**Friedman Two Way Analysis of Variance by Ranks**

**© 1998 by Dr. Thomas W. MacFarland -- All Rights Reserved**

************ friedman.doc ************ 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 ************ friedman.dat ************ 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 ************ friedman.r01 ************ 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 ************ 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 ************ friedman.con ************ 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. ************ friedman.lis ************ % 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 -------------------------- 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'