NP Tests

2. Non Parametric Presentation
• Friedman Test
&
• Quade Test
Group
Members
Iqra Tanveer (05)
Anwar Ul Haq (28)
Ameer Umar Khan (30)
Irfan Hussain (39)
Ammar Ahmad Khan (45)

3. Learning Objectives
• History
• Introduction
• Assumptions
• General Procedure
• Applications
• Advantages
• Disadvantages
• Example

4. History
• Friedman,
Milton (December 1937).
• Friedman, Milton (March
1940).
• “A comparison of
alternative tests of
significance for the problem
of m rankings“.
• Kendall, M. G. Rank
Correlation Methods. (1970)
London.

5. History
• Hollander, M., and Wolfe, D. (1973). New
York.
• Siegel, Sidney, and Castellan, N. John
Nonparametric Statistics for the Behavioral
Sciences. (1988).

6. Introduction

7. Introduction

8. Introduction
• The Friedman test is a non-parametric statistical
test developed by the U.S. economist Milton
Friedman.
• Friedman test is a non-parametric randomized
block analysis of variance.
• Similar to the parametric repeated
measures ANOVA. It is used to detect differences
in treatments across multiple test attempts.

9. Introduction
• The procedure involves ranking each row
(or block) together, then considering the values
of ranks by columns. Applicable to complete
block designs.
• It is thus a special case of the Durbin test.
• The Friedman test is used for one-way repeated
measures analysis of variance by ranks. In its use
of ranks it is similar to the Kruskal-Wallis one-
way analysis of variance by ranks.

10. Introduction
• It is non-parametric test in which we use the
f-table for critical region.

11. Assumptions
• Data should consist of three or more than three
samples.
• Data should be consist of random samples from
population.
• All samples data should be independent.
• Measurement scale should be at least ordinal.
• Variable of interest should be continuous.
• Data need not be normally distributed.
• Within each block the observation may be rank
according to some criteria of interest.

12. General Procedure
1. Null & Alternative Hypothesis.
Ho: The distributions (whatever they are)
are the same across repeated measures.
H1: The distributions across repeated
measures are different.
2. Level Of Significance.
α = 0.01, 0.05, 0.01…. (chose as required
for test)

13. General Procedure
3. Test Statistics
Step 1
If no ties occur in data.
T1 =
12
bk(k + 1)
j=1
k
Rj −
b(k + 1)
2
2

14. General Procedure
If ties occur in data
T1 =
(k − 1) j=1
k
Rj
2
− bC1
A1 − C1
Where
A1 = i=1
b
j=1
k
R(xij)
2
And
C1 =
bk(k + 1)2
4

15. General Procedure
Step 2
Put the value of 𝑇1 into this equation
𝑇2 =
(𝑏−1)𝑇1
𝑏(𝑘−1)𝑇1
Where 𝑇2 ↝ 𝐹 𝑘1, 𝑘2 𝛼
And
𝑘1=k-1
𝑘2=(k-1)(b-1)

16. General Procedure
4. Calculation
5. Critical Region
if 𝑇2> 𝐹 𝑘1, 𝑘2 𝛼 Reject Ho
6. Decision
From the provided evidence as our calculated
value is …….. So we …… and conclude
that…….

17. Applications
This can be used to perform the testing in every
field, where comparison between variables is
required.
1. Used to compare the effects of same fertilizer in
different patches of field having different
fertility levels. (In agricultural Field)
2. Comparison between different companies cold
drinks.
3. Test the equality of difference car’s engines
performance. (In industries)

18. Applications
4. Comparison between the average
performance of players. (Games)
5. Comparison of different pain-killer tablets
average effect.
So this is valuable test used as non-
parametric test of multiple comparison. Where
data is not normally distributed. That is
assumption of normality is violated.

19. Advantages
1. Since the Friedman test ranks the values in
each row, it is not affected by sources of
variability that equally affect all values in a
row (since that factor won't change the ranks
within the row).
2. The test controls experimental variability
between subjects, thus increasing the power
of the test.

20. Disadvantage
• Since this test does not make a distribution
assumption, it is not as powerful as the
ANOVA.

21. Example
A B C D
4 3 2 1
4 2 3 1
3 1.5 1.5 4
3 1 2 4
4 2 1 3
2 2 2 4
1 3 2 4
2 4 1 3
3.5 1 2 3.5
4 1 3 2
4 2 3 1
3.5 1 2 3.5
38 23.5 34.5 30
Ranked Data

22. Solution
1. Null & Alternative Hypothesis.
Ho: The distributions (whatever they are)
are the same across repeated measures.
H1: The distributions across repeated
measures are different.
2. Level Of Significance.
α=0.05

23. Solution
3. Test Statistics
Because ties occur in data so….
𝑇1 =
(𝑘 − 1) 𝑗=1
𝑘
𝑅𝑗
2
− 𝑏𝐶1
𝐴1 − 𝐶1
4. Calculation
calculate the 𝑨 𝟏 & 𝑪 𝟏

24. Solution
As we know that:
𝐴1 =
𝑖=1
𝑏
𝑗=1
𝑘
𝑅(𝑥𝑖𝑗)
2
And
𝐶1 =
𝑏𝑘(𝑘+1)2
4
By using these formulas we calculate that
𝐴1=456.5 𝐶1=300

25. Solution
𝑇1 =
(𝑘 − 1) 𝑗=1
𝑘
𝑅𝑗
2
− 𝑏𝐶1
𝐴1 − 𝐶1
𝑇1 = 8.097
Put 𝑇1 in 𝑇2 formula
𝑇2 =
(𝑏−1)𝑇1
𝑏(𝑘−1)𝑇1
=5.6006

26. Solution
Critical Value
𝑇2 ↝ 𝐹 𝑘1, 𝑘2 𝛼
And
𝑘1=k-1
𝑘2=(k-1)(b-1)
CR=2.8742

27. Solution
Decision:
From the provided evidence as our
calculated is greater than our tabulated value.
so, we will reject the H0 and hence concluded
that the distributions across repeated measures
are different.

28. Multiple Comparison Test
𝑅𝑗 − 𝑅𝑖 ≥ 𝑡1− 𝛼
2
𝐴1−𝐶1
(𝑏−1)(𝑘−1)
1 −
𝑇1
𝑏(𝑘−1)
1
2
with d.f 𝑡(𝑘−1)(𝑏−1)
• If ties Occur
𝐴1 − 𝐶1 =
𝑏𝑘(𝑘+1)(𝑘−1)
12