Non Parametric Test by Vikramjit Singh

NON-PARAMETRIC
TESTS
DR. VIKRAMJIT SINGH

NON PARAMETRIC TESTS:
1
Non-parametric tests, also known as distribution-
free tests, are used in statistics when the
assumptions of parametric tests (tests that assume
a specific probability distribution for the data) are
not met or when dealing with ordinal or nominal
data.

WHEN TO USE A
NONPARAMETRIC TEST
2
Nonparametric tests are sometimes called distribution-free tests
because they are based on fewer assumptions (e.g., they do not
assume that the outcome is approximately normally distributed).
Parametric tests involve specific probability distributions (e.g., the
normal distribution) and the tests involve estimation of the key
parameters of that distribution (e.g., the mean or difference in
means) from the sample data. The cost of fewer assumptions is that
nonparametric tests are generally less powerful than their
parametric counterparts (i.e., when the alternative is true, they may
be less likely to reject H0).

WHEN TO USE A
NONPARAMETRIC TEST
2
It can sometimes be difficult to assess whether a continuous
outcome follows a normal distribution and, thus, whether a
parametric or nonparametric test is appropriate. There are
several statistical tests that can be used to assess whether data
are likely from a normal distribution. The most popular are the
Kolmogorov-Smirnov test, the Anderson-Darling test, and the
Shapiro-Wilk test1.

WHEN TO USE A
NONPARAMETRIC TEST
2
when the outcome is an ordinal variable or a rank,
when there are definite outliers or
when the outcome has clear limits of detection.
The most practical approach to assessing normality involves investigating
the distributional form of the outcome in the sample using a histogram
and to augment that with data from other studies, if available, that may
indicate the likely distribution of the outcome in the population.
There are some situations when it is clear that the outcome does not
follow a normal distribution. These include situations:

SIGN TEST
3.1
Sign test is used to test a null hypothesis that the
median of a distribution is equal to some value it can be
used in place of one sample t test , paired sample t-test
orfor other categorical data where are numerical scale
is inappropriate but where it is possible to rank the
observation.

PROCEDURE FOR CARRYING OUT THE
SIGN TEST.
3.1 The observation in a sample of size n , are x1, x2, x3,…,xn(these
observation can be paired observation).The null hypothesis is that the
population median is equal to some value M. Suppose that r+ of the
observations are exactly greater than M and r- are smaller than M (For
a sign test in place of paired t-test, Median will be equal to zero), The
value of x which are exactly equal to M are ignored. The sum r+ +r−
may therefore be less than n—we will denote it by n′. Under the null
hypothesis we would expect half of the x’s to be above the median and
half below. therefore under the null hypothesis both r+ and r- follow a
binomial distribution with p=1/2 and n=n'.

SIGN TEST.
Example
Table describes below the level of attention span of the students
during the two periods as observed in minutes.
3.1

SIGN TEST.
Example
R+ = 5, R- =6
Min(R+,R-)= 5, n'=11
From the binomial table of Sign Test expansion for n'-11(Critical value is 1).
Is not significant at 0.05 level.
Use: To compare a continuous outcome in two matched or paired samples.
Null Hypothesis: H0: Median difference is zero
Test Statistic: The test statistic is the smaller of the number of positive or
negative signs.
Decision Rule: Reject H0 if the smaller of the number of positive or negative
signs < critical value from table.
3.1

CHI-SQUARE TEST
3.2 The chi-square test is a statistical test used to determine if there is a
significant association or relationship between two categorical
variables. It is often used to assess the independence of variables in a
contingency table, but there are different types of chi-square tests,
depending on the specific research question and the nature of the data
.
The concept of a chi-square test revolves around comparing the
observed frequencies of certain events in categorical data with the
expected frequencies under a null hypothesis of independence. If the
observed and expected frequencies are significantly different, it
suggests that there is a relationship between the variables.

CHI-SQUARE TEST
3.2
Steps for Conducting a Chi-Square Test:
1. Formulate the Null Hypothesis (H0) and Alternative Hypothesis (H1):
Start by stating the null hypothesis, which typically assumes that there
is no association or relationship between the variables. The alternative
hypothesis suggests that there is a significant relationship.
2. Create a Contingency Table: Construct a contingency table, also
known as a cross-tabulation or a two-way table. This table displays the
frequency or counts of observations for each combination of categories
of the two variables.

CHI-SQUARE TEST
3.2
Here's a simple example of a contingency table:
In this example, we'll test whether there's an association between
gender and the preference for pink or blue colour among a group of
students.
Null Hypothesis (H0):There is no association between gender and the
preference for pink or blue as their colour choice.
Alternative Hypothesis (H1):There is an association between gender
and the preference for pink or blue as their colour choice.

CHI-SQUARE TEST
3.2

CHI-SQUARE TEST
3.2
Step 3: Calculate Expected Frequencies
Calculate the expected frequencies for each cell under the assumption
of independence. You can do this by using the formula:
Expected Frequency = (row total * column total) / grand total
For example, the expected frequency for the "Boys" and "Pink" cell
would be:
Expected Frequency = (100 * 70) / 200 = 35

CHI-SQUARE TEST
3.2
Step 4: Calculate the Chi-Square Statistic
For each cell in the table, calculate the chi-square statistic using the
formula:
Chi-Square = Σ((O - E)^2 / E),
where O is the observed frequency and E is the expected frequency.
For the "Boys" and "Pink" cell, the calculation is:
Chi-Square = ((40 - 35)^2 / 35) = 0.714
Repeat this calculation for all cells and sum up the values to get the
total chi-square statistic.

CHI-SQUARE TEST
3.2 Step 5: Determine the Degrees of Freedom
Degrees of freedom (df) for a chi-square test of independence is calculated as (number of
rows - 1) * (number of columns - 1), which in this case is (2 - 1) * (2 - 1) = 1.
Step 6: Look up the Critical Value
You would use a chi-square distribution table or a statistical calculator to find the critical
chi-square value for your chosen level of significance (alpha) and degrees of freedom. Let's
assume you chose alpha = 0.05, which corresponds to a critical value of approximately 3.841.
Step 7: Compare the Calculated Chi-Square Statistic
In this step, you compare the calculated chi-square statistic (from step 4) to the critical
value (from step 6). If the calculated chi-square statistic is greater than the critical value,
you reject the null hypothesis. If it's less than or equal to the critical value, you fail to
reject the null hypothesis.
Step 8: Draw a Conclusion

THE MCNEMAR TEST
3.3 The McNemar test is a non-parametric test used to analyze paired
nominal data. It is a test on a 2 x 2 contingency table and checks the
marginal homogeneity of two dichotomous variables. The test requires
one nominal variable with two categories (dichotomous) and one
independent variable with two dependent groups.
The McNemar Test is a statistical test used to analyze the association
or dependency between two paired categorical variables. It's often
employed when you have two related groups of subjects and want to
determine if there is a significant change in the distribution of a
particular attribute between the two time points, conditions, or
treatments. Here's an overview of the concept, steps, and an example
problem with a solution:

THE MCNEMAR TEST
3.3 The McNemar Test is used to answer questions like, "Is there a
significant change in a condition from time 1 to time 2?" or "Is there a
significant difference in outcomes for two related groups?" It works
with a contingency table that displays the counts of four different
scenarios, typically denoted as a, b, c, and d.
- a: Subjects with the attribute at both time points.
- b: Subjects with the attribute at time 1 but not at time 2.
- c: Subjects with the attribute at time 2 but not at time 1.
- d: Subjects without the attribute at both time points.

THE MCNEMAR TEST
3.3 Steps:
1. Construct a 2x2 contingency table with counts for a, b, c, and d.
2. Calculate the McNemar statistic: (|b - c|)^2 / (b + c).
3. Use the McNemar statistic to obtain a p-value from a chi-squared
distribution with 1 degree of freedom.
4. Compare the p-value to your chosen significance level (e.g., α =
0.05).
Use:
- McNemar Test is often used in medical and social sciences to compare
before-and-after scenarios or matched pairs.
- It's appropriate for cases where other tests, like the chi-squared test,
might not be suitable because the data is dependent or paired.

THE MCNEMAR TEST
3.3 Suppose you are conducting a study on the effectiveness of a new teaching
method. You have 100 students, and you want to know if there's a significant
difference in the number of students who passed an exam before and after the
new method was implemented.
Contingency Table:
Before After
Pass 45 30
Fail 55 70
Calculations:
- a = 30 (students who passed both times)
- b = 45 - 30 = 15 (students who passed before but not after)
- c = 55 - 30 = 25 (students who passed after but not before)
- d = 100 - (a + b + c) = 30 (students who failed both times)

THE MCNEMAR TEST
McNemar statistic = (|15 - 25|^2) / (15 + 25) = 100 / 40 = 2.5
Using a chi-squared distribution with 1 degree of freedom, you can find the p-
value for McNemar statistic (2.5). Let's assume the p-value is 0.05.
Since p-value (0.05) < α (0.05), you would conclude that there's a significant
difference in the number of students passing before and after the new
teaching method.
3.3

MANN WHITNEY U TEST
(WILCOXON RANK SUM TEST)
Use: To compare a continuous outcome in two independent samples.
Null Hypothesis: H0: Two populations are equal
Test Statistic: The test statistic is U, the smaller of
where R1 and R2 are the sums of the ranks in groups 1 and 2, respectively.
Decision Rule: Reject H0 if U < critical value from table
3.4

MANN WHITNEY U TEST
3.4

MANN WHITNEY U TEST
3.4
TABLE VALUE FOR N = 7, CRITICAL VALUE IS 8
SO, U1 = 15, U2=34
USING THE FORMULA U1= 15, U2= 34,
CRITICAL VALUE FROM TABLE FOR N= 7 IS 8 OR OBTAINED VALUE IS
LOWER TERM OF U1 AND U2, I.E. 15
HENCE SINCE U(15) IS NOT <=8, NULL HYPOTHESIS IS ACCEPTED.

THE FRIEDMAN TEST
3.5
THE FRIEDMAN TEST IS A NON-PARAMETRIC STATISTICAL TEST
USED TO DETERMINE IF THERE ARE STATISTICALLY SIGNIFICANT
DIFFERENCES BETWEEN MULTIPLE RELATED GROUPS. IT'S
TYPICALLY USED WHEN YOU HAVE A SINGLE DEPENDENT VARIABLE
MEASURED ON THREE OR MORE RELATED GROUPS, AND YOU WANT
TO ASSESS IF THERE ARE DIFFERENCES IN THE CENTRAL
TENDENCIES (E.G., MEDIANS) OF THESE GROUPS.

THE FRIEDMAN TEST
3.5
Here are the steps to perform the Friedman Test:
1. Formulate the null and alternative hypotheses:
- Null Hypothesis (H0): There are no differences between the groups.
- Alternative Hypothesis (Ha): There are differences between the groups.
2. Collect Data: Gather data from your multiple related groups. Each group should have the
same individuals or subjects measured under different conditions or at different time
points.
3. Rank the data: Rank the data within each group from lowest to highest. Ties should be
assigned the average rank. If there are repeated values, they should have the same rank.

THE FRIEDMAN TEST
4. Calculate the Friedman statistic (χ²):This statistic is used to test the null hypothesis.
The formula is:
Where:
- N is the number of subjects or observations., - k is the number of related groups.
- (R_j) is the sum of ranks in the jth group.
5. Determine the critical value: Based on your chosen level of significance (e.g., 0.05),
consult a chi-squared distribution table to find the critical value for the degrees of
freedom (df) of (k-1).
6. Compare the calculated χ² to the critical value: If χ² is greater than the critical value,
you can reject the null hypothesis and conclude that there are statistically significant
differences between the groups.
7. Post-hoc analysis (if necessary): If you reject the null hypothesis, you can perform
post-hoc tests (e.g., Nemenyi test) to determine which specific groups differ from each
other.
3.5

THE FRIEDMAN TEST
3.5
Uses of the Friedman Test:
- It is used when you have repeated measurements on the same
subjects.
- It's appropriate when the assumptions of parametric tests like
ANOVA are violated.
- It can be applied in various fields, such as medicine, social
sciences, and business, to compare treatments or interventions.

THE FRIEDMAN TEST
Suppose you are an educator and want to determine if three different teaching
methods (A, B, and C) have a statistically significant effect on the exam scores of your
students. You collect the scores of 15 students who were taught using each method.
The data is as follows:
3.5

THE FRIEDMAN TEST
- Null Hypothesis (H0): There are no significant differences in the teaching
methods' effectiveness (median exam scores are equal for all methods).
- Alternative Hypothesis (Ha): There are significant differences in the teaching
methods' effectiveness (median exam scores are not equal for at least one pair of
methods).
Calculate the Friedman statistic (χ²):Use the formula for the Friedman statistic:
Where:
- N = Total number of students= 15
- k = Number of methods (3)
Plugging in the values:
Χ2 = 12/ 15*3*4 [ 302+152+452] – 3*15*4
= (12/180 )*3150 – 180
=210-180 = 30
3.5

THE FRIEDMAN TEST
- Determine the critical value:
At a 0.05 significance level and degrees of freedom (df) = 3 - 1 = 2, the critical
value is approximately 5.991 from the chi-squared distribution table.
Compare χ² to the critical value:
Since χ² (30) is greater than the critical value (5.991), we reject the null
hypothesis.
Conclusion:
There are significant differences in the teaching methods' effectiveness in
improving exam scores.
If you wish to identify which methods are significantly different from each other,
you can perform post-hoc tests such as the Nemenyi test.
3.5

WILCOXON SIGNED RANK TEST
-The Wilcoxon Signed Rank Test, also known as the Wilcoxon T test,
is a non-parametric statistical test used to determine if there are
significant differences between two related or paired groups. It is
often used when the data do not meet the assumptions of a
parametric test or when the data are ordinal or not normally
distributed. This test assesses whether the medians of the two
groups are significantly different.
3.6

The Wilcoxon Signed Rank Test is applied when you have
paired data, meaning that each data point in one group is
related to a specific data point in the other group. It's
commonly used in educational research to compare the
performance of students before and after an intervention,
or to compare the effectiveness of two teaching methods
on the same group of students.
3.6

1. Formulate the null and alternative hypotheses:
- Null Hypothesis (H0): There is no significant difference between the two
groups.
- Alternative Hypothesis (Ha): There is a significant difference between the
two groups.
2. Collect data: Gather data from the two related groups, typically
represented as before-and-after measurements.
3. Calculate the differences: Find the differences between the pairs of data
points. If the post-intervention score is greater, assign a positive sign to the
difference; if it's smaller, assign a negative sign.
4. Rank the absolute differences: Rank the absolute values of the
differences from smallest to largest, ignoring the signs. Ties should receive
the average rank.
3.6

5. Calculate the test statistic (W+ or W-): Depending on the number of pairs,
calculate the Wilcoxon statistic (W+ for even sample size, W- for odd sample
size). The test statistic is determined by the sum of the ranks of the positive
differences (W+) or the negative differences (W-).
6. Determine the critical value: Based on your chosen level of significance (e.g.,
0.05), consult a Wilcoxon Signed Rank Test table or calculator to find the critical
value.
7. Compare the calculated test statistic to the critical value: If the test statistic
is less than or equal to the critical value, you can reject the null hypothesis and
conclude that there is a significant difference between the two groups.
8. Interpret the results: If you reject the null hypothesis, you can conclude that
there is a significant difference between the two groups. You can also compute
the effect size to measure the practical significance of the difference.
3.6

A teacher wants to determine if a new teaching method has a significant
impact on students' test scores. The teacher collects test scores from 10
students before and after implementing the new method. The data is as
follows:
Before Teaching Method :
75, 80, 65, 90, 72, 88, 70, 68, 78, 82
After Teaching Method :
80, 85, 70, 95, 75, 92, 74, 72, 84, 88
Solution:
Step 1: Formulate the hypotheses (H0 and Ha) as described above.
Step 2: Calculate the differences between the paired data points:
3.6

3.6
Step 3: Rank the absolute differences and calculate the Wilcoxon statistic (W-):
Ranks of Absolute Differences:
|-6, -6, -5, -5, -5, -4, -4, -4, -3, -4|
W- = Sum of ranks of negative differences = -37
Step 4: Determine the critical value:
At a 0.05 significance level, consult a Wilcoxon Signed Rank Test table for n=10
(number of pairs) to find the critical value. For a two-tailed test, the critical value is
approximately 18.
Step 5: Compare the calculated test statistic to the critical value:
Since W- (-37) is less than the critical value (-18), we reject the null hypothesis.

3.6 Step 6: Interpret the results:
There is a significant difference in test scores before and after implementing the
new teaching method. This suggests that the new method has had a significant
impact on the students' performance.
Keep in mind that you can also calculate an effect size (e.g., Cohen's d) to
measure the practical significance of the difference between the two groups.Use:
To compare a continuous outcome in two matched or paired samples.
Null Hypothesis: H0: Median difference is zero
Test Statistic: The test statistic is W, defined as the smaller of W+ and W- which
are the sums of the positive and negative ranks of the difference scores,
respectively.
Decision Rule: Reject H0 if W < critical value from table.

KRUSKAL WALLIS TEST
3.7 The Kruskal-Wallis test is a non-parametric statistical test
used to determine if there are significant differences
between three or more independent groups. It's used when
the assumptions of parametric tests like one-way ANOVA
(analysis of variance) are not met, or when the data is
ordinal or not normally distributed. This test compares the
medians of the groups and can be thought of as a non-
parametric equivalent to ANOVA.

KRUSKAL WALLIS TEST
3.7 The Kruskal-Wallis test assesses whether there are
significant differences among the groups' central
tendencies (medians). It is applied when you have three or
more independent groups (or conditions) and a continuous
or ordinal dependent variable. This test is commonly used
in research to compare the effects of different treatments
or interventions on an outcome.

KRUSKAL WALLIS TEST
3.7 1. Formulate the null and alternative hypotheses:
- Null Hypothesis (H0): There are no significant differences among the
groups.
- Alternative Hypothesis (Ha): There are significant differences among
the groups.
2. Collect data: Gather data from the independent groups, making
sure the groups are mutually exclusive and independent.
3. Rank the data: Combine all data points from all groups and rank
them from lowest to highest. Ties should be assigned the average
rank.

KRUSKAL WALLIS TEST
3.7 4. Calculate the Kruskal-Wallis test statistic (H): This statistic is used to test the null
hypothesis. The formula is quite complex, but it essentially compares the observed
differences between groups to the expected differences under the assumption of no
differences. This statistic follows a chi-squared distribution.
5. Determine the critical value: Based on your chosen level of significance (e.g., 0.05) and the
degrees of freedom, consult a chi-squared distribution table to find the critical value.
6. Compare the calculated H statistic to the critical value: If H is greater than the critical
value, you can reject the null hypothesis and conclude that there are significant differences
among the groups.
7. Post-hoc analysis (if necessary): If you reject the null hypothesis, you can perform post-hoc
tests (e.g., Dunn's test or Mann-Whitney U tests) to determine which specific groups differ
from each other.

KRUSKAL WALLIS TEST
3.7 A researcher wants to determine if three different types of teaching
method (A, B, and C) have a significant impact on the achievement of
the students. s:
Method A:
23, 28, 25, 24, 26, 27, 29
Method B:
31, 33, 32, 30, 29, 34, 35
Method C:
20, 22, 21, 18, 19, 23, 24

KRUSKAL WALLIS TEST
3.7 Step 1: Formulate the hypotheses (H0 and Ha) as described above.
Step 2: Combine all the data and rank it:
All Data: 18, 19, 20, 21, 22, 23, 23, 24, 24, 25, 26, 27, 28, 29, 29, 30, 31,
32, 33, 34, 35
Step 3: Calculate the Kruskal-Wallis test statistic (H). The calculations
are complex but can be done using software or statistical calculators.
In this case, let's assume that H = 8.97.
Step 4: Determine the critical value. The degrees of freedom for the
Kruskal-Wallis test are equal to the number of groups minus 1, so df =
3 - 1 = 2. At a 0.05 significance level, the critical value for df = 2 is
approximately 5.991 from the chi-squared distribution table.

KRUSKAL WALLIS TEST
3.7 Step 5: Compare the calculated H statistic (8.97) to the critical value
(5.991). Since 8.97 is greater than 5.991, we can reject the null
hypothesis.
Step 6: Interpret the results: There are significant differences in the
achievement between at least one pair of teaching methods. Post-
hoc tests or further analysis can be conducted to identify which two
methods are significantly different.
The Kruskal-Wallis test helps determine whether there are
differences among the groups, and if it's significant, additional tests
are performed to understand where these differences lie.

KRUSKAL WALLIS TEST
To compare a continuous outcome in more than two independent
samples.
Null Hypothesis: H0: k population medians are equal
Test Statistic: The test statistic is H,
where k=the number of comparison groups, N= the total sample
size, nj is the sample size in the jth group and Rj is the sum of the
ranks in the jth group.
Decision Rule: Reject H0 if H > critical value
3.7

RUN TEST
-
Here's the concept and the steps of the runs test:
Concept:
A "run" in this context refers to a sequence of consecutive data
points with the same value. The runs test assesses whether the
pattern of runs in the data is consistent with randomness or if it
suggests a systematic trend or pattern.
3.8

RUN TEST
- Use: To test for randomness in a sequence of data.
- Example: You are analyzing the sequence of heads (H) and tails
(T) in a series of coin flips to determine if the coin is fair or biased.
A runs test, also known as the Wald-Wolfowitz runs test, is a
statistical test used to determine if a sequence of data points
exhibits a random pattern or displays a non-random pattern or
structure. This test is particularly useful when examining a
sequence of binary outcomes, like heads or tails in a coin toss, or
when checking for trends or patterns in data. The test evaluates
whether the data alternates between two states more or less
frequently than expected by chance.
3.8

Non Parametric Test by Vikramjit Singh

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