1. Republic of the PhilippinesMINDANAO STATE UNIVERSITYGeneral Santos City GRADUATE PROGRAM FACTORIAL ANALYSIS OF VARIANCE A class report to the class of Dr. Ava Clare Marie O. Robles Presented by: Chellyn Mae P. Dalut MST Elementary Math

2. LearningObjectives: At the end of this session, students are expected to:

3. What is Factorial Analysis of Variance?

4. What is Factorial Analysis of Variance?

5. Two-way ANOVA/F-Test Two Factor/ANOVA Two-Factor

6. Examples: A study on of Effects of Method and class size on Achievement Accebility of luncheon Meat from Commercial, Milkfish Bone Meal, and Goatfish Bone Meal 20 x 3=60 Luncheon Meat 2x2=4 Method

7. Assumptions (www.statford.com)

8. Steps in using Two-Way ANOVA Consider the following when answering A RESEARCH PROBLEM:

9. Steps in using Two-Way ANOVA

10. Steps in using Two-Way ANOVA

11. Illustration/Application: Statement of the problem: The researcher wishes to conduct a study on the flavor acceptability of luncheon meat from commercial, milk fish bone meal and goat fish bone meal (Hence: experimental groups) Specific Research Problem: “ Is there a significant difference on the flavor acceptability of luncheon meat from commercial, milkfish bone meal, and goat fishbone meal?” Null Hypothesis: There is no significant difference on the flavor acceptability of luncheon meat from commercial, milkfish bone meal, and goatfish bone meal. Ho: X = X2 = X3 =0 Statistical tool: Two-way ANOVA Significance Level: Alpha= 0.01 Sampling Distribution : N=20

12. Rejection Region: The null hypothesis is rejected if the computed F-value is equal to or greater that the tabular F-value. Fcomputed>Ftabular Computation: Please refer to your excel exercises. (Slide 9 & 10) Lets do step 1 & 2 Illustration/Application:

14. Step No. 3 Illustration/Application: Given: CF= (∑x)2 = (460)2 ∑x2 = 71082 3N 60 P = 20 CF=3526.66667 Compute the Sum of Squares for Samples (SSs) SSs=∑x2-CF P Where: SSS- Sum of Squares for Samples ∑x2 - Summation of X P or N- Panelist or (N) (Subject) CF - Correction Factor CF= ∑x N SSs=∑x2-CF P SSs=71082- 3526.66667 20 SSs= 3554.1-3526.66667 SSs= 27.43333

15. Given: CF= (∑x)2 = (460)2 ∑y2 = 10624 3N 60 S = 3 CF=3526.66667 Step No. 4 Illustration/Application: Compute the Sum of Squares for Panelist (SSp) SSp=∑y2-CF S Where: SSp== Sum of squares for panelist ∑y2 = Sum of squared total for treatment CF = Correction Factor S = Sample (number of experimental group SSP=∑y2-CF s SSP =10624- 3526.66667 3 SSP = 3541.33333-3526.66667 SSp= 14.66667

16. Illustration/Application: Given: CF= (∑x)2 = (460)2 ∑∑ij2= 3574 3N 60 CF = 3526.66667 Step No. 5 Compute the Sum of Squares for Total (SST) SST= ∑∑ij2- CF Where: SST = Sum of squares for total ∑∑ij2= Grand sum of each observation per treatment CF = Correction factor SST= ∑∑ij2- CF SST = 3574 - 3526.66667 SST = 47.3333

17. Illustration/Application: Given: SST = 47.3333 SSp= 14.66667 SSs= 27.43333 Step No. 6 Compute the Sum of Square for Errors (SSE) SSE=SST-(SSs+SSp) Where: SSE = Sum of squares for Errors SST = Sum of squares for total SSp= Sum of squares for panelist SSE=SST-(SSs+SSp) SSE = 47.3333 – (27.43333 + 14.66667) SSE = 47.3333 – 42.1 SSE = 5.23333

18. Illustration/Application: Given: Ns= 3 NP=20 NT = 60 Step No. 7 Get the degrees of freedom of: (dfs= N-1) and (dfP= N-1) and (dfT= N-1) and dfE= dfT- (dfS + dfP) Where: dfE=degrees of freedom of error dfs= degrees of freedom of samples dfP= degrees of freedom of panelist dfT= degrees of freedom of total Ns = Number of samples NP = Number of panelist NT= Number of Total (P x N) dfs= Ns-1 dfT = NT -1 = 3-1 = 60-1 = 2= 59 dfP= NP-1 dfE=dfT- (dfS + dfP) = 20-1 =59-(2+19) = 19 dfE=38

19. Illustration/Application: Given: SSS= 27.43333 SSE= 5.23333 SSp =14.66667 dfs = 2 ; dfP=19; dfE= 38 Step No. 8 & 9 MSS = SSS (MSE)= SSE dfsdfE = 27.43333 = 5.23333 2 38 =13.71667 = 0.137719 (MSp)=SSp dfp =14.66667 19 =0.77193 Compute the Mean Square (MS) Computation and the Mean square of error (MSS)= SSS and (MSp)=SSp and (MSE)= SSE dfsdfpdfE Where: MSS = Mean of Square for sample MSp =Mean of Square for panelist SSS = Sum of square for samples SSp = Sum of square for panelist dfs= degrees of freedom of samples dfP= degrees of freedom of panelist SSE = Sum of squares for error dfE = Degrees of freedom of error

20. Illustration/Application: Given: MSS= 13.71667 MSE=0.137719 MSp =0.77193 Step No. 10 Observe F Computation Fs = MSS and Fp = MSP MSEMSE Where Fp= F-computation for panelist Fs = F-computation for samples MSS = Mean of Square for sample MSp =Mean of Square for panelist MSE = Sum of squares for error dfE = Degrees of freedom of error FS = MSS Fp = MSP MSE MSE = 13.71667= 0.77193 0.1377190.137719 =99.59873 =5.605096 (Significant @ level 0.01) (Significant @ level 0.01)

21. Illustration/Application: Remember of our Rejection Region- Fcomputed>Ftabular Interpretation: The computed F-value obtained for samples is 99.59873 which is greater than the tabular F-value for samples of 5.21 which is significant at 0.01 level of significance with df=2,38. For panelist, the computed F-value obtained is 5.605096 also greater than the tabular F-value of 2.42 and is also significant at .01 level of confidence with df = 19,38. This means that the samples and evaluation of the panelist really differ with each other because milkfish bone meal luncheon meat is most acceptable. Hence, the null hypothesis is rejected. There is significant difference on the flavor acceptability of luncheon meat, and goatfish bone meal. FS = MSS Fp = MSP MSE MSE = 13.71667= 0.77193 0.1377190.137719 =99.59873 =5.605096 (Significant @ level 0.01) (Significant @ level 0.01) Tabular FS = Tabular Fp = df2,38 (0.01) = 5.21 df19,38 (0.01) = 5.605096 (Hence: Please see tabular value of F on your copy)

22. Illustration/Application: Remember of our Rejection Region- Fcomputed>Ftabular Interpretation: The computed F-value obtained for samples is 99.59873 which is greater than the tabular F-value for samples of 5.21 which is significant at 0.01 level of significance with df=2,38. For panelist, the computed F-value obtained is 5.605096 also greater than the tabular F-value of 2.42 and is also significant at .01 level of confidence with df = 19,38. This means that the samples and evaluation of the panelist really differ with each other because milkfish bone meal luncheon meat is most acceptable. Hence, the null hypothesis is rejected. There is significant difference on the flavor acceptability of luncheon meat, and goatfish bone meal.

23. Illustration/Application: Use the data that I gave for solving this, refer to your excel file. Using computer:

24. The computer displays as follows:

25. What is Friedman Two-Way Analysis of Variance by ranks (xr2)?

26. What is Factorial Analysis of Variance?

27. To substitute formula, the steps are as follows:

28. Illustration/Application: Statementof the problem: The researcher wishes to conduct a study on the adequacy of facilities at the Northern Ilo-Ilo Polytechnic State College as perceived by top managers, middle managers, lower managers, and professors. Specific Research Problem: “ Is there a significant difference on the adequacy of facilities at the Northern Ilo-Ilo Polytechnic State College as perceived by top managers, middle managers, lower managers, and professors?” Null Hypothesis: There is no significant difference on the on the adequacy of facilities at the Northern Ilo-Ilo Polytechnic State College as perceived by top managers, middle managers, lower managers, and professors. Ho : X = X2 = X3 =X4=0 Statistical tool: Friedman Two-way ANOVA by ranks Significance Level: Alpha= 0.01 Sampling Distribution : K= 4 N=20

29. Rejection Region: The null hypothesis is rejected if the computed Friedman (XR2) value is equal to or greater that the tabular F-value. (Refer to the chi-square (X2) (XR2) compute>(X2) tabular Computation: Please refer to your excel exercises. (Slide 26) Lets do step 1 & 2 Illustration/Application:

31. Illustration/Application: Scale : -very much adequate -adequate -fairly adequate 1 -Inadequate Friedman (XR2) Test Computation Xr2= 12 ∑(R) 2 – 3N (K +1) NK (K+1) =12 (10272.5 ) - 3(20) (4+1) 20(40) (4+1) = 0.03 (10272.5) – 300 =308.175-300 Xr2 = 8.75 (insignificant at 0.01 level) Degrees of freedom tabular value df = K-1 df3(0.01)= 11.34 Df = 4-1 Df = 3 Given: ∑(R) 2 =102722.5 N =20 K =4 Where: Xr2 = Friedman 2-way ANOVA by rank N =Number of rows K =Number of columns

32. Illustration/Application: Remember of our Rejection Region- (XR2) compute> (X2) tabular Interpretation: The computed Friedman test XR2 value obtained of 8.175 is insignificant because it is lesser than the tabular value of 11.34 with df = 3 at 0.01 level of confidence. This means that the adequacy of facilities at the Northern Ilo-Ilo Polytechnic State College as perceived by top managers, middle managers, lower managers, and professors are almost the same. ACCEPTANCE OF NULL HYPOTHESIS: The null hypothesis is accepted because there is no significant difference on the adequacy of facilities at the Northern Ilo-Ilo Polytechnic State College as perceived by top managers, middle managers, lower managers and professor. Xr2 = 8.75 (insignificant at 0.01 level) Degrees of freedom Tabular value df = K-1 df3(0.01)= 11.34 Df = 4-1 Df = 3 (Hence: Please see critical values of Chi-square)

33. References: Online Resources: Books : www. statford.com www.psych.nyu.edu www.mathworks.com www.statsoft.com Calmorin, Laurentina Paler (2010). Reaserch and Statistics with computer. National BookStore, Mandaluyong City, Metro Manila Fraenkel, J and Nancy Wallen (2007). How to Design and Evaluate Research in Education,3rd Edition, McGraw Hills Companies, Inc. New York Robles, Ava Clare Marie (2011).Parametric Statistics Made Easy using MS Excel (2011). MECS Publishing House, Inc.,LeonLlido St., General Santos City