ANALYSIS OF VARIANCE (ANOVA)

PA 298 RESEARCH for SOCIAL SCIENCE

PRESENTED BY: MS. MANNIELYN MENONG

PRESENTED TO: DR. MARIA THERESA P. PELONES

A METHOD FOR DIVIDING THE VARIATION OBSERVED INTO DIFFERENT PARTS, EACH PART ASSIGNABLE TO A KNOWN SOURCE, CAUSE, OR FACTOR.

USED WHEN WE WISH TO TEST THE SIGNIFICANCE OF THE DFFERENCES BETWEEN THE TWO OR MORE MEANS OBTAINED FROM INDEPENDENT SAMPLES.

F test for differences among more than two variables.

UNDERSTANDING ANALYSIS OF VARIANCE

Sum ofSquares. A. Deviation Method

Within-GroupsSum of Squares(SSw)

BetweenGroupsSum ofSquares(SSb)

Total Sum of Squares (SSt)

MeanSquare(MS)

THE F-TEST OR RATIO

X X x² X²

9 4 16 81

8 3 9 64

7 2 4 49

6 1 1 36

5 0 0 25

4 -1 1 16

3 -2 4 9

2 -3 9 4

1 -4 16 1

ΣX=45 Σx=0 Σx²=60 ΣX²=285

(Deviation Method) (Raw Score Method)

SSw=Σx²+Σx²+∑x²+∑xWhere:

x-= a deviation score (X-X)

X1 x x²

2

4

3

3

4

∑X=16 X= ∑x=

X2 x x²

2

3

3

3

4

∑X=15 X= ∑x

Group1 (N=5) (Superior)

Group2 (N=5)(Above Average)

X3 x x²

2

3

2

3

4

∑X=14 X ∑x

X4 x x²

4

3

2

2

2

∑X=13 X ∑x

Group3 (N=5)(Average)

Group4 (N=5(Below Average)

Plugging the formula:SSw=2.8+2.0+2.8+3.2=10.8 Answer

Xtotal=58/20=2.9

SSb=∑ (X – Xt)N

Where:X= mean of any groupXt= the mean of the total distributionN= the number of scores in any groupSSb= between -groups sum of squares

Applying the formula,

SSb=∑ (X – Xt)N = (3.2 – 2.9) 5 + (3 – 2.9) 5 + (2.8 – 2.9) 5 + (2.6 – 2.9) 5 = (.3) 5 + (.1) 5 + (-.1) 5 + (-.3) 5 = .09(5) + .01(5) + .01(5) + .09(5) = .45 + .05 + .05 + .45SSb= 1 Answer

4. Total Sum of Squares (SSt)

By formula,

SSt= SSb + SSwSSt= 1 + 10.8SSt= 11.8 Answer

By definition, you can solve the sum of squares through this formula.SSt= ∑( X - X t)

Where:X- z raw score in any groupXt= the mean from all the groups combinedSSt= the total sum of squares

Example 4| For the data in Example 2,Solve the sum of squares

SSt= (2-2.9)+(4-2.9)+(3-.29) +(3-2.9)+(4-2.9)+(2-2.9) +(3-2.9)+(3-2.9)+(3-2.9) +(4-2.9)+(2-2.9)+(3-2.9) +(2-2.9)+(3-2.9)+(4-2.9) +(4-2.9)+(3-2.9)+(2-2.9) +(2-2.9)+(2-2.9) =(-.9)+(1.1)+(.1)+(.1)+(1.1)+ (-.9)+(.1)+(.1)+(.1)+(1.1)+ (-.9)+(.1)+(-.9)+(.1)+(1.1)+ (1.1)+(.1)+(-.9)+(-.9)+(-.9) =.81 + 1.21 + .01 + .01 + 1.21 + .81+ .81 + .01 + .81 + .01 + 1.21 +1.21 + .01 + .81 + .81 + .81 = 11.8 Answer

By formula,

MSb= SSb / dfb

Where:MSb= Mean Square of bet.- GroupsSSb=Sum of Squares of the bet.-groupsdfb=degrees of freedom of the bet.-groups

AndMSw= SSw / dfwWhere:MSw=Mean Square of the within-groupsSSw=the Sum of Squares of the within-groupsdfw=degrees of freedom within-groups

In determining the MSb and MSw, you need to solve

the appropriate degrees of freedom between andwithin. The formulas to use are:

dfb= k -1Dfw= Nt – k

Where:k = the number of groupsNt = total number of scores in all groups combined

Using the data in ex. 2 & 3, solve the MSb & MSw. Since the SSw is 10.8 and SSb is 1, the correspondingdegrees of freedom are as follows:

dfb = 4-1 dfw= 20 -4 = 3 Answer = 16 Answer

So, MSw = 10.8 / 16 = .675 Answer MSb= 1/ 3 = .333 Answer

F = MSb/ MSw

For the data in example 5, the F-ratio is,

F = .333 / .675 = .493 Answer

To determine if the computed F-ratio is sig-nificant, we shall look at the two df values, bet. Groups degrees of freedom and within-groups degrees of freedom in the table. WeShould not forget that the degrees of freedomWith the numerator (df bet.) are located accross the top of the page, while the degrees of Feedom associated with the denoinator (df within) are found on the left column of the table. The body of the table shows significant F-ratios at .05 and .01 confidence levels. The computed F-ratio shown in ex. 6 is .493. Thus using the table, the tabular value at .05 is 3.24 and at .01 is 5.29. This is done by tracing the four degrees of freedom between (numerator), and the 16 degrees of freedom within (denominator) in the table. Since the F-ratio is .493, we have no choice but to accept the

null hypothesis and attribute the group mean differences to sampling error rather than to a true difference in the populations of the groups (superior, above average, average, and below average.

Example A. Deviation Method The following academic grades are:

X1 x x²

90 -1.8 3.24

92 .2 .04

89 -2.8 7.84

395 3.2 10.24

493 1.2 1.44

∑X=459 ∑x=22.80

X1 x x²

78 1 1

75 -2 4

77 0 0

79 2 4

76 -1 1

∑X=385 X= ∑x=10

X1 x x²

85 -1.2 1.44

87 3.2 10.24

83 -.8 .64

84 .2 .04

80 -3.8 14.44

∑X=419 X= ∑x=26.80

Above Average (N=5)

Below Average (N=5)

Average (N=5)

Xt= 84.2

Step by step procedure for testing the statistical significance of the three teaching aptitude groups’

mean difference.

Step 1. Compute the Mean for Each Group X=∑X1/N=459/5=91.8 Answer X=∑X2/N=419/5=83.8 Answer X=∑Xз/N=385/5=77.0 Answer

Step 2. Compute the Within-Groups Sum of Squares SSw=∑x+∑x+∑x+∑x =22.8+26.8+10 =59.6 Answer

Step 3. Compute the Between-Groups Sum of Squares SSb=∑(X-Xt)N =(91.8-84.2)5+(83.8-84.2)5+(77-84.2)5 =(7.6)5+(-.4)5+(-7.2)5 =(57.76)5+(.16)5+(51.84)5 =288.8+.8+259.2 =548.8 Answer

Step 4. Compute the Total Sum of Squares SSt=SSbetween+Sswithin =548.8+59.6 =608.4 Answer

Step 5. Compute the Between-Groups and Within-Groups Degrees of Freedom dfb=k-1 =3-1 =2 Answer

dfw=Ntotal-k =15-3 =12 Answer

Step 6.Compute the Between-Groups and Within-Groups Mean Square MSb=Ssbetween/dfbetween =548.8/2 =274.4 Answer MSw=Sswithin/dfwithin =59.6/12 =4.97 Answer

Step 7. Compute the F-ratio F=MSb/MSw =274.4/4.97 =55.21 Answer

Step 8. Enter your computed data in a table

Summary Table for the ANOVA

Source of Variation

df Sums of Squares

Mean of Squares

Between groups 2 548.8 274.4

Within groups 12 59.6 4.97

Total 14 608.4

Step 9. Interpret the result by comparing the computed F ration with the appropriate tabular value.

Computed F ratio = 55.21

Tabular values = 3.88 at .05 = 6.93 at .01

Degrees of freedom= 2/12

Since the computed F ratio of 55.21 is greater than 3.88 at the .05 level, the null hypothesis is rejected and the research or alternative hypothesis is accepted. Specifically, we can conclude that the above average, average, and below average teaching aptitude groups differ significantly with respect to academic performance.

EXERCISE:

1. Compute the F-ratio among the following three groups of scores:

Group I Group II Group III17 9 811 10 713 12 1012 10 811 6 6

2. On the following four groups of teaching attitude, test the null hypothesis that academic performance does not vary due to teaching attitude.

Superior Above Average Below Average Average

90 85 80 7889 86 82 7688 84 83 7594 83 81 7793 88 80 75

REPORTED BY: MANNIELYN C. MENONG