One way ANOVA balanced design

1. One Way Analysis of Variance
ANOVA (Balanced)
Presented by : Eng. Waleed Alzaghal
You Tube Channel : Waleed Alzaghal

2. Assumptions to use one-way ANOVA.

3. In an assembly line in a factory, the number of daily assembled products
were recorded for three operators on five randomly selected days. Below are the
number of the products assembled by each operator.
Carry out the one way ANOVA by hand to test the hypothesis that the
productivity of the three operators is the same (α = 0.05). Consider all assumption
are fulfilled.
Productivity Records / day
Operator
1
Operator
2
Operator
3
7 8 10
10 10 11
8 12 12
10 11 11
7 10 13

4. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups
Within Groups
(Error)
Total
One Way ANOVA table

5. 1) Write the Hypothesises to be tested :
H0 : µ1 = µ2 = µ3
Ha : µ1 ≠ µ2 ≠ µ3 or
Ha : µ1 ≠ µ2 = µ3 or
Ha : µ1 = µ2 ≠ µ3
H0 : All three means are same
Ha : At least one mean is different from
other means.

6. 2) Calculate Degree of Freedom Between Groups (dfB) :
dfB = k – 1 , ( k : number of Groups = 3)
dfB = 3 – 1
dfB = 2
Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups ?
Within Groups
(Error)
Total

7. 2) Calculate Degree of Freedom Between Groups (dfB) :
dfB = k – 1 , ( k : number of Groups = 3)
dfB = 3 – 1
dfB = 2
Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2
Within Groups
(Error)
Total

8. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2
Within Groups
(Error) ?
Total
3) Calculate Degree of Freedom Within Groups (dfw) :
dfW = N – k , ( N = n1 + n2 + n3 = 5 + 5 + 5 = 15 )
dfW = 15 - 3
dfW = 12

9. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2
Within Groups
(Error)
12
Total
3) Calculate Degree of Freedom Within Groups (dfw) :
dfW = N – k , ( N = n1 + n2 + n3 = 5 + 5 + 5 = 15 )
dfW = 15 - 3
dfW = 12

10. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2
Within Groups
(Error)
12
Total ?
4) Calculate Total Degree of Freedom dftotal
dftotal = dfB + dfw
dftotal = 2 + 12
dftotal = 14

11. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2
Within Groups
(Error)
12
Total 14
4) Calculate Total Degree of Freedom dftotal
dfTotal = dfB + dfw
dfTotal = 2 + 12
dfTotal = 14

12. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2
Within Groups
(Error)
12
Total 14 ?
5) Calculate the Correction Factor (CF) :
CF = ( ∑ xi )2 / N = (∑ x1 + ∑ x2 + ∑ x3 )2 / N
CF = (42 + 51 + 57)2 / 15
CF = (150)2 / 15
CF = 1500
x1 x1
2
x2 x2
2
x3 x3
2
7 49 8 64 10 100
10 100 10 100 11 121
8 64 12 144 12 144
10 100 11 121 11 121
7 49 10 100 13 169
n 5 5 5
∑ xi 42 51 57
Mean 8.4 10.2 11.4
∑ xi
2
362 529 655
Observations

13. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2
Within Groups
(Error)
12
Total 14 ?
6) Calculate the Total Sum Square (SSTotal) :
SSTotal = ( ∑ x1
2 + ∑ x2
2 + ∑ x3
2 ) – CF
SSTotal = ( 362 + 529 + 655 ) – 1500
SSTotal = 1546 – 1500
SSTotal = 46
x1 x1
2
x2 x2
2
x3 x3
2
7 49 8 64 10 100
10 100 10 100 11 121
8 64 12 144 12 144
10 100 11 121 11 121
7 49 10 100 13 169
n 5 5 5
∑ xi 42 51 57
Mean 8.4 10.2 11.4
∑ xi
2
362 529 655
Observations

14. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2
Within Groups
(Error)
12
Total 14 46
6) Calculate the Total Sum Square (SSTotal) :
SSTotal = ( ∑ x1
2 + ∑ x2
2 + ∑ x3
2 ) – CF
SSTotal = ( 362 + 529 + 655 ) – 1500
SSTotal = 1546 – 1500
SSTotal = 46
x1 x1
2
x2 x2
2
x3 x3
2
7 49 8 64 10 100
10 100 10 100 11 121
8 64 12 144 12 144
10 100 11 121 11 121
7 49 10 100 13 169
n 5 5 5
∑ xi 42 51 57
Mean 8.4 10.2 11.4
∑ xi
2
362 529 655
Observations

15. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 ?
Within Groups
(Error)
12
Total 14 46
7) Calculate the SS Between Groups (SSB) :
SSB = (( ∑ x1)2/n1 + (∑ x2)2 / n2 + (∑ x3 )2 / n3 ) – CF
SSB = (422/5 + 512/5 + 572/5) – 1500
SSB = ( 352.8 + 520.2 + 649.8) – 1500
SSB = 1522.8 – 1500
SSB = 22.8
x1 x1
2
x2 x2
2
x3 x3
2
7 49 8 64 10 100
10 100 10 100 11 121
8 64 12 144 12 144
10 100 11 121 11 121
7 49 10 100 13 169
n 5 5 5
∑ xi 42 51 57
Mean 8.4 10.2 11.4
∑ xi
2
362 529 655
Observations

16. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8
Within Groups
(Error)
12
Total 14 46
7) Calculate the SS Between Groups (SSB) :
SSB = (( ∑ x1)2/n1 + (∑ x2)2 / n2 + (∑ x3 )2 / n3 ) – CF
SSB = (422/5 + 512/5 + 572/5) – 1500
SSB = ( 352.8 + 520.2 + 649.8) – 1500
SSB = 1522.8 – 1500
SSB = 22.8
x1 x1
2
x2 x2
2
x3 x3
2
7 49 8 64 10 100
10 100 10 100 11 121
8 64 12 144 12 144
10 100 11 121 11 121
7 49 10 100 13 169
n 5 5 5
∑ xi 42 51 57
Mean 8.4 10.2 11.4
∑ xi
2
362 529 655
Observations

17. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8
Within Groups
(Error)
12 ?
Total 14 46
8) Calculate SS Within groups (SSW):
SSTotal = SSB + SSW
SSW = SSTotal – SSB
SSW = 46 – 22.8
SSW = 23.2

18. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8
Within Groups
(Error)
12 23.2
Total 14 46
8) Calculate SS Within groups (SSW):
SSTotal = SSB + SSW
SSW = SSTotal – SSB
SSW = 46 – 22.8
SSW = 23.2

19. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8 ?
Within Groups
(Error)
12 23.2
Total 14 46
9) Calculate the Mean Square Between groups (MSB) :
MSB = SSB / dfB
MSB = 22.8 / 2
MSB = 11.4

20. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8 11.4
Within Groups
(Error)
12 23.2
Total 14 46
9) Calculate the Mean Square Between groups (MSB) :
MSB = SSB / dfB
MSB = 22.8 / 2
MSB = 11.4

21. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8 11.4
Within Groups
(Error)
12 23.2 ?
Total 14 46
10) Calculate the Mean Square Within groups (MSW) :
MSW = SSW / dfW
MSW = 23.2 / 12
MSW = 1.93

22. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8 11.4
Within Groups
(Error)
12 23.2 1.93
Total 14 46
10) Calculate the Mean Square Within groups (MSW) :
MSW = SSW / dfW
MSW = 23.2 / 12
MSW = 1.93

23. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8 11.4 ?
Within Groups
(Error)
12 23.2 1.93
Total 14 46
11) Calculate the test statistic ( F ) :
F = MSB / MSW
F = 11.4 / 1.93
F = 5.907

24. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8 11.4 5.907
Within Groups
(Error)
12 23.2 1.93
Total 14 46
11) Calculate the test statistic ( F ) :
F = MSB / MSW
F = 11.4 / 1.93
F = 5.907

25. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8 11.4 5.907 ?
Within Groups
(Error)
12 23.2 1.93
Total 14 46
12) Find Critical Value for F-Test :
∝ = 0.05
df numerator = dfB = 2
df denominator = dfw = 12
F(0.05, 2 , 12) = ?

26. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8 11.4 5.907 ?
Within Groups
(Error)
12 23.2 1.93
Total 14 46
F(0.05, 2 , 12) = 3.8853

27. Source of
Variation
Degree
of Freedom
(df)
Sum of
Squares
(SS)
Mean
Squares
(MS)
F – Stat. F – Crit.
Between Groups 2 22.8 11.4 5.907 3.885
Within Groups
(Error)
12 23.2 1.93
Total 14 46
F(0.05, 2 , 12) = 3.8853

28. 13) Make conclusion :
As Fstat (5.907) > Fc (3.885 ), we reject H0 which states that µ 1 = µ 2 = µ 3 at 5%
significance level.
We accept Ha : There is a significant difference between the three workers means.
At least one is different from other means.