yes, a non-parametric test is appropriate here.

2. Sign Test★★☆☆☆⏱ 4 min

📘 Definition

Sign Test

$S$ = number of positive signs, $n$ = number of non-zero differences

A simple non-parametric test used to test claims about a population median, or median difference for paired data.

Example:

Testing whether median customer satisfaction score is above 3/5.

The test works by calculating the difference between each observation and the null median. Values above the null get a $+$ sign, values below get a $-$ , values equal are discarded. Under the null hypothesis, $S \sim B (n, 0.5)$ .

📐 Worked Example

Test at 5% significance whether the median mass of a rabbit is greater than 1.5kg. A sample of 8 rabbits gives masses: [1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9]

1
State hypotheses: $H_{0} : \tilde{μ} = 1.5$ , $H_{1} : \tilde{μ} > 1.5$ , where $\tilde{μ}$ is population median.
2
Calculate differences and signs: 1 zero difference (discarded), 3 negative differences, 4 positive differences. $n = 7$ , $S = 4$ .
3
Under $H_{0}$ , $S \sim B (7, 0.5)$ . Calculate $P (S \geq 4) = 0.5$ .
4
Compare to significance: $0.5 > 0.05$ .
5
Conclusion: Insufficient evidence to reject $H_{0}$ , cannot conclude median mass is greater than 1.5kg.

3. Wilcoxon Signed-Rank Test★★★☆☆⏱ 4 min

The Wilcoxon signed-rank test is a more powerful alternative to the sign test for paired data or single sample median tests, when differences are approximately symmetric. It accounts for the magnitude of differences, not just their sign.

📘 Definition

Wilcoxon Signed-Rank Test

Non-parametric test for median difference in paired data that uses rank of difference magnitudes to calculate the test statistic.

Example:

Testing whether a new fertilizer increases median crop yield compared to an old fertilizer, paired by field.

📐 Worked Example

6 paired observations have differences, with positive differences having ranks 2, 3, 5, 6 and negative differences having ranks 1, 4. Calculate the test statistic $T_{+}$ and test at 5% one-tailed, critical value = 1.

1
Sum the ranks of all positive differences to get $T_{+}$ : $T_{+} = 2 + 3 + 5 + 6 = 16$ .
2
Recall the rejection rule for CIE exams: reject $H_{0}$ if $T_{+} \leq$ critical value.
3
Compare: $16 > 1$ , so we fail to reject the null hypothesis.
4
Conclusion: There is insufficient evidence at 5% significance that median difference is greater than zero.

4. Mann-Whitney U Test★★★★☆⏱ 4 min

The Mann-Whitney U test is the non-parametric alternative to the unpaired two-sample t-test, used to test for a difference in medians between two independent samples.

📘 Definition

Mann-Whitney U Test

For sample sizes $n_{1}, n_{2}$ , $U_{1} = R_{1} - \frac{n _{1} ( n _{1} + 1 )}{2}$ , $U_{2} = n_{1} n_{2} - U_{1}$ , $U = min (U_{1}, U_{2})$

Non-parametric test for difference in medians between two independent samples.

📐 Worked Example

Two independent samples have sizes $n_{1} = 4$ , $n_{2} = 5$ . The sum of ranks for sample 1 is $R_{1} = 14$ . Calculate the test statistic U, for a two-tailed 5% test with critical value 2.

1
Calculate $U_{1}$ using the formula: $U_{1} = R_{1} - \frac{n _{1} ( n _{1} + 1 )}{2} = 14 - \frac{4 \times 5}{2} = 14 - 10 = 4$ .
2
Calculate $U_{2} = n_{1} n_{2} - U_{1} = (4 \times 5) - 4 = 20 - 4 = 16$ .
3
Test statistic is the smaller of $U_{1}$ and $U_{2}$ : $U = min (4, 16) = 4$ .
4
Reject $H_{0}$ if $U \leq$ critical value: $4 > 2$ , so fail to reject $H_{0}$ .
5
Conclusion: No significant difference between the two population medians at 5% significance.

5. Common Pitfalls

Wrong move:

Using a non-parametric test when parametric test assumptions are satisfied.

Why:

Non-parametric tests have lower statistical power than equivalent parametric tests, so you are more likely to miss a true effect.

Correct move:

Only use non-parametric tests when parametric assumptions are clearly violated, or data is ordinal.

Wrong move:

Keeping zero differences in the sample size n for sign or signed-rank tests.

Why:

Zero differences give no information about whether the median is above or below the null value, so they must be excluded.

Correct move:

Discard all zero differences, and reduce n by the number of discarded values before calculating the test statistic.

Wrong move:

Rejecting the null hypothesis when the test statistic is greater than the critical value for Wilcoxon or Mann-Whitney tests.

Why:

For all rank-based non-parametric tests in CIE 9231, smaller test statistics provide stronger evidence against the null.

Correct move:

Always reject $H_{0}$ if test statistic $\leq$ critical value, for both tests.

Wrong move:

Using the sign test instead of Wilcoxon signed-rank when differences are symmetric.

Why:

The sign test only uses the direction of differences, not their size, so it is less powerful than Wilcoxon when symmetry holds.

Correct move:

Use Wilcoxon signed-rank for paired data when differences are approximately symmetric.

Wrong move:

Assigning different ranks to tied values instead of the average rank.

Why:

Incorrect ranking leads to an incorrect test statistic and wrong conclusion about the null hypothesis.

Correct move:

Assign the average of the relevant ranks to all tied values, then proceed with the test as normal.

6. Quick Reference Cheatsheet

Test Name	Use Case	Test Statistic	Rejection Rule
Sign Test	Single median / paired median difference	$S$ = number of positive differences	Reject $H_{0}$ if p-value $\leq α$
Wilcoxon Signed-Rank	Paired median difference (symmetric)	$T_{+}$ = sum of ranks of positive differences	Reject $H_{0}$ if $T_{+} \leq T_{cr i t}$
Mann-Whitney U	Two independent samples, difference in medians	$U = min (U_{1}, U_{2})$	Reject $H_{0}$ if $U \leq U_{cr i t}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Mann-Whitney U test question
2023 · 2
Wilcoxon signed-rank test
2021 · 2
Justify non-parametric test use

Going deeper

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What's Next

Non-parametric tests fill a key gap in your hypothesis testing toolkit, allowing you to answer research questions when the strict assumptions of parametric tests cannot be met. For CIE 9231 exams, you will often be asked to both justify the use of a non-parametric test and carry out the full test calculation, so practicing ranking and test statistic calculation is critical. The concepts of ranking you learned here also underpin other non-parametric methods you will study next. This sub-topic connects closely to core hypothesis testing concepts you have already mastered, so revisiting those will help you consolidate your understanding of non-parametric tests.