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1258 LearnersLast updated on November 28, 2025

A non parametric test is a statistical approach that does not assume a specific probability distribution for the population. It is used in comparing two quantities, such as the test results of students in two different schools. In this topic, we will explore more about Non-parametric tests in detail.
A non parametric test is a type of statistical hypothesis test that does not depend on assumptions about the population’s distribution. Since these tests are distribution-free, they are useful when data do not meet the requirements of parametric tests or when the sample size is small.
For example, a teacher wants to compare the performance of students taught using Method A and Method B. The class has very few students, and their scores are not normally distributed. Instead of using a t-test, the teacher can use a Mann–Whitney U test, a non-parametric test, to compare the two groups safely.
A non parametric test should be used when the assumptions required for parametric tests are not satisfied. Specifically, you should consider using a non-parametric test in these situations:
Skewed Data: If the data are not normally distributed, the mean may not accurately reflect the data. Parametric tests may give misleading results, so non-parametric tests are preferred.
Small Sample Size: When the dataset is very small, it is difficult to verify the distribution. Non parametric tests are suitable for analyzing such data.
Categorical or Ranked Data: If the data is nominal (names, labels) or ordinal (ranked), parametric tests cannot be applied. Non parametric tests can handle these types of data effectively.
Parametric and Non parametric are the two types of hypothesis testing. Here, we will discuss the key differences between the two methods:
| Parametric Method | Non Parametric Method |
| Assumes a certain probability distribution |
Makes no assumptions about the population distribution. |
| Often tests numerical or normally distributed data. |
Suitable for ordinal, ranked, or skewed data. |
| Analyzes data using key parameters such as mean and standard deviation. |
Does not rely on pre-determined parameters. |
| Due to the rigid assumptions, it is limited to specific datasets. |
More flexible and are applied to real-world data. |
| Examples: ANOVA, t-test, linear regression. |
Examples: Mann-Whitney U test, Kruskal-Wallis test, Signed-rank test. |


The statistical tests that do not rely on a specific distribution for the data are non parametric tests. They are handy for ordinal data, small sample sizes, or data that violate parametric assumptions. The types of non parametric tests are:
Non parametric tests are statistical methods that do not assume a specific data distribution. They are useful for ordinal data or when normality assumptions are violated. The Kruskal-Wallis H-test is a popular non-parametric test used to compare the medians of three or more independent groups. Its formula is:
\(H = \frac{12}{N(N+1)} \sum \frac{R_j^2}{n_j} - 3(N+1) \)
Where H is the Kruskal-Wallis test
N is the total number of observations across all groups
\(n_i\) is the number of observations in group i
\(R_i\) is the sum of ranks for observations in group i
Non parametric tests are statistical methods that do not assume any specific distribution of data. They are especially useful for ordinal or ranked data and are widely applied in social sciences, education, and psychology. Here are the advantages and disadvantages of non parametric tests.
| Advantages | Disadvantages |
| Do not require prior knowledge or assumptions about the population distribution. | Generally, it is less efficient than a parametric test when the parametric assumptions hold. |
| Involve simpler and shorter calculations. | Benign distribution-free, they may produce results with lower accuracy. |
| Easy to understand and interpret. | May not provide precise or reliable estimates in cases where stronger measurement scales are available. |
| Applicable to all types of data, including nominal, ordinal, ranked, or non-nominal data. | It can yield less powerful conclusions than parametric techniques. |
Mastering non parametric tests helps you analyze data that doesn't follow normal distribution. These tips make it easier to choose the right test and interpret results accurately.
Parents can help children understand non-parametric ideas by encouraging simple ranking tasks at home, such as ordering objects by size, height, or preference.
To introduce non parametric concepts, teachers can use real-life examples like survey results or classroom feedback.
Teachers can incorporate visual aids such as charts, box plots, diagrams, and flashcards to help students recognize data patterns.
Students tend to make mistakes while understanding the concept of non-parametric tests. Let us see some of the common mistakes and how to avoid them in non parametric tests:
Non parametric tests are useful when data doesn't fit normal distribution assumptions. They help analyze ranked, categorical, or skewed data across various fields effectively.
Healthcare research: Used to compare patient recovery rates when the data doesn't follow a normal distribution.
Marketing analysis: Helps businesses analyze customer satisfaction surveys where responses are in ranks or categories.
Education: Applied to compare student performance based on grades or rankings rather than exact scores.
Finance: Used to study stock market trends and investment returns when data is not normally distributed.
Psychology: Helps analyze behavioral or emotional response data collected through ordinal scales or observations.
Angel wants to find out whether a new cooking method has improved her business. The sales from four customers were recorded before and after introducing the new method. The data are as follows: the first customer’s sales were 60 before and 90 after, the second customer’s sales were 52 before and 85 after, the third customer’s sales were 43 before and 95 after, and the fourth customer’s sales were 40 before and 82 after. The differences in sales (after minus before) are +30, +33, +52, and +42, respectively. Determine whether the new cooking method has significantly increased sales.
The introduction of the new cooking method has positively impacted business performance.
|
Customer |
Sales Before |
Sales After |
Difference (After-Before) |
Absolute Difference |
Rank |
|
1 |
60 |
90 |
+30 |
30 |
1 |
|
2 |
52 |
85 |
+33 |
33 |
2 |
|
3 |
43 |
95 |
+52 |
52 |
4 |
|
4 |
40 |
82 |
+42 |
42 |
3 |
Here, every rank is given a positive value as each difference is positive.
Calculating Test Statistic:
Sum of positive ranks \((W⁺) = 1 + 2 + 3 + 4 = 10\)
Sum of negative ranks \((W⁻) = 0\) (no negative differences)
The test statistic W is the smaller of \(W⁺\) and \(W⁻\), so W = 0
Comparing with Critical Value
For \(n = 4\) at \(α = 0.05\) (one-tailed test), the critical value from the Wilcoxon table is 0.
Since W (0) ≤ Critical Value (0), we reject the null hypothesis (H₀).
Hence, we can conclude that:
There has been a significant increase in sales because of the application of the new cooking method. This tells us that the technique has positively impacted business performance.
A psychologist wants to study whether screen time affects students’ attention spans. Five students were observed, and their attention spans (in minutes) before and after screen time were recorded as follows: the first student had 15 minutes before and 30 minutes after, the second had 25 minutes before and 45 minutes after, the third had 30 minutes before and 46 minutes after, the fourth had 23 minutes before and 34 minutes after, and the fifth had 35 minutes before and 45 minutes after. Determine whether screen time has a significant effect on students’ attention spans.
If H₀ is rejected, reducing screen time significantly improves attention span.
| Student | Attention Span Before | Attention Span After | Difference (After - Before) |
| 1 | 15 | 30 | 15 |
| 2 | 25 | 45 | 20 |
| 3 | 30 | 46 | 16 |
| 4 | 23 | 34 | 11 |
| 5 | 35 | 45 | 10 |
The Wilcoxon Signed-Rank Test is used.
H₀: No significant change in attention span.
H₁: Reducing screen time will increase attention span.
Here, we rank the differences, and the test statistic (W) is calculated.
If calculated is a value that is below the critical value, reject H₀.
Conclusion: If H₀ is rejected, reducing screen time will significantly improve attention span.
An athlete wants to find out whether a new training method has improved the players’ performance. The performance scores of four players were recorded before and after the new training method. The data are as follows: the first player scored 30 before and 55 after, the second player scored 40 before and 63 after, the third player scored 60 before and 82 after, and the fourth player scored 55 before and 75 after. Determine whether the new training method has significantly increased the players’ performance.
If H₀ is rejected, it means that the training method significantly improves player performance.
| Player | Performance Before | Performance After |
Difference |
| 1 | 30 | 55 | 25 |
| 2 | 40 | 63 | 23 |
| 3 | 60 | 82 | 22 |
| 4 | 55 | 75 | 20 |
We will apply the Wilcoxon Signed-Rank Test:
H₀ (Null Hypothesis): The new training method does not significantly affect performance.
H₁ (Alternative Hypothesis): The new training method improves performance.
Here, the test statistic (W) is compared to the critical value, and the differences are ranked.
Therefore, if H₀ is rejected, it means that the training method significantly improves player performance.
A teacher wants to find out whether a new teaching method improves student understanding based on their feedback. The understanding scores of six students were recorded before and after the method was introduced. The data are as follows: the first student scored 6 before and 9 after, the second scored 4 before and 8 after, the third scored 5 before and 7 after, the fourth scored 7 before and 8 after, the fifth scored 5 before and 6 after, and the sixth scored 8 before and 10 after. The differences in scores (after minus before) and their ranks are also recorded. Determine whether the new teaching method has significantly enhanced student understanding.
The smaller sum of ranks (W) is 0. Since \(W = 0\) is less than the critical value of 2 for \(n = 6\) at \(α = 0.05\), the null hypothesis is rejected. Therefore, the new teaching strategy greatly enhances student comprehension.
| Student | Understanding Score Before | Understanding Score After | Differnce (After - Before) |
Rank |
| 1 | 6 | 9 | 3 | 3 |
| 2 | 4 | 8 | 4 | 5 |
| 3 | 5 | 7 | 2 | 2 |
| 4 | 7 | 8 | 1 | 1 |
| 5 | 3 | 6 | 3 | 3 |
| 6 | 8 | 10 | 2 | 2 |
Sum of positive ranks: 16
Sum of negative ranks: 0
The Wilcoxon Signed-Rank Test is used to evaluate the efficacy of the new teaching strategy.
Significance level \((α) = 0.05 \)
Compare the smaller sum of ranks (W = 0) to the critical value from the Wilcoxon table.
H₀ (Null Hypothesis): No significant improvement in students’ comprehension.
H1 (Alternative Hypothesis): The comprehension of the particular subject has greatly increased as a result of the new teaching strategy.
A fitness trainer wants to find out whether a new workout routine improves participants’ flexibility. The flexibility scores (in centimeters) of six participants were recorded before and after the training program. The data are as follows: the first participant scored 12 before and 18 after, the second scored 15 before and 17 after, the third scored 10 before and 14 after, the fourth scored 14 before and 19 after, the fifth scored 13 before and 16 after, and the sixth scored 11 before and 15 after. Determine whether the new workout routine has significantly increased the participants’ flexibility.
In the below given table we can see that the new workout routine significantly improves flexibility
| Players | Flexibility Score Before | Flexibility Score After |
Difference (After - Before) | Absolute Difference | Rank |
| 1 | 12 | 18 | 6 | 6 | 6 |
| 2 | 15 | 17 | 2 | 2 | 1 |
| 3 | 10 | 14 | 4 | 4 | 3.5 |
| 4 | 14 | 19 | 5 | 5 | 5 |
| 5 | 13 | 16 | 3 | 3 | 2 |
| 6 | 11 | 15 | 4 | 4 | 3.5 |
As all the differences are positive, the sum of positive ranks is
\(W^+ = 1 + 2 + 3.5 + 3.5 + 5 + 6 = 21\)
The sum of the negative rank is 0
Compare with the critical value; the smaller sum of rank(W) = 0
From the Wilcoxon Signed Rank Test Table, the critical value for n = 6 at α = 0.05 = 2
As, \(W = 0 < 2\), we reject the null hypothesis
Since the test statistic W = 0 is smaller than the critical value 2, we reject the null hypothesis. This means there is significant evidence that the new workout routine improves flexibility.
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!






