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# Dalut ppt. of factorial analysis of variance-b

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• ### Dalut ppt. of factorial analysis of variance-b

1. 1. Republic of the PhilippinesMINDANAO STATE UNIVERSITYGeneral Santos City<br />GRADUATE PROGRAM<br />FACTORIAL ANALYSIS OF VARIANCE<br />A class report to the class of Dr. Ava Clare Marie O. Robles<br />Presented by:<br />Chellyn Mae P. Dalut<br /> MST Elementary Math<br />
2. 2. LearningObjectives: At the end of this session, students are expected to:<br />
3. 3. What is Factorial Analysis of Variance?<br />
4. 4. What is Factorial Analysis of Variance?<br />
5. 5. Two-way ANOVA/F-Test Two Factor/ANOVA Two-Factor<br />
6. 6. Examples:<br />A study on of Effects of Method and class size on Achievement <br />Accebility of luncheon Meat from Commercial, Milkfish Bone Meal, and Goatfish Bone Meal<br />20 x 3=60 Luncheon Meat<br />2x2=4 Method<br />
7. 7. Assumptions<br /> (www.statford.com)<br />
8. 8. Steps in using Two-Way ANOVA<br />Consider the following when answering A RESEARCH PROBLEM:<br />
9. 9. Steps in using Two-Way ANOVA<br />
10. 10. Steps in using Two-Way ANOVA<br />
11. 11. Illustration/Application:<br />Statement of the problem:<br />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)<br />Specific Research Problem: “ Is there a significant difference on the flavor acceptability of luncheon meat from commercial, milkfish bone meal, and goat fishbone meal?”<br />Null Hypothesis:<br />There is no significant difference on the flavor acceptability of luncheon meat from commercial, milkfish bone meal, and goatfish bone meal.<br />Ho: X = X2 = X3 =0 <br />Statistical tool: Two-way ANOVA<br />Significance Level: Alpha= 0.01<br />Sampling Distribution : N=20<br />
12. 12. Rejection Region:<br />The null hypothesis is rejected if the computed F-value is equal to or greater that the tabular F-value.<br />Fcomputed>Ftabular<br />Computation:<br />Please refer to your excel exercises. (Slide 9 & 10)<br /> Lets do step 1 & 2<br />Illustration/Application:<br />
13. 13.
14. 14. Step No. 3<br />Illustration/Application:<br />Given: CF= (∑x)2 = (460)2<br />∑x2 = 71082 3N 60<br />P = 20 CF=3526.66667<br />Compute the Sum of Squares for Samples (SSs)<br />SSs=∑x2-CF<br /> P<br />Where:<br />SSS- Sum of Squares for Samples<br />∑x2 - Summation of X<br />P or N- Panelist or (N) (Subject)<br />CF - Correction Factor CF= ∑x<br /> N<br />SSs=∑x2-CF<br /> P<br />SSs=71082- 3526.66667 <br /> 20<br />SSs= 3554.1-3526.66667<br />SSs= 27.43333<br />
15. 15. Given: CF= (∑x)2 = (460)2<br />∑y2 = 10624 3N 60<br />S = 3 CF=3526.66667<br />Step No. 4<br />Illustration/Application:<br />Compute the Sum of Squares for Panelist (SSp)<br />SSp=∑y2-CF<br /> S<br />Where:<br />SSp== Sum of squares for panelist<br /> ∑y2 = Sum of squared total for treatment<br /> CF = Correction Factor<br /> S = Sample (number of experimental group<br />SSP=∑y2-CF<br /> s<br />SSP =10624- 3526.66667 <br /> 3<br />SSP = 3541.33333-3526.66667<br />SSp= 14.66667<br />
16. 16. Illustration/Application:<br />Given: CF= (∑x)2 = (460)2<br />∑∑ij2= 3574 3N 60<br />CF = 3526.66667<br /> Step No. 5<br />Compute the Sum of Squares for Total (SST)<br />SST= ∑∑ij2- CF<br />Where:<br />SST = Sum of squares for total<br /> ∑∑ij2= Grand sum of each observation per treatment<br /> CF = Correction factor<br />SST= ∑∑ij2- CF<br />SST = 3574 - 3526.66667<br />SST = 47.3333<br />
17. 17. Illustration/Application:<br />Given: <br />SST = 47.3333 SSp= 14.66667<br />SSs= 27.43333<br /> Step No. 6<br />Compute the Sum of Square for Errors (SSE)<br />SSE=SST-(SSs+SSp)<br />Where:<br />SSE = Sum of squares for Errors<br /> SST = Sum of squares for total<br />SSp= Sum of squares for panelist<br />SSE=SST-(SSs+SSp)<br />SSE = 47.3333 – (27.43333 + 14.66667)<br />SSE = 47.3333 – 42.1<br />SSE = 5.23333<br />
18. 18. Illustration/Application:<br />Given: <br />Ns= 3<br /> NP=20<br /> NT = 60<br /> Step No. 7<br />Get the degrees of freedom of:<br />(dfs= N-1) and (dfP= N-1) and<br />(dfT= N-1) and dfE= dfT- (dfS + dfP)<br />Where:<br />dfE=degrees of freedom of error <br />dfs= degrees of freedom of samples<br />dfP= degrees of freedom of panelist<br />dfT= degrees of freedom of total<br />Ns = Number of samples<br /> NP = Number of panelist<br /> NT= Number of Total (P x N) <br />dfs= Ns-1 dfT = NT -1<br /> = 3-1 = 60-1<br />= 2= 59<br />dfP= NP-1 dfE=dfT- (dfS + dfP)<br /> = 20-1 =59-(2+19)<br />= 19 dfE=38<br />
19. 19. Illustration/Application:<br />Given: <br />SSS= 27.43333 SSE= 5.23333<br />SSp =14.66667<br />dfs = 2 ; dfP=19; dfE= 38 <br /> Step No. 8 & 9<br />MSS = SSS (MSE)= SSE<br />dfsdfE<br />= 27.43333 = 5.23333<br /> 2 38 <br />=13.71667 = 0.137719<br /> (MSp)=SSp<br />dfp<br />=14.66667<br /> 19<br />=0.77193<br />Compute the Mean Square (MS) Computation and the Mean square of error <br />(MSS)= SSS and (MSp)=SSp and (MSE)= SSE<br />dfsdfpdfE<br />Where:<br /> MSS = Mean of Square for sample<br />MSp =Mean of Square for panelist<br /> SSS = Sum of square for samples<br />SSp = Sum of square for panelist<br />dfs= degrees of freedom of samples<br />dfP= degrees of freedom of panelist<br /> SSE = Sum of squares for error<br />dfE = Degrees of freedom of error<br />
20. 20. Illustration/Application:<br />Given: <br />MSS= 13.71667 MSE=0.137719<br />MSp =0.77193<br /> Step No. 10<br />Observe F Computation<br />Fs = MSS and Fp = MSP<br />MSEMSE<br />Where<br />Fp= F-computation for panelist<br /> Fs = F-computation for samples<br /> MSS = Mean of Square for sample<br />MSp =Mean of Square for panelist<br /> MSE = Sum of squares for error<br />dfE = Degrees of freedom of error<br />FS = MSS Fp = MSP<br />MSE MSE<br />= 13.71667= 0.77193<br />0.1377190.137719<br />=99.59873 =5.605096<br />(Significant @ level 0.01) (Significant @ level 0.01)<br />
21. 21. Illustration/Application:<br />Remember of our Rejection Region-<br />Fcomputed>Ftabular<br /> Interpretation:<br />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.<br />For panelist, the computed F-value obtained is 5.605096<br />also greater than the tabular F-value of 2.42 and is also significant at .01 level of confidence with df = 19,38.<br />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.<br />There is significant difference on the flavor acceptability of luncheon meat, and goatfish bone meal.<br />FS = MSS Fp = MSP<br />MSE MSE<br />= 13.71667= 0.77193<br />0.1377190.137719<br />=99.59873 =5.605096<br />(Significant @ level 0.01) (Significant @ level 0.01)<br /> Tabular FS = Tabular Fp =<br />df2,38 (0.01) = 5.21 df19,38 (0.01) = 5.605096<br /> (Hence: Please see tabular value of F on your copy)<br />
22. 22. Illustration/Application:<br />Remember of our Rejection Region-<br />Fcomputed>Ftabular<br /> Interpretation:<br />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.<br />For panelist, the computed F-value obtained is 5.605096<br />also greater than the tabular F-value of 2.42 and is also significant at .01 level of confidence with df = 19,38.<br />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.<br />There is significant difference on the flavor acceptability of luncheon meat, and goatfish bone meal.<br />
23. 23. Illustration/Application:<br />Use the data that I gave for solving this, refer to your excel file.<br /> Using computer:<br />
24. 24. The computer displays as follows:<br />
25. 25. What is Friedman Two-Way Analysis of Variance by ranks (xr2)?<br />
26. 26. What is Factorial Analysis of Variance?<br />
27. 27. To substitute formula, the steps are as follows:<br />
28. 28. Illustration/Application:<br />Statementof the problem:<br />The researcher wishes to conduct a study on the adequacy of facilities at the Northern Ilo-Ilo Polytechnic<br /> State College as perceived by top managers, middle managers, lower managers, and professors. <br />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?”<br />Null Hypothesis:<br />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.<br />Ho : X = X2 = X3 =X4=0 <br />Statistical tool: Friedman Two-way ANOVA by ranks<br />Significance Level: Alpha= 0.01<br />Sampling Distribution : K= 4 N=20<br />
29. 29. Rejection Region:<br />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) <br />(XR2) compute>(X2) tabular<br />Computation:<br />Please refer to your excel exercises. (Slide 26)<br />Lets do step 1 & 2<br />Illustration/Application:<br />
30. 30.
31. 31. Illustration/Application:<br />Scale :<br />-very much adequate<br />-adequate<br />-fairly adequate<br />1 -Inadequate<br />Friedman (XR2) Test Computation<br />Xr2= 12 ∑(R) 2 – 3N (K +1) <br /> NK (K+1)<br />=12 (10272.5 ) - 3(20) (4+1)<br /> 20(40) (4+1)<br /> = 0.03 (10272.5) – 300<br /> =308.175-300<br />Xr2 = 8.75 (insignificant at 0.01 level)<br />Degrees of freedom tabular value<br />df = K-1 df3(0.01)= 11.34<br />Df = 4-1<br />Df = 3<br />Given:<br /> ∑(R) 2 =102722.5<br /> N =20<br /> K =4<br />Where:<br /> Xr2 = Friedman 2-way ANOVA by rank<br /> N =Number of rows<br /> K =Number of columns<br />
32. 32. Illustration/Application:<br />Remember of our Rejection Region-<br /> (XR2) compute> (X2) tabular<br /> Interpretation:<br />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.<br />ACCEPTANCE OF NULL HYPOTHESIS:<br />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<br /> by top managers, middle managers, lower managers and professor.<br />Xr2 = 8.75 (insignificant at 0.01 level)<br />Degrees of freedom Tabular value<br />df = K-1 df3(0.01)= 11.34<br />Df = 4-1<br />Df = 3<br /> (Hence: Please see critical values of Chi-square)<br />
33. 33. References:<br />Online Resources:<br />Books :<br />www. statford.com<br />www.psych.nyu.edu<br />www.mathworks.com<br />www.statsoft.com<br />Calmorin, Laurentina Paler (2010). Reaserch and Statistics with computer. National BookStore, Mandaluyong City, Metro Manila<br />Fraenkel, J and Nancy Wallen (2007). How to Design and Evaluate Research in Education,3rd Edition, McGraw Hills Companies, Inc. New York<br />Robles, Ava Clare Marie (2011).Parametric Statistics Made Easy using MS Excel (2011). MECS Publishing House, Inc.,LeonLlido St., General Santos City<br />