F test Analysis of Variance (ANOVA)

Prepared by:MARIANNE G. MALUYO

F-Test or Analysis of Variance (ANOVA)

- An inferential statistics used to determine the significant difference of three or more variables or multivariate collected from experimental research.

F-test: single factor analysis of variance involves the independent variables as basis for classification. This is usually applied in single-group design and complete randomized design (CRD).

Step 1. Partition the sum of squares for treatment, error, and total by using the appropriate formula:

Computation for the Sum of Squares for Treatment (SS )

SS = ΣY²- CF

NWhere:

SS = Sum of squares for treatment

ΣY² = Sum of squared total for treatment

CF = Correction factor or CF = (ΣX)²

N

rtΤ

rtΤ

rtΤ

Computation for the Sum of Squares for Total (SS )

SS = ΣΣY ij² - CF

NWhere:

SS = Sum of squares for treatment

ΣΣYij = Sum of squared total for treatment

CF = Correction factor

Τ

T

Τ

Computation for the Sum of Squares for Error (SS )

SS = SS - SS

Ε

Ε Τ rtΤ

Step 2. Divide the sum of squares for treatment, error, and total with the corresponding degrees of freedom (df), df=N-1, to get the mean square by using this formula, MS=SS/df

Step 3. Divide the mean square treatment by mean square to get the F-value. The formula in getting F equals mean square for treatment

(MS ) divided by mean square for error (MS ) which is expressed as follows:

F= MS MS

rtΤ

rtΤ

Ε

Ε

Step 4. Refer to the F –distribution table in the appendix to determine if the computed F value is significant or not at .01 or .05 level of confidence

Step 5. Prepare F-test or ANOVA Table by entering the values of Steps 1, 2 and 3.

Illustration:

Supposed the researcher wishes to determine the effectiveness of teaching Biology using Method 1, Method 2, Method 3 and Method 4 to B.S. Biology students at the University of Sto. Tomas. The specific research problem – “Is there a significant difference on the Effectiveness of Teaching Biology using Method 1, Method 2, Method 3, and Method 4 to BS Biology Students at the University of Sto. Tomas?”

Independent Variables Dependent VariablesMethod of Teaching I

Method of Teaching II

Method of Teaching III

Method of Teaching IV

Mean Grade in:

Preliminary period

Mid-Term period

Final period

To answer the problem, consider the following:

0===== XXXXH 0 432

1

1

6. Computation of F-test or ANOVA: Single Factor on the Effectiveness of Teaching Biology using Method 1, Method 2, Method 3, and Method 4 to BS Biology Students at the University of Sto. Tomas

Mean Grade (Period)

Treatment Prelim Midterm Final Total

Method 1 2.5 2.4 2.3 7.2

Method 2 2.0 1.9 1.9 5.8

Method 3 2.8 2.7 2.5 8.0

Method 4 1.5 1.4 1.3 4.2

Total 25.2

Computation for the Sum of Squares for Treatment (SS )SS = ΣY²- CF CF = (∑X)²

N N =7.2² + 5.8² + 8² + 4.2² - (25.2)²

3 3 3 3 12 = 51.84 + 33.64 + 64 + 17.64 - 635.04

3 3 3 3 12 = 17.28 + 11.2133333 + 21.3333333 + 5.88

– 52.92 = 55.7066666 – 52.92

SS = 2.786667

rtΤ

rtΤ

rtΤ

Computation for the Sum of Squares for Total (SS )

SS = ΣΣY ij² - CF

N

= 2.5² + 2.4² + 2.3² + 2² + 1.9² + 1.9² +2.8² + 2.7² + 2.5² +

1.5² + 1.4² + 1.3² -52.92

= 6.25 + 5.76 + 5.29 + 4 + 3.61 + 3.61 + 7.84 + 7.29

+ 6.25 + 2.25 + 1.96 + 1.69 – 52.92

= 55.8 – 52.92

SS = 2.88

Τ

T

Τ

Computation for the Sum of Squares for Error (SS )

SS = SS - SS= 2.88 – 2.786667

SS = 0.093333

Ε

Ε Τ rtΤ

Ε

degrees of freedom (df) computationdf =N – 1 df =df - df

= 4 – 1 =11-3df =3 df = 8

df = N – 1= 12 – 1

df = 11

rtΤ rtΤ

Τ

E

Τ

rtΤ E

Τ

Mean Square (MS) ComputationMS = SS MS = SS

df df= 2.786667 = 0.09334

3 8MS = 0.928889 MS = 0.011667

rtΤ

rtΤ

rtΤ

rtΤ E

E

E

E

Observed F Computation

F = MS Tabular FMS df₃ ₈ ₍.₀₁₎ˌ = 7.59**

= 0.928889 Computed F 0.011667 df₃ ₈ ₍.₀₁₎ˌ = 79.61678**

F = 79.6167824**(highly significant at .01 level)

rtΤ

Ε

F-test or ANOVA Single Factor Table Effectiveness of Teaching Biology using Method 1, Method 2, Method 3, and Method 4 to BS Biology Students at the University of Sto. Tomas

Source of Variance

Degrees of Freedom

Sum of Squares

Mean Square

Observed F Tabular F 1%

Treatment 3 2.786667 0.928889 79.61678 7.59

Error 8 0.93333 0.11667

Total 11 2.88

7. Interpretation: The computed F-value obtained is 79.6178 which is greater than the F-tabular value of 7.59 with df 3,8 at .01 level of confidence, hence significant. This means that the teaching Biology using Method 1, Method 2, Method 3, and Method 4 to BS Biology Students at the University of Sto. Tomas really differ with each other because Method 4 is more effective in teaching Biology. Thus, the null hypothesis is rejected.

A B C D Cell1 2.5 2.0 2.8 1.5 Cell2 2.4 1.9 2.7 1.4 Cell3 2.3 1.9 2.5 1.3

Anova: Single Factor SUMMARY

ANOVA

Groups Count Sum Average Variance

Column 1 3 7.2 2.4 0.01

Column 2 3 5.8 1.933333 0.003333

Column 3 3 8 2.666667 0.023333

Column 4 3 4.2 1.4 0.01

Source of Variation

SS df MS F F crit

Between Groups 2.786667 3 0.928889 79.61905 7.590992

Within Groups 0.093333 8 0.011667

Total 2.88 11

F-test Two Factor or ANOVA Two Factor Involves three or more independent

variables as basis for classificationIs appropriate for parallel group

design. In this design, three or more groups are used at the same time with one variable is manipulated or changed.

IllustrationSupposed the researcher wishes to conduct a

study on the flavor acceptability of luncheon meat from commercial, milkfish bone meal, and goatfish bone meal. Commercial luncheon meat is the control group while milkfish bone meal and goatfish bone meal luncheon meat are experimental groups. The specific research problem-”Is there a significant difference on the flavor acceptability of luncheon meat from commercial, milkfish bone meal, and goatfish bone meal?”

To answer the research problem, consider the following:1. Null Hypothesis: There is no significant difference on the flavor acceptability of luncheon meat from commercial, milkfish bone meal, and goatfish bone meal.

2. Statistical Tool: F-test two-factor or ANOVA two-factor3. Significance level: 0.014. Sampling distribution: N = 205. Rejection section: The null hypothesis is rejected if the computed F-value is equal to or greater than the tabular F-value.

2

0==== XXXH 1 30 2

6. Computation of F-test Two Factor or ANOVA Two Factor on the flavor acceptability of luncheon meat from commercial, milkfish bone meal, and goatfish bone meal.

Scale:9 like extremely8 like very much7 like moderately6 like slightly

PanelistLuncheon Meat Total

Commercial Milkfish Bone Meal Goatfish Bone Meal

1 8 8 7 23

2 8 9 7 24

3 7 8 6 21

4 7 8 6 21

5 8 9 8 25

6 8 8 7 23

7 8 9 7 24

8 7 9 6 22

9 9 9 8 26

10 8 8 7 23

11 9 9 7 25

12 8 9 7 24

13 8 9 7 24

14 7 8 6 21

15 8 9 7 24

16 8 8 7 23

17 7 8 6 21

18 7 8 7 22

19 7 8 6 21

20 8 8 7 23

Total 155 169 136 460

Mean 7.75 8.45 6.8

Sum of Squares for Samples (SS ) ComputationSS = ΣX²- CF

P

=155² + 169² + 136² - (460)²

20 60

= 24025 + 28561 + 18496 - 211600

20 60

= 71082 – 3526.66667

20

= 3554.1 – 3526.66667

SS = 27.43333S

S

SS = ΣY²- CF

S

=23² + 24² + 21² + 21² + 25² + 23² + 24² + 22² + 26² + 23² + 25² + 24² +

3

24² + 21² + 24² + 23² + 21² + 22² + 21² + 23² - 3526.66667

3

= 529 + 576 + 441 + 441 + 625 + 529 + 576 + 484 + 676 + 529 + 625 +

3

576 + 576 + 441 + 576 + 529 + 441 + 484 + 441 + 529 - 3526.66667

3

= 10624 – 3526.66667

3

= 3541.33333 – 3526.66667

SS = 14.66667P

P

Sum of Squares for Total (SS ) ComputationSS = ΣΣY ij² - CF

= 8²+8²+7²+8²+9²+7²+7²+8²+6²+7²+8²+6²+8²+9²+8²+8²+

8²+7²+8²+9²+7²+7²+9²+6²+9²+9²+8²+8²+8²+7²+9²+9²+7²

+8²+9²+7²+8²+9²+7²+7²+8²+6²+8²+9²+7²+8²+8²+7²+7²+

8²+6²+7²+8²+7²+7²+8²+6²+8²+8²+7² - 3526.66667

=64+64+49+64+81+49+49+64+36+49+64+36+64+81+64+ 64+64+49+64+81+49+49+81+36+81+81+64+64+64+49+

81+81+49+64+81+49+64+81+49+49+64+36+64+81+49+

64+64+49+49+64+36+49+64+49+49+64+36+64+64+49 - 3526.66667

= 3574 - 3526.66667

SS = 47.3333

Τ

T

Τ

Sum of Squares for Error (SS ) Computation

SS = SS - (SS + SS )

= 47.33333 – (27.43333 + 14.66667)

SS = 47.33333 – 42.1

Ε

Ε Τ S

Ε

P

degrees of freedom (df) computationdf = N – 1 df = N-1

= 3 – 1 =60-1df = 2 df = 59

df = N – 1 df =df – (df +df )= 20 – 1 =59 – (2+9)

df = 19 =59- 21 df =38

S

P

P

T

P

S Τ

ΤE S

E

Mean Square (MS) ComputationMS = SS MS = SS

df df=27.43333 = 5.233333

2 38MS =13.71667 MS = 0.137719

MS = SS df= 13.71667 19

MS = 0.77193

SS

S

S

E

E

E

E

P P

P

P

Observed F Computation

F = MS Tabular FMS df₂ ₃₈ ₍.₀₁₎ˌ = 5.21**

= 13.71667 Computed F 0.137719 df ₂ ₃₈ ₍.₀₁₎ ˌ = 99.59873**

F =99.59873**(significant at .01 level)

S

Ε

S

S

F = MS Tabular F (Panelists) MS df₁₉ ₃₈ ₍.₀₁₎ˌ = 2.42**

= 0.77193 Computed F 0.137719 df ₁₉ ₃₈ ₍.₀₁) = 5.605096**ˌ

F =5.605096**(significant at .01 level)

P

Ε

P

S

F-test Two Factor or ANOVA Two Factor Table Effectiveness on the acceptability of luncheon meat from commercial, milkfish bone meal, and goatfish bone meal.

Source of Variance

Degrees of Freedom

Sum of Squares

Mean Square

Observed F Tabular F 1%

Samples 2 27.43333 13.716667 99.59873** 5.21**

Panelists 19 14.66667 0.77193 5.605096** 2.42

Error 38 5.233333 0.137719

Total 59

7. 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 .01 level of significance with df = 2.38. For panelists, the computed F-value obtained is 5.605096 also greater than the tabular F-value of 2.42 and also significant at .01 level of confidence with df = 19,38. This means that the samples and evaluation of the panelists really differ with each other because milkfish bone meal luncheon meat is most acceptable. Hence, the null hypothesis is rejected.

Thank You!

Illustration:Suppose the researcher wishes to determine the

effect of chicken dung as organic fertilizer upon the yield of tomatoes. There are three plots planted with tomatoes. First plot has 5% chicken dung as organic fertilizer; the second plot, 8%; and third plot, 10%. The specific research problem is “Is there a significant difference on the effect of chicken dung as organic fertilizer upon the yield of tomatoes planted in plots?”

Independent Variables Dependent Variables

Treatment(Chicken Dung)

T1 5%T2 8%

T3 10%

Yield(Weight of Tomatoes)

Figure 8.1 Independent and Dependent Variables on the effect of Chicken Dung as Organic Fertilizer Upon the Yield of Tomatoes Planted in Plots.

To answer the aforementioned specific research problem, consider the ff:

1. Null Hypothesis: there is no significant difference on the effect of chicken dung as organic fertilizer upon the yield of tomatoes planted in plots.

2. Statistical Tool: F-Test or ANOVA Single Factor3. Significant level: Alpha = 0.014. Sampling Distribution: Treatment = 35. Rejection section: The null hypothesis is rejected

if the computed F-Value is equal to or greater than tabular F-value.

6. Computation of F-Test or ANOVA Single Factor on the Effect of Chicken Dung as Organic Fertilizer upon the Yield of Tomatoes Planted in Plots

Treatment (T) Weight of Tomatoes (kg)Sampling

1 2 3 4 5

Total

T1 5 4 3 2 1 15

T2 8 7 6 5 4 30

T3 10 9 8 7 6 40

Total 85

Computation for the Sum of Squares for Treatment (SS )SS = ΣY²- CF CF = (∑)²

N N

=15² + 30² + 40² - (85)²

5 5 5 15

= 225 + 900 + 1600 - 7225

5 5 5 15

= 45 + 180 + 320 – 481.666667

SS = 63.33333

rtΤ

rtΤ

rtΤ

Computation for the Sum of Squares for Total (SS )

SS = ΣΣY ij² - CF

N

= 5² + 4² + 3² + 2² + 1² + 8² + 7² + 6² + 5² + 4² + 10² + 9²

+ 8² + 7² + 6² - 481.666667

= 25 + 16 + 9 + 4 + 1 + 64 + 49 + 36 + 25 + 16 + 100 + 81 + 64 + 81 + 64 + 49 + 36 – 481.666667

= 575 - 481.666667

SS = 93.33333

Τ

T

Τ

Computation for the Sum of Squares for Error (SS )

SS = SS - SS= 93.3333 – 63.33333

SS = 30

Ε

Ε Τ rtΤ

Ε

degrees of freedom (df) computationdf =N – 1 df =df - df

= 3 – 1 =14-2df =2 df = 12

df = N – 1= 15 – 1

df = 14

rtΤ rtΤ

Τ

E

Τ

rtΤ E

Τ

Mean Square (MS) ComputationMS = SS MS = SS

df df= 63.33333 = 30 2 12

MS = 31.666665 MS =2.5

rtΤ

rtΤ

rtΤ

rtΤ E

E

E

E

Observed F Computation

F = MS Tabular FMS df₂ ₁₂ ₍.₀₁₎ ˌ = 6.93**

= 31.66665 Computed F2.5 df₂ ₁₂ ₍.₀₁₎ ˌ = 2.66667**

F = 12.666667**(significant at .01 level)

rtΤ

Ε

F-test or ANOVA Single Factor Table on the Effect of Chicken Dung as Organic Fertilizer Upon the Yield of Tomatoes Planted in Plots Source of Variance

Degrees of Freedom

Sum of Squares

Mean Square

Observed F Tabular F 1%

Treatment 2 63.3333 31.66665 12.66667 6.93

Error 12 30 2.5

Total 14 93.3333

7. Interpretation: The computed F-value obtained is greater than the F-tabular value. Thus the computed F-value is significant at 1 percent level of confidence. This means that the yield of tomatoes using chicken dung as organic fertilizer really differ with each other because tomatoes treated with 10% chicken dung is most effective as organic fertilizer in planting tomatoes in plots. Hence the null hypothesis is rejected.

Cell A Cell B Cell C5 8 10 4 7 9 3 6 8 2 5 71 4 6

Anova: Single Factor SUMMARY

ANOVA

Groups Count Sum Average Variance

Column 1 5 15 3 2.5

Column 2 5 30 6 2.5

Column 3 5 40 8 2.5

Source of Variation

SS df MS F F crit

Between Groups 63.33333 2 31.66667 12.66667 6.9266

Within Groups 30 12 2.5

Total 93.33333 14