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

1. ANALYSIS OF VARIANCE (ANOVA) PA 298 RESEARCH for SOCIAL SCIENCE PRESENTED BY: MS. MANNIELYN MENONG PRESENTED TO: DR. MARIA THERESA P. PELONES

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

5. 1. Sum of Squares. A) Deviation Method (Deviation Method) (Raw Score Method)

6. 2. Within-Groups Sum of Squares (SSw) SSw=Σx²+Σx²+∑x²+∑x Where: x-= a deviation score (X-X) Group1 (N=5) (Superior) Group2 (N=5) (Above Average)

7. 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

8. 3. Between Groups Sum of Squares (SSb) SSb=∑ (X – Xt)N Where: X= mean of any group Xt= the mean of the total distribution N= the number of scores in any group SSb= 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 + .45 SSb= 1 Answer

9. Example 4| For the data in Example 2, Solve the sum of squares 4. Total Sum of Squares (SSt) 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, SSt= SSb + SSw SSt= 1 + 10.8 SSt= 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 group Xt= the mean from all the groups combined SSt= the total sum of squares

10. 5. Mean Square (MS) By formula, MSb= SSb / dfb Where: MSb= Mean Square of bet.- Groups SSb=Sum of Squares of the bet.-groups dfb=degrees of freedom of the bet.-groups And MSw= SSw / dfw Where: MSw=Mean Square of the within-groups SSw=the Sum of Squares of the within-groups dfw=degrees of freedom within-groups In determining the MSb and MSw, you need to solve the appropriate degrees of freedom between and within. The formulas to use are: dfb= k -1 Dfw= Nt – k Where: k = the number of groups Nt = 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 corresponding degrees of freedom are as follows: dfb = 4-1 dfw= 20 -4 = 3 Answer = 16 Answer

11. So, MSw = 10.8 / 16 = .675 Answer MSb= 1/ 3 = .333 Answer 6. The F-test Ratio F = MSb/ MSw For the data in example 5, the F-ratio is, F = .333 / .675 = .493 Answer 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. 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. We Should not forget that the degrees of freedomWith the numerator (df bet.) are located accro ss the top of the page, while the degrees of Feedom associated with the denoinator (df with in) 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

12. Procedure in Solving one-Way Analysis of Variance Example A. Deviation Method The following academic grades are: Above Average (N=5) Below Average (N=5) Xt= 84.2 Average (N=5)

13. 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

14. 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

15. 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

16. 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.

17. EXERCISE: 1. Compute the F-ratio among the following three groups of scores: Group I Group II Group III 17 9 8 11 10 7 13 12 10 12 10 8 11 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 78 89 86 82 76 88 84 83 75 94 83 81 77 93 88 80 75

18. Reported by: MANNIELYN C. MENONG THANK YOU AND GODBLESS!