Statistics

1. ONE– WAY AND
TWO – WAY
CLASSIFICATION
ANALYSIS OF
VARIANCE (ANOVA)
Adam Julian L. Che, Ph.D.

2. WHAT IS ANOVA?
• AnoVa stands for analysis of
variance.
• Statistical Analysis of at least two
population means can be performed
by conducting an ANOVA.
• This method partitions the total
variance of the variables of interest
into several components of sources.

3. A single factor or one-way ANOVA
is used to test the null hypothesis
that the means of several
populations are all equal.
ANOVA: Single Factor

4. PROCEDURES FOR ONE-WAY ANOVA

5. PROCEDURES:
1. Ho: μ1= μ2=…= μp , That the p population
means are equal
Ha: μi ≠ μk , At least two means are not equal.
2. Test-Statistic: Use F-Test at α level of
significance.

6. PROCEDURES:
3. Decision Criterion:
Reject Ho if
4. Computations:
Notations:
yij = the ith observation in the jth group
= total of the observations in the ith group
y.. = the grand total
= the grand mean
)]
),(
1
[( p
n
p
c F
F −
−
 
.
i
y
..
y

7. Source of
Variation
(SV)
Degrees of
Freedom
(DF)
Sum of
Squares
(SS)
Mean
Square
(MS)
Computed
F or F-
Ratio
Between Groups
Within Groups
p – 1
n – p
BSS
WSS
MSB
MSW
Total n - 1 TSS
The ANOVA Table
c
F
ns= not significant * = significant at α level

8. PROCEDURES
5. State your decision based on the
decision criterion and computed F.
6. State your conclusion based on
your decision.

9. ILLUSTRATIVE EXAMPLE
FOR THE ONE-WAY ANOVA

10. PROBLEM
Three sections of the same
Mathematics subjects for
Grade 4 pupils are taught
by three different teachers.
The first quarter grades of
the pupils were presented
in the following table:

11. Students
1st Quarter Grades
Adonis Bong Carlito
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
73
89
82
43
80
73
66
60
45
93
36
77
88
78
48
91
51
85
74
77
31
73
62
76
96
80
56
68
79
56
91
71
71
87
41
59
68
53
79
15
Is there a
significant
difference in the
average grades
given by the three
teachers? Use
0.01 level of
significance.

12. SOLUTION:
1. Ho: μA= μB= μC , That the average grades given by
the three teachers do not differ significantly
Ha: μi ≠ μk , At least two of the average grades given
by the three teachers differ significantly.

13. SOLUTION:
2. Test-Statistic: Use F-Test at 1% level of significance.
3. Decision Criterion:
a. Reject Ho if:
18
.
5
)]
3
40
),(
1
3
[(
01
.
0 =
 −
−
F
Fc

14. SOLUTION:
4. Computation:
Students
Final Grades
TOTAL
Adonis Bong Carlito
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
73
89
82
43
80
73
66
60
45
93
36
77
88
78
48
91
51
85
74
77
31
73
62
76
96
80
56
68
79
56
91
71
71
87
41
59
68
53
79
15
n 12 15 13 40
817 1066 838 2721
59407 80306 58994 198707
68.08 71.07 64.46 68.03
Y
2
Y
Y

15. SOLUTION:
Computation Formulas:
a.
b.
03
.
185096
40
)
2721
(
.. 2
2
=
=
=
n
y
CF

= =
−
=
p
i
n
j
ij
i
CF
y
TSS
1 1
2
97
.
13610
03
.
185096
198707
=
−
=
03
.
185096
)
15
79
...
89
73
( 2
2
2
2
−
+
+
+
+
=

16. SOLUTION:
CF
n
y
BSS
p
i i
i
−
= 
=1
2
.
89
.
303
97
.
13610 −
=
−
= BSS
TSS
WSS
03
.
185096
13
838
15
1066
12
817 2
2
2
−








+
+
=
89
.
303
03
.
185096
92
.
185399
=
−
=
08
.
13307
=
• c.
• d.

17. SOLUTION:
Computation Formulas:
e.
f.
95
.
151
)
1
3
(
89
.
303
)
1
(
=
−
=
−
=
p
BSS
MSB
65
.
359
)
3
40
(
08
.
13307
)
(
=
−
=
−
=
p
n
WSS
MSW

18. SOLUTION:
Computation Formulas:
g. 4225
.
0
65
.
359
95
.
151
=
=
=
MSW
MSB
Fc

19. Source of
Variation
(SV)
Degrees of
Freedom
(DF)
Sum of
Squares
(SS)
Mean
Square
(MS)
Computed
F or F-
Ratio
Between Groups
Within Groups
p – 1
n – p
BSS
WSS
MSB
MSW
Total n - 1 TSS
The ANOVA Table
c
F
ns= not significant * = significant at α level

20. Source of
Variation
(SV)
Degrees of
Freedom
(DF)
Sum of
Squares
(SS)
Mean
Square
(MS)
Computed
F or F-
Ratio
Between Groups
Within Groups
2
37
303.89
13307.08
151.95
359.65
Total 39 13610.97
The ANOVA Table
ns
42
.
0
ns= not significant * = significant at α level

21. SOLUTION:
5. Decision:
a. Since , we failed to reject Ho .
6. Conclusion:
Therefore, the average grades given by the three teachers
do not differ significantly.
18
.
5
4225
.
0 ]
37
,
2
[
01
.
0 =

= F
Fc

30. Roger is a pilot and he does extensive bad weather
flights. He decides to buy a battery-powered radio as
an independent back-up for his regular radios which
depend on the airplane’s electrical system. He has a
choice of three brands of rechargeable batteries that
vary in cost. He randomly selects four batteries for
each brand and tests them for operating time (in hours)
before recharging is necessary. He obtains the sample
data in the following table.
Problem
Set # 10.A.
(MS EXCEL)
“One-Way
Classification ANOVA”
BRAND OPERATING TIME (In Hours)
Evereday 20.7 21.9 20.9 22.2
Energeezer 21.0 25.6 24.7 24.5
Imarfleex 26.5 26.7 25.0 24.6
Do the three brands have
the same mean usable
time before recharging is
required? Use the level
of significance 𝜶 = 0.01.

31. Roger is a pilot and he does extensive bad weather
flights. He decides to buy a battery-powered radio as
an independent back-up for his regular radios which
depend on the airplane’s electrical system. He has a
choice of three brands of rechargeable batteries that
vary in cost. He randomly selects four batteries for
each brand and tests them for operating time (in hours)
before recharging is necessary. He obtains the sample
data in the following table.
Problem
Set # 10.A.
(Jamovi)
“One-Way
Classification ANOVA”
BRAND OPERATING TIME (In Hours)
Evereday 20.7 21.9 20.9 22.2
Energeezer 21.0 25.6 24.7 24.5
Imarfleex 26.5 26.7 25.0 24.6
Do the three brands have
the same mean usable
time before recharging is
required? Use the level
of significance 𝜶 = 0.01.

32. This tool is used when the variance
depends on two factors and if we
are collecting only a single data
point for a specified condition.
ANOVA: Two Factor without Replication

33. HYPOTHESIS TESTING
PROCEDURES FOR 2-WAY ANOVA
WITHOUT REPLICATION

34. PROCEDURES:
1. Ho’: α1=α2=…=αr , That the row means are equal
Ha’: At least two of the αi’s differ significantly
Ho’’: β1=β2=…=βc , That the column means are equal
Ha’’: At least two of the βj’s differ significantly
2. Test-Statistic: Use F-Test at α level of
significance.

35. PROCEDURES:
3. Decision Criterion:
a. Reject Ho’ if
b. Reject Ho’’ if
4. Computation
Fc(row )
³Fa[(r-1),(r-1)(c-1)]
Fc(column)
³Fa[(c-1),(r-1)(c-1)]

36. Source of
Variation
Degrees of
Freedom
Sum of
Squares
Mean
Square
Computed
F
Row
Column
Error
r – 1
c – 1
(r – 1)(c – 1)
SSRow
SSColumn
SSError
MSRow
MSColumn
MSError
Fc(row)
Fc(column)
Total rc - 1 SSTotal
The ANOVA Table
ns= not significant * = significant at 5% level

37. PROCEDURES
5. State your decision based on the decision
criterion and computed F.
6. State your conclusion based on your
decision.

38. ILLUSTRATIVE EXAMPLE
FOR THE TWO-WAY ANOVA

39. PROBLEM
A study is made to determine
the force required to pull apart
pieces of glued plastic. Three
types of plastics were tested
using four different levels of
humidity. The results, in
kilogram, are given as follows:

40. Plastic
Type
Humidity
30% 50% 70% 90%
A
B
C
38.2
30.9
27.3
32.5
27.8
30.4
34.2
28.4
27.3
33.0
30.5
31.6
Use a 0.05 level of significance to test the hypothesis
that there is no significant difference in the mean force
required to pull the glued plastic apart:
a. When different types of plastic are used;
b. For different humidity conditions.

41. SOLUTION:
1. Ho’: αA=αB=αC, that there is no significant difference in
the mean force required to pull the glued plastic apart
when different types of plastic are used.
Ha’: αi ≠ αk ,At least two of the plastic types differ
significantly in terms of the mean force required to pull
the glued plastic apart.

42. SOLUTION:
1. Ho’’: β30%=β50%=β70%=β90%, that there is no significant
difference in the mean force required to pull the
glued plastic apart based on humidity conditions.
Ha’’: βi≠βk ,At least two of the humidity conditions differ
significantly in terms of the mean force required to
pull the glued plastic apart.

43. SOLUTION:
2. Test –Statistic: Use F-Test at α=0.05 level of
significance
3. Decision Criterion:
a. Reject Ho’ if:
b. Reject Ho” if:
Fc(plastic)
³F0.05[(3-1),(3-1)(4-1)]
= 5.14
Fc(humidity )
³ F0.05[(4-1),(3-1)(4-1)]
= 4.76

45. 4. COMPUTATION:
Plastic
Type
Humidity
Total Mean
30% 50% 70% 90%
A
B
C
38.2
30.9
27.3
32.5
27.8
30.4
34.2
28.4
27.3
33.0
30.5
31.6
137.9
117.6
116.6
34.48
29.40
29.15
Total 96.4 90.7 89.9 95.1 372.1
Mean 32.13 30.23 29.97 31.7 31.01

46. SOLUTION:
a.
CF =
x..2
rc
=
372.1
( )
2
3
( ) 4
( )
=11538.20083

47. SOLUTION:
Plastic
Type
Humidity
Total Mean
30% 50% 70% 90%
A
B
C
38.2
30.9
27.3
32.5
27.8
30.4
34.2
28.4
27.3
33.0
30.5
31.6
137.9
117.6
116.6
34.48
29.40
29.15
Total 96.4 90.7 89.9 95.1 372.1
Mean 31.01
r = 3
c = 4

48. SOLUTION:
b.
SSTotal = xij
2
j=1
c
å
i=1
r
å -CF
= 38.22
+32.52
+...+31.62
é
ë
ù
û-11538.20083
=11651.89-11538.20083
SSTotal =113.6892

49. SOLUTION:
Plastic
Type
Humidity
Total Mean
30% 50% 70% 90%
A
B
C
38.2
30.9
27.3
32.5
27.8
30.4
34.2
28.4
27.3
33.0
30.5
31.6
137.9
117.6
116.6
34.48
29.40
29.15
Total 96.4 90.7 89.9 95.1 372.1
Mean 31.01

50. SOLUTION:
c.
SSPlastic =
xi.
2
i=1
r
å
c
-CF =
137.92
+117.62
+116.62
4
é
ë
ê
ù
û
ú-11538.20083
=
46441.73
4
-11538.20083
SSPlastic = 72.23167

51. SOLUTION:
Plastic
Type
Humidity
Total Mean
30% 50% 70% 90%
A
B
C
38.2
30.9
27.3
32.5
27.8
30.4
34.2
28.4
27.3
33.0
30.5
31.6
137.9
117.6
116.6
34.48
29.40
29.15
Total 96.4 90.7 89.9 95.1 372.1
Mean 31.01
c = 4

52. SOLUTION:
d.
SSHumidity =
x. j
2
j=1
c
å
r
-CF=
96.42
+90.72
+89.92
+95.12
3
é
ë
ê
ù
û
ú-11538.20083
=
34645.47
3
é
ë
ê
ù
û
ú -11538.20083
SSHumidity =10.28917

53. SOLUTION:
Plastic
Type
Humidity
Total Mean
30% 50% 70% 90%
A
B
C
38.2
30.9
27.3
32.5
27.8
30.4
34.2
28.4
27.3
33.0
30.5
31.6
137.9
117.6
116.6
34.48
29.40
29.15
Total 96.4 90.7 89.9 95.1 372.1
Mean 31.01
r = 3

54. SOLUTION:
e. SSError =SSTotal -SSPlastic-SSHumidity
=113.6892-72.2317-10.2892
SSError = 31.1683

55. SOLUTION:
f.
MSPlastic =
SSPlastic
r -1
=
72.2317
3-1
MSPlastic = 36.1159

56. SOLUTION:
g.
MSHumidity =
SSHumidity
c -1
=
10.2892
4-1
MSHumidity = 3.4297

57. SOLUTION:
h.
MSError =
SSError
r -1
( ) c -1
( ) =
31.1683
3-1
( ) 4-1
( )
MSError = 5.1947

58. SOLUTION:
i.
Fc(Plastic)
=
MSPlastic
MSError
=
36.1159
5.1947
Fc(Plastic)
= 6.9525

59. SOLUTION:
j.
Fc(Humidity )
=
MSHumidity
MSError
=
3.4297
5.1947
Fc(Humidity )
= 0.6602

60. Source of
Variation
Degrees of
Freedom
Sum of
Squares
Mean
Square
Computed
F
Plastic Type
Humidity
Error
2
3
6
72.2317
10.2892
31.1683
36.1159
3.4297
5.1947
6.95*
0.66ns
Total 11 113.6892
The ANOVA Table
ns= not significant * = significant at 5% level

61. SOLUTION:
5. Decision
a. Since , reject Ho’.
b. Since , we fail to
reject Ho’’
Fc(plastic)
= 6.95³F0.05[(3-1),(3-1)(4-1)]
= 5.14
Fc(humidity )
= 0.66<F0.05[(4-1),(3-1)(4-1)]
= 4.76

62. SOLUTION:
6. Conclusion:
a. At least two of the plastic types differ significantly at 5% level in
terms of the mean force required to pull glued plastic apart.
b. The mean force required to pull the glued plastic apart do not
differ significantly at 5% level of significance among humidity
conditions.

63. ANOVA IN EXCEL: IT’S YOUR TURN AGAIN!
To start with the ANOVA function, open the workbook
containing the data you want to run the test on. Then, follow these
steps:
1. Click in a cell on your spreadsheet where your output will begin. The
results, of course, will cover a range of cells.
2. Click on the Data tab from the main ribbon and select data analysis,
which should be in the analysis menu on the right.
3. Select the appropriate ANOVA test from the options in the Data
Analysis menu.

64. The following table presents the final grades obtained
by five students in Mathematics, English, ESP, and
Science:
Problem
Set # 11.A.
(MS EXCEL)
– “Two-Way
Classification
ANOVA w/o
Replication”
SUBJECTS
STUDENTS Math English ESP Science
1 80 78 85 79
2 86 90 93 91
3 75 78 83 79
4 78 80 84 76
5 83 87 85 87
Use 0.05 level of significance to test the ff. hypotheses:
a. Students have equal ability;
b. Subjects are of equal difficulty.

65. The following table presents the final grades obtained
by five students in Mathematics, English, ESP, and
Science:
Problem
Set # 11.B.
(JAMOVI)
– “Two-Way
Classification
ANOVA w/o
Replication”
SUBJECTS
STUDENTS Math English ESP Science
1 80 78 85 79
2 86 90 93 91
3 75 78 83 79
4 78 80 84 76
5 83 87 85 87
Use 0.05 level of significance to test the ff. hypotheses:
a. Students have equal ability;
b. Subjects are of equal difficulty.

66. This tool is used when we have two
factors on which the variance
depends and if we are collecting
multiple data points for a specified
condition.
ANOVA: Two Factor with Replication

67. HYPOTHESIS TESTING
PROCEDURES FOR 2-WAY ANOVA
WITH REPLICATION

69. PROCEDURES:
1. Ho’’’: There is no interaction between the two factors
Ha’’’: There is interaction between the two factors
Ho’: α1=α2=…=αr , That the row means are equal
Ha’: At least two of the αi’s differ significantly
Ho’’: β1=β2=…=βc , That the column means are equal
Ha’’: At least two of the βj’s differ significantly (p.689)
2. Test-Statistic: Use F-Test at α level of
significance.

70. PROCEDURES:
3. Decision Criterion:
a. Reject Ho’’’ if
b. Reject Ho’ if
c. Reject Ho’’ if
4. Computation using Excel:
Fc(row )
³ F
a [(r-1),(r-1)(c-1)]
Fc(column)
³ Fa [(c-1),(r-1)(c-1)]
𝐹𝑐(𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛) ≥ 𝐹𝛼[ 𝑟−1 ,(𝑟−1)(𝑐−1]

71. PROCEDURES
5. State your decision based on the decision
criterion and computed F.
6. State your conclusion based on your
decision.

72. SAMPLE PROBLEM:

73. SOLUTION:

74. Marathon Times. Listed below are New York Marathon
running times (in seconds) for randomly selected
runners who completed the marathon. At 0.05 level of
significance, are the running times affected by an
interaction between gender and age bracket? Are
running times affected by gender? Are running times
affected by age bracket?
Problem
Set #12.A.
(MS
EXCEL)–
“Two-Way
Classification
ANOVA with
Replication”

75. Marathon Times. Listed below are New York Marathon
running times (in seconds) for randomly selected
runners who completed the marathon. At 0.05 level of
significance, are the running times affected by an
interaction between gender and age bracket? Are
running times affected by gender? Are running times
affected by age bracket?
Problem
Set #12.B.
(JAMOVI)–
“Two-Way
Classification
ANOVA with
Replication”

76. REFERENCES
• Statistics Made Easy by Dr. Consuelo A. Tagaro
and Engr. Alipio T. Tagaro
• Statistics for Managers by Levine, et. al
• Slides by Dr. Joel E. Genzon

77. Thank you for
listening!
☺ ☺ ☺