Friedman Two Way Analysis of Variance by Ranks
© 1998 by Dr. Thomas W. MacFarland -- All Rights Reserved
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friedman.doc
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Background: Friedman Two Way Analysis of Variance is a useful
test when:
1. it is desirable to know if differences exist
between three or more samples
2. data are ordinal (i.e., ranked)
It is generally desirable to structure a design so
that a parametric analysis is used. But, as the later
sample demonstrates, there are many times when it is
only possible to use ordinal data. When that is the
case, the Friedman test may be an appropriate
alternate statistical aid when looking at multiple
analyses and possible interaction.
Scenario: This file examines possible differences in graded
performance to three separate activities (e.g.,
final examination score, composite score for all
homework problems, final project score) in a high
school Logo programming language class.
To add greater precision to this analysis, this
analysis will further determine:
-- differences by gender
1 = Female
2 = Male
-- adherence to the prerequisite BASIC programming
course
1 = Yes, Prerequisite Passed
2 = No, Prerequisite Waived
Because the teacher conducting this analysis has a
concern that homework scores and final project scores
are ordinal data (data are ordered, but not with the
precision of interval data), it is best to use the non-
parametric Friedman test instead of the Twoway
Analysis of Variance (ANOVA) test, which is based on
the use of interval data.
A visual presentation of the data is provided in
Table 1.
Table 1
Graded Activities (Final Examination, Homework,
Project) in a High School Logo Programming Course
by Gender and by Adherence to a Prerequisite
BASIC Programming Course
=========================================================
Gender Pre_Req
---------- ---------
Student 1 = Female 1 = Yes
Number 2 = Male 2 = No Final Home_Wk Project
---------------------------------------------------------
01 1 1 086 091 067
02 2 1 088 089 074
03 1 1 067 073 057
04 1 2 083 091 093
05 2 1 082 085 080
06 2 1 083 078 092
07 2 2 055 071 065
08 1 2 087 093 072
09 1 2 088 083 081
10 1 1 093 096 084
11 1 2 086 075 065
12 2 2 087 094 100
13 1 1 088 097 093
14 1 1 082 078 075
15 1 1 088 083 094
16 1 1 074 079 098
17 2 2 069 082 076
18 1 1 094 087 095
19 2 2 075 072 084
20 1 2 074 076 068
21 1 2 085 078 093
22 1 2 087 087 082
23 1 1 082 092 076
24 1 1 094 096 090
25 2 1 068 073 082
---------------------------------------------------------
Ho: Null Hypothesis: There is no difference in graded
performance to three separate activities (e.g.,
final examination score, composite score for all
homework problems, final project score) between
students in a high school Logo programming language
class by gender (male/female) and adherence to a
prerequisite BASIC programming course (yes/no)
(p = .05).
Files: 1. friedman.doc
2. friedman.dat
3. friedman.r01
4. friedman.o01
5. friedman.con
6. friedman.lis
Command: At the Unix prompt (%), key:
%spss -m < friedman.r01 > friedman.o01
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friedman.dat
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01 1 1 086 091 067
02 2 1 088 089 074
03 1 1 067 073 057
04 1 2 083 091 093
05 2 1 082 085 080
06 2 1 083 078 092
07 2 2 055 071 065
08 1 2 087 093 072
09 1 2 088 083 081
10 1 1 093 096 084
11 1 2 086 075 065
12 2 2 087 094 100
13 1 1 088 097 093
14 1 1 082 078 075
15 1 1 088 083 094
16 1 1 074 079 098
17 2 2 069 082 076
18 1 1 094 087 095
19 2 2 075 072 084
20 1 2 074 076 068
21 1 2 085 078 093
22 1 2 087 087 082
23 1 1 082 092 076
24 1 1 094 096 090
25 2 1 068 073 082
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friedman.r01
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SET WIDTH = 80
SET LENGTH = NONE
SET CASE = UPLOW
SET HEADER = NO
TITLE = Friedman Twoway ANOVA by Ranks
COMMENT = This file examines possible differences
in graded performance to three separate
activities (e.g., final examination score,
composite score for all homework problems,
final project score) in a high school Logo
programming language class. To add greater
precision to this analysis, this analysis
will further determine differences by gender
(1 = Female and 2 = Male) and adherence to
the prerequisite BASIC programming course
(1 = Yes, Prerequisite Passed and 2 = No,
Prerequisite Waived).
Because the teacher conducting this analysis
has a concern that homework scores and final
project scores are ordinal data (data are ordered,
but not with the precision of interval data), it
is best to use the non-parametric Friedman test
instead of the Twoway Analysis of Variance (ANOVA)
based on the use of interval data.
DATA LIST FILE = 'friedman.dat' FIXED
/ Stu_Code 20-21
Gender 30
Pre_Req 40
Final 50-52
Home_Wk 56-58
Project 65-67
Variable Lables
Stu_Code "Student Code"
/ Gender "Gender"
/ Pre_Req "Prerequisite Status"
Value Labels
Gender 1 'Female'
2 'Male'
/ Pre_Req 1 "Yes, Passed"
2 "No, Waived"
NPAR TESTS FRIEDMAN = Final Home_Wk Project
************
friedman.o01
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2 SET WIDTH = 80
3 SET LENGTH = NONE
4 SET CASE = UPLOW
5 SET HEADER = NO
6 TITLE = Friedman Twoway ANOVA by Ranks
7 COMMENT = This file examines possible differences
8 in graded performance to three separate
9 activities (e.g., final examination score,
10 composite score for all homework problems,
11 final project score) in a high school Logo
12 programming language class. To add greater
13 precision to this analysis, this analysis
14 will further determine differences by gender
15 (1 = Female and 2 = Male and adherence to
16 the prerequisite BASIC programming course
17 (1 = Yes, Prerequisite Passed and 2 = No,
18 Prerequisite Waived).
19
20 Because the teacher conducting this analysis
21 has a concern that homework scores and final
22 project scores are ordinal data (data are ordered,
23 but not with the precision of interval data), it
24 is best to use the non-parametric Friedman test
25 instead of the Twoway Analysis of Variance (ANOVA)
26 based on the use of interval data.
27 DATA LIST FILE = 'friedman.dat' FIXED
28 / Stu_Code 20-21
29 Gender 30
30 Pre_Req 40
31 Final 50-52
32 Home_Wk 56-58
33 Project 65-67
34
This command will read 1 records from friedman.dat
Variable Rec Start End Format
STU_CODE 1 20 21 F2.0
GENDER 1 30 30 F1.0
PRE_REQ 1 40 40 F1.0
FINAL 1 50 52 F3.0
HOME_WK 1 56 58 F3.0
PROJECT 1 65 67 F3.0
35 Variable Lables
36 Stu_Code "Student Code"
37 / Gender "Gender"
38 / Pre_Req "Prerequisite Status"
39
40 Value Labels
41 Gender 1 'Female'
42 2 'Male'
43
44 / Pre_Req 1 "Yes, Passed"
45 2 "No, Waived"
46
47 NPAR TESTS FRIEDMAN = Final Home_Wk Project
***** Workspace allows for 16384 cases for NPAR tests *****
- - - - - Friedman Two-Way Anova
Mean Rank Variable
1.86 FINAL
2.30 HOME_WK
1.84 PROJECT
Cases Chi-Square D.F. Significance
25 3.3800 2 .1845
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friedman.con
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Outcome: Computed Fr = 3.3800 (Fr approximates Chi-square)
df = k-1 = 3-1 = 2
Criterion Fr (alpha = .05, df = 2) = 5.9915
Computed Fr (3.3800) < Criterion Fr (5.9915)
Therefore, the null hypothesis is accepted and
it can be claimed that there is no difference in
the graded performance to three separate activities
(e.g., final examination score, composite score for
all homework problems, final project score) in a
high school Logo programming language class by
gender and by adherence to a prerequisite BASIC
programming course (p = .05).
The p value is another way to view differences in
the three graded activities:
-- The calculated p value is .1845.
-- The delcared p value is .05.
The calculated p value exceeds the declared p value
and there is, accordingly, no difference in scores
of the three graded activities at this level of
significance by gender and by adherence to a
prerequisite BASIC programming course (p = .05).
Differences in mean rankings of scores for all three
graded activities are due only to chance.
Note: Although the test statistic for Friedman Two Way
Anova is "Fr," you will notice that chi-square
values are used for data analysis. Fr approximates
the chi-square distribution.
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friedman.lis
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% minitab
MTB > outfile 'friedman.lis'
Collecting Minitab session in file: friedman.lis
MTB > # MINITAB addendum to friedman.dat
MTB > read 'friedman.dat' c1-c6
Entering data from file: friedman.dat
25 rows read.
MTB > print c1-c6
ROW C1 C2 C3 C4 C5 C6
1 1 1 1 86 91 67
2 2 2 1 88 89 74
3 3 1 1 67 73 57
4 4 1 2 83 91 93
5 5 2 1 82 85 80
6 6 2 1 83 78 92
7 7 2 2 55 71 65
8 8 1 2 87 93 72
9 9 1 2 88 83 81
10 10 1 1 93 96 84
11 11 1 2 86 75 65
12 12 2 2 87 94 100
13 13 1 1 88 97 93
14 14 1 1 82 78 75
15 15 1 1 88 83 94
16 16 1 1 74 79 98
17 17 2 2 69 82 76
18 18 1 1 94 87 95
Continue? y
19 19 2 2 75 72 84
20 20 1 2 74 76 68
21 21 1 2 85 78 93
22 22 1 2 87 87 82
23 23 1 1 82 92 76
24 24 1 1 94 96 90
25 25 2 1 68 73 82
MTB > # The series of MINITAB commands to conduct the
MTB > # Friedman Twoway ANOVA statistic is far too
MTB > # complex for our needs.
MTB > #
MTB > # Fortunately, the Chi-square statistic approximates
MTB > # the Friedman statistic.
MTB > # I will use the Chi-square test to obtain a
MTB > # reasonable estimate of Friedman.
MTB > chisquare c4-c6
Continue? y
Expected counts are printed below observed counts
C4 C5 C6 Total
1 86 91 67 244
80.74 82.87 80.39
2 88 89 74 251
83.06 85.25 82.69
3 67 73 57 197
65.19 66.91 64.90
4 83 91 93 267
88.35 90.68 87.96
5 82 85 80 247
81.73 83.89 81.37
6 83 78 92 253
83.72 85.93 83.35
7 55 71 65 191
63.20 64.87 62.92
Continue? y
8 87 93 72 252
83.39 85.59 83.02
9 88 83 81 252
83.39 85.59 83.02
10 93 96 84 273
90.34 92.72 89.94
11 86 75 65 226
74.78 76.76 74.46
12 87 94 100 281
92.98 95.44 92.58
13 88 97 93 278
91.99 94.42 91.59
14 82 78 75 235
77.76 79.82 77.42
15 88 83 94 265
87.69 90.01 87.30
Continue? y
16 74 79 98 251
83.06 85.25 82.69
17 69 82 76 227
75.12 77.10 74.79
18 94 87 95 276
91.33 93.74 90.93
19 75 72 84 231
76.44 78.46 76.10
20 74 76 68 218
72.14 74.04 71.82
21 85 78 93 256
84.71 86.95 84.34
22 87 87 82 256
84.71 86.95 84.34
23 82 92 76 250
82.73 84.91 82.36
Continue? y
24 94 96 90 280
92.65 95.10 92.25
25 68 73 82 223
73.79 75.74 73.47
Total 2045 2099 2036 6180
Continue? y
ChiSq = 0.343 + 0.797 + 2.229 +
0.294 + 0.165 + 0.914 +
0.050 + 0.554 + 0.962 +
0.324 + 0.001 + 0.288 +
0.001 + 0.015 + 0.023 +
0.006 + 0.732 + 0.898 +
1.065 + 0.579 + 0.068 +
0.156 + 0.641 + 1.463 +
0.255 + 0.078 + 0.049 +
0.078 + 0.116 + 0.392 +
1.682 + 0.040 + 1.201 +
0.385 + 0.022 + 0.595 +
0.173 + 0.070 + 0.022 +
0.231 + 0.041 + 0.076 +
0.001 + 0.545 + 0.514 +
0.988 + 0.458 + 2.834 +
0.498 + 0.312 + 0.020 +
0.078 + 0.485 + 0.182 +
0.027 + 0.532 + 0.819 +
0.048 + 0.052 + 0.203 +
0.001 + 0.921 + 0.889 +
0.062 + 0.000 + 0.065 +
0.006 + 0.592 + 0.491 +
Continue? y
0.020 + 0.009 + 0.055 +
0.455 + 0.099 + 0.991 = 31.327
df = 48
MTB > # Computed Chi-square = 31.327 (df = 48)
MTB > #
MTB > # Criterion Chi-square = 67.505 (p <= .05 and df = 50)
MTB > #
MTB > # Note how I used the table value for Chi-square with
MTB > # df = 50, not df = 48. Most tables present criterion
MTB > # values for Chi-square from 1 to 30, and then increments
MTB > # from 30 to 40, 40 to 50, etc.
MTB > #
MTB > # Because the computed Chi-square statistic (31.327) is
MTB > # less than the criterion Chi-square statistic (67.505),
MTB > # accept the Null Hypothesis. There is no difference in
MTB > # grades on the three activities: final examination,
MTB > # homework, project.
MTB > stop
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Disclaimer: All care was used to prepare the information in this
tutorial. Even so, the author does not and cannot guarantee the
accuracy of this information. The author disclaims any and all
injury that may come about from the use of this tutorial. As
always, students and all others should check with their advisor(s)
and/or other appropriate professionals for any and all assistance
on research design, analysis, selected levels of significance, and
interpretation of output file(s).
The author is entitled to exclusive distribution of this tutorial.
Readers have permission to print this tutorial for individual use,
provided that the copyright statement appears and that there is no
redistribution of this tutorial without permission.
Prepared 980316
Revised 980914
end-of-file 'friedman.ssi'