Non-parametric test.

3. Introduction

4. •Formula
)1(
6
1 2
1
2



nn
d
r
n
i
i
s

5. •Formula Explanation

6. Six steps for test.
(1) Check the null and alternative
hypothesis.
Ho: there is a relation ship.
Hi: There is no relation ship.
(2) Level of significance. α=0.05 and 0.01%.
(3) Test Statistics.
Spearman's rank correlation.
(4) Calculation.
Arrange the values in ascending order.

7. (5) Critical region.
If Dcal ≥ Table values
DCAL =? , Table value =?

8. (6) Conclusion
• Check the result and interpret.

9. •Introduction to charles
Spearmans
• Born
• Charles Edward Spearman
(1863-09-10)10 September 1863
London, United Kingdom
• Died
• 17 September 1945(1945-09-17)
(aged 82)
London, United Kingdom
• Institutions
• University College London

10. Charles Spearmans Picter

11. Charles Edward Spearman, (10
September 1863 – 17 September
1945) was an English psychologist
known for work in statistics, as a
pioneer of factor analysis, and for
Spearman's rank correlation
coefficient. He also did seminal work
on models for human intelligence,
including his theory that disparate
cognitive test scores reflect a single
General intelligence factor and
coining the term g factor.

12. •Biography
• Spearman had an unusual background
for a psychologist. In his childhood he
was ambitious to follow an academic
career. He first joined the army as a
regular officer of engineers in August
1883, and was promoted to captain on 8
July 1893, serving in the Munster
Fusiliers. After 15 years he resigned in
1897 to study for a PhD in experimental
psychology. In Britain, psychology was
generally seen as a branch of philosophy
and Spearman chose to study in Leipzig
under Wilhelm Wundt, because
Spearman had no conventional
qualifications and Leipzig had liberal
entrance requirements. There he met
Krueger and Wirth, both of whom he
admired.

13. He started in 1897, and after some interruption
(he was recalled to the army during the Second
Boer war, and served as a Deputy Assistant
Adjutant General from February 1900) he
obtained his degree in 1906. He had already
published his seminal paper on the factor
analysis of intelligence (1904). Spearman met
and impressed the psychologist William
McDougall who arranged for Spearman to
replace him when he left his position at
University College London. Spearman stayed at
University College until he retired in 1931.
Initially he was Reader and head of the small
psychological laboratory. In 1911 he was
promoted to the Grote professorship of the
Philosophy of Mind and Logic. His title changed
to Professor of Psychology in 1928 when a
separate Department of Psychology was
created.

14. •Example
In this example, the raw data
in the table below is used to
calculate the correlation
between the IQ of a person
with the number of hours
spent in front of TV per
week.

15. •Solution
S.no Xi(IQ) Yi(Hour of TV per week)
1 106 7
2 86 0
3 100 27
4 101 50
5 99 28
6 103 29
7 97 20
8 113 12
9 112 6
10 110 17

16. •Solution steps
(1) The data by the first column
(Xi). Create a new
column(xi)and assign it the
ranked values 1,2,3,...n.
(2) Next the data by the second
column (Yi). Create a fourth
column and similarly assign it
the ranked values 1,2,3,...n.
(3) Create a fifth column (di) to
hold the differences between
the two rank columns (xi and
yi).
(4) Create one final column(di²)to
hold the value of
column(di)squared.

17. •Six steps
(1) Ho: There is relation ship.
HI: There is no relation ship.
(2) Level of significance. α=0.05
(3) Test Statistics.
Spearman's rank correlation.
(4) Calculation.
Arrange the data in ascending
order.
Then calculate differences and
calculate (Di²)

18. •Calculated Table
S.no X Y X(Rank) Y(Rank) d d²
1 86 0 1 1 0 0
2 97 20 2 6 -4 16
3 99 28 3 8 -5 25
4 100 27 4 7 -3 9
5 101 50 5 10 -5 25
6 103 29 6 9 -3 9
7 106 7 7 3 4 16
8 110 17 8 5 3 9
9 112 6 9 2 7 49
10 113 12 10 4 6 36
∑= 1027 196 ∑=194

19. •∑di² = 194
n = 10
)1(
6
1 2
1
2



nn
d
r
n
i
i
s

20. •Putting the value in
formula

21. The calculated value is
-0.1757

22. (5) CRITICAL REGION:
If Dcal ≥ Table values
DCAL =-0.17 , Table value =0.632

23. (6) CONCLUSION
Since the calculated value falls in
rejection region so we reject our
null hypothesis, because their have
no relation ship between these
two variables.

24. Table for Spearmans
ranks correlation

26. Question no 1.
S.no Height(feet) Chest size (inches)
1 6.4 32.0
2 6.6 30.3
3 6.9 28.9
4 5.4 32.0
5 4.3 22.8
6 7.0 34.6
7 6.6 30.6
8 5.0 25.0

27. If you have any
question, you may ask.
Thank you