ANNOVA

TWO WAY ANALYSIS OF VARIANCE

Ò  The one way analysis of variance technique is used to analyse the effect of one independent variable or one type of treatment. It is used to study the main effect of one variable only.

Ò  In two way classification two independent variables are taken simultaneously.  It has two main effects and one interaction effect of two variables on the dependent variable.

Ò  Usually in two way classification 3 F – values are calculated. Two F values for two main effects and one F value for interaction effect.

Example for two way classification data

Levels (B) Intelligence		Treatment   (A)   (Methods)
		A1	A2
	B1	M11	M12	MB1
	B2	M21	M22	MB2
		MA1	MA2	T

Types of effects

Main Effects (two)

             Main effect of A = M A1 – M A2 = DA

             Main effect of B = M B1 – M B2 =DB

 Simple Effects (four)

     Simple effect at B1 level = M 11 – M 12 = d1

             Simple effect at B2 level = M 21 – M 22 = d2

             Simple effect at A1 level = M 11 – M 21 = d3

             Simple effect at A2 level = M 12 – M 22 = d4

Interaction Effects (two)

First Interaction effect = d1 – d2 

Second Interaction effect = d3 - d4

Steps involved in two way ANOVA

Ò  First Step - Finding Correction factor C =T²/N

Ò  Second Step - Sum of Square Total (SS_T)=∑(X₁²+X₂²+X₃²+…..X_n²) –  C

Ò  Third Step – Sum of square of A treatments

    (SS_A) = (∑A₁)²+ (∑A₂)²/2n – C

Ò  Fourth Step – Sum of square of level ‘B’ intelligence

    (SS_B) = (∑B₁)²+ (∑B₂)²/2n – C

Ò  Fifth Step - Sum of Square Cells

    SS_Cell = (∑A₁B₁)²+ (∑B₁A₂)²+ (∑B₂A₁)² +(∑B₂A₂)² /n – C

Ò  Sixth Step

    Sum of Square A X B (SS_AB)= SS_Cell – (SS_A +SS_B )      

Ò  Seventh Step – Sum of Square within subject    SS_W = SS_T – SS_Cell

Ò  Eighth Step – Analysis of Variance Table for 2 X 2 design and finding the value of F

Ò  Ninth Step – Interpretation

Example sum – Method  x  Intelligence

Levels Intelligence	Treatment Methods
	A1		A2
High  B1	12		14
	13	∑A1B1=68	14	∑A2B1=69
	14	n=5	13	n=5	∑B1=137
	15		13
	14		15
Low B2	14		11
	16	∑A1B2=74	10	∑A2B2=57	∑B1=131
	16	n=5	12	n=5
	15		13
	13		11
		∑A1=142		∑A2=126	T=268

 

The obtained scores are modified by subtracting a constant 10 from each score.  The Variance will not change by subtracting a constant 10 from each score but it facilitates the computational process. 

            The procedure of two way analysis of variance is used on the modified scores given in the following table.  There are four groups and four cells in 2 X 2 design.

Modified Score

(Subtracting a constant 10)

Levels Intelligence	Treatment Methods
	A1		A12	A2		A22
High  B1	2		4	4		16
	3	∑A1B1=18	9	4	∑A2B1=19	16
	4	n=5	16	3	n=5	9	∑B1=37
	5		25	3		9
	4		16	5		25
	18		70	19		75
Low B2	4		16	1		1
	6	∑A1B2=24	36	0	∑A2B2=7	0	∑B2=31
	6	n=5	36	2	n=5	4
	5		25	3		9
	3		9	1		1
	24		122	7		15	∑ X2=282
		∑A1=42			∑A2=26		T=68

First Step

Correction Factor (c) = T²/N = 68 X 68/20 = 4624/20 = 231

Second Step

Sum of Square Total (SS_T) = ∑ X² – C = 282 – 231 = 51

Third Step

Sum of Square of ‘A’ treatments

(SS_A) = (∑A₁)²+ (∑A₂)²/2n – C = (42² + 26²)/2X5 - 231

=(1764+676/10) - 231 = 244 – 231 = 13

Fourth Step

Sum of square of level ‘B’ intelligence

(SS_B) = (∑B₁)²+ (∑B₂)²/2n – C = (37² + 31²)/2X5 - 231

= (1369+961/10)  - 231 = 2330/10  - 231 = 233 – 231 = 2

Fifth Step

Sum of Square Cells

    SS_Cell = (∑A₁B₁)²+ (∑B₁A₂)²+ (∑B₂A₁)² +(∑B₂A₂)² /n – C

              = (18² + 19² +24² + 7²)/5 – 231 

              = (324+361+576+49)/5 – 231

              = 1310/5 – 231 = 262 – 231 = 31

Sixth Step

Sum of Square A X B (SS_AB)= SS_Cell – (SS_A +SS_B )  

                                              = 31- (13+2)  = 31 – 15 = 16

Seventh Step

Sum of Square within subject    SS_W = SS_T – SS_Cell 

                                                                                        = 51 – 31 = 20

        

Eighth Step

Analysis of Variance Table 2X2 Design

(Two ways classification)

Sources	df	Sum of Square (SS)	Mean Sum of Square(MS)
Methods (A)	(2-1)=1	13	13
Levels (B)	(2-1)=1	2	2
Interaction (AXB)	(2-1)(2-1)=1X1=1	16	16
MSV Between	3	31	10.33
MSV Within	16	20	1.25

Main effect (A) F_A = MS_A/MSVW = 13/1.25 = 10.4 df (1,16)

Main effect (B) F_B = MS_B/MSVW = 2/1.25 = 1.6 df (1,16)

Interaction effect of (AXB) F_AB = MS_AB/MSVW = 16/1.25 = 12.8 df (1,16)

Table value of F with df (1,16)

Levels of Significance                                          0.05(or) 5 %           0.01 (or) 1%

F Values                                                                   4.49                       8.53

The F_A value 10.4 with df (1,16) for the difference between methods is higher than the table value even at 0.01 level of significance.  The null hypothesis is rejected.  It may be stated that the difference between methods is highly significant.

            The F_B value 1.6 with df (1,16) for the difference between levels is not significant at any level.  The null hypothesis is not rejected.

            The F_AB value 12.8 with df (1,16) for the interaction effect is highly significant, because the F value is greater than table value even at 0.01 level of significant.  It may be interpreted that the joint effect of method of teaching and intelligence on criterion variable is significant.