Spearman Rank Correlation

A measure of Rank Correlation

Group 3

The Spearman Correlation

• Spearman’s correlation is designed to measure the relationship between variables measured on an ordinal scale of measurement.

• Similar to Pearson’s Correlation, however it uses ranks as opposed to actual values.

Assumptions

• The data is a bivariate random variable.

• The measurement scale is at least ordinal.

• Xi ,Yi is independent of Xj ,Yj where i ≠ j

Advantages

1. Less sensitive to bias due to the effect of outliers - Can be used to reduce the weight of outliers (large distances get treated as a one-rank difference)

2. Does not require assumption of normality.

3. When the intervals between data points are problematic, it is advisable to study the rankings rather than the actual values.

Disadvantages

1.Calculations may become tedious. Additionally ties are important and must be factored into computation.

Steps in Calculating Spearman’s Rho

1. Convert the observed values to ranks (accounting for ties)

2. Find the difference between the ranks, square them and sum the squared differences.

3. Set up hypothesis, carry out test and conclude based on findings.

Steps in Calculating Spearman’s Rho

4. If the null is rejected then calculate the Spearman correlation coefficient to measure the strength of the relationship between the variables.

Hypothesis: IHypothesis: I

A. (Two-Tailed)

Ho : There is no correlation between the Xs and the Ys.

(there is mutual independence between the Xs and the Ys)

H1 : There is a correlation between the Xs and the Ys.

(there is mutual dependence between the Xs and the Ys)

Hypothesis: IIHypothesis: II

B. (One-Tailed - Lower)

Ho : There is no correlation between the Xs and the Ys.

(there is mutual independence between the Xs and the Ys)

H1 : There is a negative correlation between the Xs and the Ys.

Hypothesis: IIIHypothesis: III

C. (One-Tailed - Upper)

Ho : There is no correlation between the Xs and the Ys.

(there is mutual independence between the Xs and the Ys)

H1 : There is a positive correlation between the Xs and the Ys.

Test StatisticTest Statistic

For small samples (N < 40):

T=Σdi2 = Σ[R(Xi) - R(Yi)]2

For large samples:

(Reject using the appropriate Z critical value)

)1()1(361

)(161

22

3

*

nnn

nnTZ

Test StatisticTest Statistic

• In the case of a large sample:

)1()1(

36

1);(

16

1~ 223 nnnnnNT

Decision RulesDecision RulesA. Two-tailed:

Reject H0 if T≤ Sα/2 or T > S1- α/2 .

Do not reject otherwise.

B. One-tailed - Lower:

Reject H0 if T > S1- α .

Do not reject otherwise.

C. One-tailed- Upper:

Reject H0 if T≤ Sα.

Do not reject otherwise.

In the case of few ties (less than 5% of the sample):

Where di is the difference in the ranks of each pair and N is the number of pairs

Spearman’s Rho

)1(

61

21

2

NN

n

iid

Spearman’s RhoSpearman’s Rho

If there are numerous ties:

5.0

1

22

5.0

1

22

1

2

21

)(2

1)(

21

)()(

n

ii

n

ii

n

iii

nnyR

nnxR

nnyRxR

Spearman’s RhoAssumes values between -1 and +1

-1 ≤ ρ ≤ 0 ≤ ρ ≤ +1

Perfectly Negative Correlation

Perfectly Positive Correlation

Example 1Example 1The ICC rankings for One Day International (ODI) and Test matches for nine teams are shown below.

Test whether there is correlation between the ranks

Team Test Rank ODI RankAustralia 1 1India 2 3South Africa 3 2Sri Lanka 4 7England 5 6Pakistan 6 4New Zealand 7 5West Indies 8 8Bangladesh 9 9

Example 1Example 1Team Test Rank ODI Rank d d2

Australia 1 1 0 0India 2 3 1 1South Africa 3 2 1 1Sri Lanka 4 7 3 9England 5 6 1 1Pakistan 6 4 2 4New Zealand 7 5 2 4West Indies 8 8 0 0Bangladesh 9 9 0 0

Total 20

Answer:

202 i

dT 3

Example 2Example 2

A composite rating is given by executives to each college graduate joining a plastic manufacturing firm. The executive ratings represent the future potential of the college graduate. The graduates then enter an in-plant training programme and are given another composite rating. The executive ratings and the in-plant ratings are as follows:

Graduate Executive rating (X) Training rating (Y)

A 8 4

B 10 4

C 9 4

D 4 3

E 12 6

F 11 9

G 11 9

H 7 6

I 8 6

J 13 9

K 10 5

L 12 9

A) At the 5% level of significance, determine if there is a positive correlation between the variables

B) Find the rank correlation coefficient if the null is rejected