Spearman Rank Correlation Coefficient
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Transcript of Spearman Rank Correlation Coefficient
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Statistical Methodand Advanced Application
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SPEARMAN RANK
CORRELATION COEFFICIENT
SMALL SAMPLELARGE SAMPLE
APPROXIMATION
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SMALL SAMPLE
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Spearman Rank Correlation
Coefficient
Learning Outcome
Student should be able to make
a relationship between two
variable by using Spearman rankcorrelation coefficient.
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Assumptions
a)The data consist of a random sample ofn
pairs of numeric or non-numeric
observations.
b)Each pair of observations represents two
measurements taken on the same object orindividual, called the unit of association.
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Procedures
1)If the data consist of observations from a bivariate population, wedesignate the n pairs of observations (X1,Y1), (X2,Y2),, (Xn,Yn).
2)EachXis ranked relative to all other observed values ofX, from
smallest to largest in order of magnitude. The rank of the ith value of
Xis denoted by R(Xi )=1 ifXi is the smallest observed value ofX.
3) Each Yis ranked relative to all other observed values ofY, from
smallest to largest in order of magnitude. The rank of the ith value of
Yis denoted by R(Yi )=1 ifYi is the smallest observed value ofY.
4)If ties occur among theXs or among the Ys, each tied value is
assigned the mean of the rank positions for which it is tied.
5)If the data consist of nonnumeric observations, they must be capable
of being ranked as described.
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Hypotheses
Case A
(two-sided)
Case B
(one-sided)
Case C
(one-sided)
H0 : X and Y are
independent.H0 : X and Y are
independent.
H0 : X and Y are
independent.
H1 : X and Y are
either directly
or inversely
related.
H1 : There is a
direct
relationship
between X andY.
H1 : There is an
inverse
relationship
between Xand Y.
= 0
0
= 0 = 0sH :0 sH :0 sH :0
sH :1 0:H s1 > 0:H s1 <
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Test Statistic
1) Rank all sample observations from smallest to
largest.
2)Find di and di2where di=R(Xi)R(Yi), and di is
the difference in the ranks given to the variablesXi and Yi.
3)Sum the square of the difference in the ranks
observations from populationXi and Yi, di2
4)The test statistic is,)1(
61
2
2
nn
dr
i
s
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5) rsis to measure the strength of the relationship
between the sampleXand Yvalues and as an
estimate of the strength of the relationship
betweenXand Yin the sampled population.
6) When the rank ofXis the same as the rank ofY
for every pair of observations (perfect direct
relationship), all the differences di will be equalto zero, and rs will be equal to +1.
7) Kendall(T3)* has shown that in general rs = -1
when the rank of one variable within each pair of
observations (Xi, Yi) is the reverse of the other
(perfect inverse relationship).
*Kendall, M.G.,Rank Correlation Methods, fourth edition, London: Griffin, 1970.
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Thus if
[R(X) = 1, R(Y) = n]
[R(X) = 2, R(Y) = n-1],.,[R(X) = n, R(Y) = 1]
for n pairs of observations, rs = -1. This may be illustrated
by means of a simple example. Suppose have the followingpairs of observations of (Xi, Yi) : (0,10),(8,3),(2,9),(5,6).
The ranks are
So, the di2=(-3) 2 +(3) 2 + (-1) 2 + (1) 2 = 20. Then
rs = 1[6(20)/4(16-1) = -1
8) Kendall also shows that rs can never be greater than +1 or
less than -1.
Xi 0 8 2 5
R(Xi) 1 4 2 3
Yi 10 3 9 6
R(Yi) 4 1 3 2
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Decision Rule
Case Reject H0 if
A
(Two-sided)
B
(One-sided positive)
C
(One-sided negative)
2rr
s
rr
s
rr
s
2rr
s
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Example 9.1(pg 360)
Pincherle and Robinson(E1)* note a marked interobservervariation in blood pressure. They found that doctors whoread high on systolic tended to read high on diastolic.Table 9.1 shows mean systolic and diastolic bloodpressure readings by 14 doctors. We wish to compute a
measure of the strength of the relationship between thetwo variables. Under the consumption that these 14doctors constitute a random sample from a population ofdoctors, we wish to know whether we may conclude fromthe data that there is a direct relationship between systolicand diastolic readings. Suppose we let = 0.05
*G.Pincherle and D.Robinson, Mean Blood Pressure and its Relation to Other Factors
Determined at a Routine Executive Health Examination, J.Choronic Dis.,27 (1974),245-260;used with permission of Pergamon Press.
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Mean blood pressure readings, millimeters mercury, by doctor
Doctor Systolic Diastolic
1 141.8 89.72 140.2 74.4
3 131.8 83.5
4 132.5 77.8
5 135.7 85.8
6 141.2 86.5
7 143.9 89.4
8 140.2 89.3
9 140.8 88.0
10 131.7 82.2
11 130.8 84.6
12 135.6 84.4
13 143.6 86.3
14 133.2 85.9
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1) HYPOTHESES
Systolic and diastolic blood pressure readings by
doctors are independent.
There is a direct relationship between systolic and
diastolic blood pressure readings by doctors
(claim).
:0H
:1H
2) TEST STATISTIC
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2) TEST STATISTICSYSTOLIC
( )
DIASTOLIC
( )R( ) R( ) = R ( )R( )
141.8 89.7 12 14 - 2 4
140.2 74.4 8.5 1 7.5 56.25131.8 83.5 3 4 -1 1
132.5 77.8 4 2 2 4
135.7 85.8 7 7 0 0
141.2 86.5 11 10 1 1
143.9 89.4 14 13 1 1
140.2 89.3 8.5 12 - 3.5 12.25
140.8 88.0 10 11 -1 1
131.7 82.2 2 3 -1 1
130.8 84.6 1 6 -5 25
135.8 84.4 6 5 1 1
143.6 86.3 13 9 4 16
133.2 85.9 5 8 -3 9
iX
iX iYi
Y ix iYid2id
50.132=2id
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By using the formula of test static:
From the table, we get the value ofand n = 14. Then, substitute the values into the
equation above.
50.1322
id
)1-14(14
)50.132(6-12
sr
7088.0
)1(
6-1
2
2
nn
dr is
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3) CRITICAL VALUE
Table A.21 reveals that, for n = 14 and (1) = 0.05,
the critical value of is 0.464.s
r
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4) DECISION
Since 0.7088 > 0.464, we reject .
5) CONCLUSION
There is enough evidence to support the claim
that the doctors who read high on systolic
tend to read high on diastolic blood pressure.
0H
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Exercise 1
Table 9.4 shows the serum and bone magnesium levels of 14 patients are reported by
Alfrey et al.*
. Can we conclude from these data that a relationship exist betweenserum magnesium and bone magnesium in the sample population?
Serum Mg ( m Eq./L.) Bone Mg ( m Eq./kg ash )3.60 6722.85 6102.80 6212.70 5672.60 5702.55 6382.55 6122.45 5522.25 5241.80 4001.45 2771.35 2941.40 3380.90 230
*Alfrey, Allen C.,Nancy L. Miller, and Donald Butkus, Evaluation Of Body Magnesium
Stores,J.Lab. Clin. Med.,84 (1974), 153-1
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Exercise 2
Ten seventh-grade children randomly selected from a certain public school
system were ranked according to the quality of their home environment and thequality of their performance in school. The result are shown in table 9.45.
compute rs and determine whether one can conclude that the two variable are
directly related.
Ten seventh-grade children ranked according to quality of home
environment and quality of performance in school.
Child Home environment Performance in school1 3 12 7 93 10 84 9 105 2 36 1 47 6 58 4 29 8 610 5 7
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Exercise 3
The Spearman's Rank Correlation Coefficient is used to discover the strength of a link
between two sets of data. This example looks at the strength of the link between the
price of a convenience item (a 50cl bottle of water) and distance from the
Contemporary Art Museum (CAM ) in El Raval, Barcelona. compute rs and
determine whether one can conclude that the two variable are inversly related.
Convinence store Distance from CAM ( m) Price of 50cl bottle ()1 50 1.802 175 1.203 270 2.004 375 1.005 425 1.006 580 1.207 710 0.808 790 0.609 890 1.00
10
980
0.85
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Large-Sample Approximation
When the sample size is greater than 100, we
cannot use Table A.21 to test the significance
ofrs. Then we may compute
which is distributed approximately as thestandard normal.
1 nrzs
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LINK OF YOUTUBE
http://www.youtube.com/watch?v=Eu_XOoFNfR4&feature=mfu_in_order&list=UL
http://www.youtube.com/watch?v=Eu_XOoFNfR4&feature=mfu_in_order&list=ULhttp://www.youtube.com/watch?v=Eu_XOoFNfR4&feature=mfu_in_order&list=ULhttp://www.youtube.com/watch?v=Eu_XOoFNfR4&feature=mfu_in_order&list=ULhttp://www.youtube.com/watch?v=Eu_XOoFNfR4&feature=mfu_in_order&list=UL -
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Thank You