Dalut ppt. of factorial analysis of variance-b

Republic of the PhilippinesMINDANAO STATE UNIVERSITY

General 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

Learning Objectives: At the end of this session, students are expected to:

•Understand what is Factorial Analysis of Variance 1•Apply the Steps using Two-Way ANOVA2•Analyze the Visual Analysis of Variance3

What is Factorial Analysis of Variance?

1• According to Dr . Thomas W. MacFarland (1998• Two way ANOVA is to determine differences (and

possible interactions) when variables have two or more categories.

2 • The study of Factorial Analyses of Variance leads us into some of the most powerful methods in Statistics for analyzing data. Most experimental research in psychology and other social sciences use either variants of analysis of variance or other procedures which are derived from them and from correlational analyses. ( Erwin Segal, www.statsoft.com)

3 • The two-way analysis of variance is an extension to the one-way analysis of variance. There are two independent variables (hence the name two-way). (

What is Factorial Analysis of Variance?• Factorial Designs extend the

number of relationships that may be examined in an experimental study. The value of this, is that it allows a researcher to study the interaction of an independent variable with one or more other variables, sometimes called moderator variables. (Moderator Variables may be either treatment variables or subject characteristic variables.

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

• F-Test two-factor or ANOVA two-factor involves three or more independent variables as basis for classification.

•The two-way analysis of variance is used when two variables are involved..the column and the row.

Examples:A study on of Effects of Method and class size on Achievement

Class size Inquiry(x1) Lecture(x2)

Small (Y1)

Large (Y2)

Accebility of luncheon Meat from Commercial, Milkfish Bone Meal, and Goatfish Bone Meal

Panelist C MB GF

1

2

3

4

5

6

7

8

….

20

2x2=4 Method

20 x 3=60 Luncheon Meat

Assumptions (www.statford.com)

•The populations from which the samples were obtained must be normally or approximately normally distributed. 1•The samples must be independent. •The groups size must have the same sample size.2•The variances of the populations must be equal. 3

Steps in using Two-Way ANOVAConsider the following when answering A RESEARCH PROBLEM:

1 •Null Hypothesis•Ho: X1 = X2 = X3 =0

2 •Statistical Tool•Two-Way Anova

3 •Significance level•Alpha = 0.01

4• Sampling

Disribution• N=20

5 •Rejection Region•The null hypothesis is rejected if the computed F-value is equal to or greater than the tabular value

6• Computation

(see example on the next slide)

Steps in using Two-Way ANOVA

Step 1• Let us make a table

with column and row and plot all data.

Step 2• Get the arithmetic mean of

(∑x )vertically and square them . Do the same on (∑y).

• Get the CF=(∑x)2

• N

Step 3• Compute the Sum of

Squares for Samples (SSs)• (SSs)=∑x2-CF• P

Step 4 • Compute the Sum of Squares for Panelist (SSp)• (SSp)=∑y2-CF• S

Step 5 •Compute the Sum of Squares for Total (SST)•(SST)= ∑∑ij

2)- CF

Step 6 •Compute the Sum of Square for Errors (SSE)•(SSE)=SST-(SSs+SSp)

Steps in using Two-Way ANOVA

Step 7

• Get the degrees of freedom of:

• (dfs= NS-1) and

• (dfP= NP-1) and

• (dfT= NT-1)

Step 8 • Compute the Mean Square (MS) Computation• (MSS)= SSS and (MSp)=SSp

• dfs dfp

Step 9 •Get the MSE = SSE

• dfT

Step 10

•Observe F Computation•Fs = MSS and Fp = MSP

• MSE MSE

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 ANOVASignificance Level: Alpha= 0.01Sampling Distribution : N=20

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


Commercial (X1) Bone Meal (X2) Goat Fish Bone Meal (X3)

1 8 8 7 23 529 1772 8 9 7 24 576 1943 7 8 6 21 441 1494 7 8 6 21 441 1495 8 9 8 25 625 2096 8 8 7 23 529 1777 8 9 7 24 576 1948 7 9 6 22 484 1669 9 9 8 26 676 226

10 8 8 7 23 529 17711 9 9 7 25 625 21112 8 9 7 24 576 19413 8 9 7 24 576 19414 7 8 6 21 441 14915 8 9 7 24 576 19416 8 8 7 23 529 17717 7 8 6 21 441 14918 7 8 7 22 484 16219 7 8 6 21 441 14920 8 8 7 23 529 177∑X 155 169 136 460 10624 3574

Mean 7.75 8.45 6.8∑X2 24025 28561 18496 71082 ∑y2

Scale:9 -like extremely8 -like very much ∑X7 -like moderately6 -like slightly

(Calmorin, 2011)

∑∑ij2

PanelistLuncheon Meat

∑y ∑y2 ∑∑ij2

Step No. 3

Compute the Sum of Squares for Samples (SSs)

SSs=∑x2-CF P

Where:

SSS- Sum of Squares for Samples

∑x2 - Summation of XP or N- Panelist or (N) (Subject)

CF - Correction Factor CF= ∑x

N

Given: CF= (∑x)2 = (460)2

∑x2 = 71082 3N 60P = 20 CF=3526.66667

SSs=∑x2-CF P

SSs=71082- 3526.66667 20

SSs= 3554.1-3526.66667

SSs= 27.43333


Given: CF= (∑x)2 = (460)2

∑y2 = 10624 3N 60S = 3 CF=3526.66667Step No. 4

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 treatmentCF = Correction FactorS = Sample (number of experimental group


SSP=∑y2-CF s

SSP =10624- 3526.66667

3SSP = 3541.33333-3526.66667

SSp= 14.66667

Given: CF= (∑x)2 = (460)2

∑∑ij2= 3574 3N 60

CF = 3526.66667 Step No. 5Compute the Sum of Squares for Total

(SST)

SST= ∑∑ij2- CF

Where: SST = Sum of squares for total∑∑ij

2= Grand sum of each observation per treatmentCF = Correction factor


SST= ∑∑ij2- CF

SST = 3574 - 3526.66667

SST = 47.3333

Given:SST = 47.3333 SSp= 14.66667

SSs= 27.43333 Step No. 6Compute 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

Given: Ns= 3

NP=20

NT = 60 Step No. 7Get the degrees of freedom of:(dfs= N-1) and (dfP= N-1) and

(dfT= N-1) and dfE= dfT- (dfS + df P)

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 + df P)

= 20-1 =59-(2+19) = 19 dfE =38

Given: SSS= 27.43333 SSE= 5.23333

SSp =14.66667 dfs = 2 ; dfP = 19; dfE= 38 Step No. 8 & 9

Compute the Mean Square (MS) Computation and the Mean square of error

(MSS)= SSS and (MSp)=SSp and (MSE)= SSE

dfs dfp dfE

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


MSS = SSS (MSE)= SSE

dfs dfE

= 27.43333 = 5.23333

2 38

=13.71667 = 0.137719

(MSp)=SSp

dfp

=14.66667 19

=0.77193

Given:MSS= 13.71667 MSE=0.137719

MSp =0.77193 Step No. 10

Observe F ComputationFs = MSS and Fp = MSP

MSE MSE

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

0.137719 0.137719

=99.59873 =5.605096(Significant @ level 0.01) (Significant @ level 0.01)

Remember of our Rejection Region-


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

0.137719 0.137719

=99.59873 =5.605096(Significant @ level 0.01) (Significant @ level 0.01)

Tabular FS = Tabular Fp =

df 2,38 (0.01) = 5.21 df 19,38 (0.01) = 5.605096

(Hence: Please see tabular value of F on your copy)

Remember of our Rejection Region-


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.


ANOVA Source of Variation SS df MS F P-value F crit

Rows 14.667 19 0.77193 5.605096 3.3E-06 2.421474Columns 27.433 2 13.71667 99.59873 7.7E-16 5.211225Error 5.2333 38 0.137719 Total 47.333 59

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.

Use the data that I gave for solving this, refer to your excel file.

Using computer:


•On the microsoft excel 2007,Highlight the data. Click the Data then Click Data Analysis.1•Click ANOVA: Two-factor without replication2•Type in the Input Range (x1,x2,x3)•Type 0.01 Alpha then ok3

The computer displays as follows:

WHAT IS FRIEDMAN TWO-WAY ANALYSIS OF VARIANCE BY RANKS (XR

2)?

According to Dr. Laurentina Paler-Calmorin (2011)

• It is another inferential statistics used to both descriptive and experimental researches when data from K related samples consist o atleast an ordinal scale have been drawn from the same population.

What is Factorial Analysis of Variance?

Where:Xr

2 = Friedman two-way ANOVA by ranks

N = Number of rowsK = Number of columns

• In Friedman’s two-way ANOVA by ranks, the data are arranged in two-way table, the rows and the columns. The formula is expressed as follows:

• Xr2= 12 ∑(R) 2 – 3N (K + 1)

• NK (K+1)

To substitute formula, the steps are as follows:

•Prepare a two-way table consists of rows and columns.1

•Enter the data in steps 1 and rank horizontally and the lowest rank is 1;second lowest value,2; third lowest value,3 and so on.(refer to the scale)2•Add the rank in each column. Apply the formula.•Xr

2= 12 ∑(R) 2 – 3N (K + 1)• NK (K+1)3


Statement of 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 ranksSignificance Level: Alpha= 0.01Sampling Distribution : K= 4 N=20

Rejection Region:

The null hypothesis is rejected if the computed Friedman (XR

2) 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


Facilities Top Middle Managers Lower Managers Professors

X1 Fr X2 Fr X3 Fr X4 Fr

1 Classrooms 3.2 3 3.2 3 3.2 3 3.1 1

2 Air Conditioned Offices 3.3 2.5 3.3 2.5 3.3 2.5 3.3 2.5

3 Books 3.1 3 3.1 3 3.1 3 2.8 1

4 Reference books 2.9 2.5 2.9 2.5 2.9 2.5 2.9 2.5

5 Buildings 3.4 2.5 3.4 2.5 3.4 2.5 3.4 2.5

6 Library holdings 2.6 2.5 2.6 2.5 2.6 2.5 2.6 2.5

7 Laboratory rooms 3.2 4 3.1 3 3 2 2.9 1

8 Equipment 3.1 3.5 3.1 3.5 2.9 2 2.8 1

9 Apparatus 3 3.5 3 3.5 2.9 1.5 2.9 1.5

10 Computers 3.5 2.5 3.5 2.5 3.5 2.5 3.5 2.5

11 Internet Café 3 2.5 3 2.5 3 2.5 3 2.5

12 Ventilation 2.5 4 2.4 3 2.3 2 2.2 1

13 Comfort rooms for professors 3.2 2.5 3.2 2.5 3.2 2.5 3.2 2.5

14 Comfort rooms for employees 3 2.5 3 2.5 3 2.5 3 2.5

15 Comfort rooms for sudents 2.8 2.5 2.8 2.5 2.8 2.5 2.8 2.5

16 Drinking fountains 1 2.5 1 2.5 1 2.5 1 2.5

17 Water supply 2.9 2.9 2.8 3 2.7 2 2.5 1

18 Intercom 3.1 3.1 3.1 3 3.1 3 2.4 1

19 Canteen 3 3 3 2.5 3 2.5 3 2.5

20 Vehicles 2.8 2.8 2.8 3 2.8 3 2.6 1

Total Rank 58.5 55.5 49 37

Total Rank2 3422.25 3080.25 2401 1369 10272.5

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 valuedf = K-1 df 3(0.01) = 11.34

Df = 4-1Df = 3

Scale :4 -very much adequate5 -adequate6 -fairly adequate1 -Inadequate

Given:∑(R) 2 =102722.5N =20K =4


Where:Xr

2 = Friedman 2-way ANOVA by rank

N =Number of rowsK =Number of columns

Remember of our Rejection Region- (XR

2) 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 perceivedby top managers, middle managers, lower managers and professor.


Xr2 = 8.75 (insignificant at 0.01 level)

Degrees of freedom Tabular valuedf = K-1 df 3(0.01) = 11.34

Df = 4-1Df = 3

(Hence: Please see critical values of Chi-square)

Books :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.,Leon Llido St., General Santos City

Online Resources:www. statford.comwww.psych.nyu.eduwww.mathworks.comwww.statsoft.com

References:

http://www.mathworks.com/