Spearman Rank Order Correlation
Spearman Rank Order Correlation
The sixth grade creative writing contest was scored by two different teachers without the use of a scoring rubric. Of a possible 45 points, the work of the nine writers was awarded in the following manner.
Score A Score B
Adams, Larry 28 Adams, Larry 41
Burns, Kate 35 Burns, Kate 26
Conners, Chuck 31 Conners, Chuck 34
Due, Donna 36 Due, Donna 38
Edwards, Ed 26 Edwards, Ed 28
Hall, Mary 25 Hall, Mary 30
Jones, Betty 37 Jones, Betty 39
Miller, Art 40 Miller, Art 36
Smith, Dan 27 Smith, Dan 29
N = 9 N = 9
Range = 15 Range = 15
How closely do the two scores seem to be in agreement about the merits of the group’s creative writing skills?
Developing a rank order correlation will help us to find the “degree” of agreement.
Perfect Perfect
Negative Positive
Correlation Correlation
-1.0 .80 .60 .20 ?? .20 .40 .60 .80 +1.0
Rank order correlations are useful for teachers using single class and a pair of variables… be very careful … the correlation may not mean cause & effect!
We would expect certain variables to usually be positively correlated.
We would expect a usually negative correlation with certain variables; as one increases, the other decreases.
- academic achievement + hours watching TV
- time practicing keyboarding & typing errors
We would expect some traits are not correlated.
Spearman Rank Order Correlation
Student ranking highest in I.Q. also highest in math.
Rank IQ Rank Math
Alex 1 çè 1 a perfect positive
Bill 2 çè 2 correlation p = +1.0
Chuck 3 çè 3 highest I.Q. always
Dan 4 çè 4 gets highest math score.
Ed 5 çè 5
Student spending most practice time has fewest number of missed baskets.
Time shooting Number of
Baskets Misses
Tom 1 5 a perfect negative correlation
Sue 2 4 p = -1.0
Jeff 3 3 Tom had most practice and least
Diane 4 2 misses. Bob least practice…most
Bob 5 1 misses.
We can compare almost any two discrete measures.
Height I.Q.
Rank Rank
Alice 1 3 no clear correlation between ranks but
Sue 2 4 Ann happens to have lowest I.Q. and is
Mary 3 2 shortest
Beth 4 1
Ann 5 5
Correlations
Scorer Scorer Difference Sum Difference
A B in Rank Squared
------------------------------------------------------------------------------------------
Miller 1 _______ _________ ________________
Jones 2 _______ _________ ________________
Doe 3 _______ _________ ________________
Burns 4 _______ _________ ________________
Conners 5 _______ _________ ________________
Adams 6 _______ _________ ________________
Smith 7 _______ _________ ________________
Edwards 8 _______ _________ ________________
Hall 9 _______ _________ ________________
Spearman Rank Order Correlation
P = 1 – 6ED2
N(N2 – 1)
1 - ____________
1 - ____________
1 –
P =
Use this “rule of thumb” correlation guide to measure
the “degree” or extent of agreement
Minus No Plus
-1.0 .80 .60 .40 .20 Correlation .20 .40 .60 .80 +1.0
Answers for Spearman Rank Order Correlation Exercise
Scorer Scorer Difference Sum Difference
A B in Rank Squared
_____ ____ _______ ____________
Miller, Art 1 4 3 9
Jones, Betty 2 2 0 0
Doe, Donna 3 3 0 0
Burns, Kate 4 9 5 25
Conners, Chuck 5 5 0 0
Adams, Larry 6 1 5 25
Smith, Don 7 7 0 0
Edwards, Ed 8 8 0 0
Hall, Mary 9 6 3 9
N = 9 _____ ED2 68
Spearman Rank Order Correlation
P = 1 – 6 ED2
N (N2 – 1)
1 – 6 (68) 6 x 68 = 408
9 (80) 9 x 80 = 720
1 - 408 408 .567
720 - 720 720 408.000
1 - .567 1,000
- .567
______
.433
P = .433
Correlations
Judge Judge Difference Sum Difference
A B in Rank Squared
Bill 1 4 3 9
Sue 2 3 1 1
Tom 3 1 2 4
Jack 4 2 2 4
Paul 5 5 0 0
Mary 6 6 0 0
Ellen 7 8 1 1
Bob 8 9 1 1
Tami 9 7 2 4
Joe 10 11 1 1
Fay 11 10 1 1
N = 11 _______ ED2
F = 1 – 6ED2
N (N2 – 1)
1 – 6 x 26 6 x 26 = 156
11 (121 – 1) 11 x 11 = 121
1 – 156 156
1320 1320 = .117
1 - .117 1.000
-.117
______
.883
P = .88
Correlations
The first diving meet in the school’s new pool is next weekend. The P. E. Department has four experienced judges but needs one more. Which of the three new judges would be the best choice for the event.
Expert New New New
Judge Judge A Judge B Judge C
Angela 1 4 5 3
Susana 2 3 3 5
Graciela 3 1 6 6
Elena 4 2 2 1
Maria 5 5 1 2
Gaby 6 6 9 7
Victoria 7 8 7 9
Evette 8 9 8 4
Ana 9 7 4 8
Rosa 10 11 11 10
Yolanda 11 10 10 11
N = 11
Based on the degree of correlation between the “expert” judge and each of the possible new judges, I would select Judge ____ as my first choice, Judge ____ as my next choice and Judge ____ only as a last resort.
Perfect Perfect
Negative Positive
Correlation No Correlation
-1.0 .80 .60 .40 .20 Correlation .20 .40 .60 .80 +1.0
Enter each degree of correlation check P = 1 – 6ED2/N (N2-1)
Using Basic Statistics
Why … to see overall patterns … to see the “big picture” … to display overall shape of the data … see where the center lies and major deviations.
How … by constructing bar graphs, pie charts, Histograms, frequency polygons, etc.
Computing basic measurements from a distribution of test scores
31 24 33 21 27 25
22 26 19 29 37 38
29 28 34 30 32 35
33 32 31 29 35 25
26 29 32 28 25 27
N = 30
To get a “good look” to display the “overall shape” of these data we must:
- arrange scores in rank order
- develop a frequency distribution
- develop a numeric frequency
- develop a cumulative frequency
Numeric
Rank Frequency Frequency Cumulative
Order Distribution Distribution Frequency
38 1 1 30
37 1 1 29
36 0 0 28
35 11 2 28
34 1 1 26
33 11 2 25
32 111 3 23
31 11 2 20
30 1 1 18
29 1111 4 17
28 11 2 13
27 11 2 11
26 11 2 9
25 111 3 7
24 1 1 4
23 0 0 3
22 1 1 3
21 1 1 2
20 0 0 1
19 1 1 1
N = 30 N = 30
o Rank order lists scores from high to low.
o Frequency distribution accounts for all scores.
o Numeric frequency distribution substitutes “workable” numerals from hashmarks.
o Cumulative frequency accumulates scores bottom to top.
Experimental Research Designs
One Shot
Case Study X O1
Least adequate design
Baseline ?? Control group??
We can’t “generalize” from results.
One Group
Pre-Test Post-Test O1 X O2
An improved design
Can compute - + difference between scores
Cannot compare to similar group not involved in the experiment.
Quasi-experimental
Non-equivalent group O1 X O2
Pre-test Post-test O3 C O4
Often used by teachers
Real randomness not possible
Interpret results accordingly
Do not generalize too much
Post-test Only R X O1
Equivalent Groups R C O2
Compares randomly selected X and C groups…can analyze difference in scores … was difference result of sampling error or result of X treatment. This is a true experimental design.
Pre-test Post-test R O1 X O2
Equivalent Groups R O3 C O4
Similar to above, but both groups tested pre and post treatment … it is possible to determine if treatment yielded significantly different scores. This is the strongest experimental design.
Static Group X O1
Comparison C O2
Compare groups after the event no evidence groups were ever alike cannot claim equivalence, so we cannot claim difference in scores can be attributed to the treatment.
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