NON PARAMETRIC TESTSSPEARMANS RANK CORRELATION TEST

Introduction

• A non parametric measure of association between two variables X and Y is given by the rank correlation coefficient.

•Formula

)1(

61 2

1

2

nn

dr

n

ii

s

•Formula Explanation

• Where di is the difference between the ranks assigned to Xi and Yi, and n is the number of pairs of data.

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.

(5) Critical region.

If Dcal ≥ Table values DCAL =? , Table value =?

(6) Conclusion

• Check the result and interpret.

•Introduction to charles Spearmans• Born• Charles Edward Spearman

(1863-09-10)10 September 1863London, United Kingdom

• Died• 17 September 1945(1945-09-17)

(aged 82)London, United Kingdom

• Institutions• University College London

Charles Spearmans Picter

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.

•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.

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.

•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.

•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

•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.

•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)

•Calculated TableS.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

•∑di² = 194n = 10

)1(

61 2

1

2

nn

dr

n

ii

s

•Putting the value in formula

The calculated value is

-0.1757

(5) CRITICAL REGION:

If Dcal ≥ Table values DCAL =-0.17 , Table value =0.632

(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.

Table for Spearmans ranks correlation

For Practice Questions

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

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