Analysis of variance

1. ANALYSIS OF VARIANCE
Analysis of variance (Abbreviated by ANOVA) is a procedure to test
more than two means .This technique is introduced by Sir R. A. Fisher
in 1923. This technique compares two different estimates of variance by
using F-distribution to determine whether the population means are
equal.
ASSUMPTION FOR ANOVA
1- The samples are drawn randomly, and each sample is independent
of the other samples.
2- The population from which the sample values are obtained all have
the same unknown population variance.
3- The population under consideration are normally distributed.

2. For example-1
Consider the three machines in operation are performing with equal
efficiency. Random samples have been drawn from the machines, and
the deviations of samples from specifications have been recorded in
millimeters as shown below
Machine-A Machine-B Machine-C
55 60 45
55 55 70
58 50 60
50 60 55
60 57 50
∑ 𝑥𝐴 = 278 ∑ 𝑥𝐵 = 282 ∑ 𝑥𝐶 = 280
Hypotheses
𝐻0 = 𝜇𝐴 = 𝜇𝐵 = 𝜇𝐶 (Likely)
𝐻𝐴 = 𝐴𝑡𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑠 𝑓𝑟𝑜𝑚 𝑜𝑡ℎ𝑒𝑟𝑠.
𝑥̅𝐴 =
∑ 𝑥𝐴
𝑛
=
278
5
= 55.6
𝑥̅𝐵 =
∑ 𝑥𝐵
𝑛
=
282
5
= 56.4
𝑥̅𝐶 =
∑ 𝑥𝐶
𝑛
=
280
5
= 56
Conclusion: 𝐻0 is accepted because the means are close to each
other.

3. For example-2
Consider the three machines in operation are performing with equal
efficiency. Random samples have been drawn from the machines, and
the deviations of samples from specifications have been recorded in
millimeters as shown below.
Machine-A Machine-B Machine-C
38 58 45
40 52 70
37 50 60
39 63 55
35 60 50
∑ 𝑥𝐴 = 189 ∑ 𝑥𝐵 = 283 ∑ 𝑥𝐶 = 280
Solution
Hypotheses
𝐻0 = 𝜇𝐴 = 𝜇𝐵 = 𝜇𝐶 (UnLikely)
𝐻𝐴 = 𝐴𝑡𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑠 𝑓𝑟𝑜𝑚 𝑜𝑡ℎ𝑒𝑟𝑠.
𝑥̅𝐴 =
∑ 𝑥𝐴
𝑛
=
189
5
= 37.8
𝑥̅𝐵 =
∑ 𝑥𝐵
𝑛
=
283
5
= 56.6
𝑥̅𝐶 =
∑ 𝑥𝐶
𝑛
=
280
5
= 56
Conclusion: 𝐻0 is rejected because 𝑥̅𝐴 is far away from others
means. Or the means are not close to each other.

4. Now we solve the same examples by the following two
computing techniques for testing hypotheses.
Computing technique - I
Example-1
Consider the three machines in operation are performing with equal
efficiency. Random samples have been drawn from the machines, and
the deviations of samples from specifications have been recorded in
millimeters as shown below.
Machine-A Machine-B Machine-C
55 60 45
55 55 70
58 50 60
50 60 55
60 57 50
Solution
1- Hypotheses
𝐻0 = 𝜇𝐴 = 𝜇𝐵 = 𝜇𝐶
𝐻𝐴 = 𝐴𝑡𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑠 𝑓𝑟𝑜𝑚 𝑜𝑡ℎ𝑒𝑟𝑠.
2- Level of significance α = 0.05
3- Computing estimates of 𝜎2
: 𝜎
̂𝑏𝑒𝑡𝑤𝑒𝑒𝑛
2
and 𝜎
̂𝑤𝑖𝑡ℎ𝑖𝑛
2

5. Machine-A Machine-B Machine-C
55 60 45
55 55 70
58 50 60
50 60 55
60 62 50
∑ 𝑥𝐴 = 278 ∑ 𝑥𝐵 = 282 ∑ 𝑥𝐶 = 280
𝑥̅𝐴 =
∑ 𝑥𝐴
𝑛
=
278
5
= 55.6
𝑥̅𝐵 =
∑ 𝑥𝐵
𝑛
=
282
5
= 56.4
𝑥̅𝐶 =
∑ 𝑥𝐶
𝑛
=
280
5
= 56
For variance within groups, find sum of squares within groups
Sum of square within groups = 57.2 + 69.2 + 370 = 496.40
M.A (𝑥 − 𝑥̅𝐴)2 M.B (𝑥 − 𝑥̅𝐵)2 M.C (𝑥 − 𝑥̅𝐶)2
55 (55 − 55.6)2
= 0.36
60 (60 − 56.4)2
= 12.96
45 (45 − 56)2
= 121
55 0.36 55 1.96 70 196
58 5.76 50 40.96 60 16
50 31.36 60 12.96 55 01
60 19.36 57 0.36 50 36
∑(𝑥 − 𝑥̅𝐴)2
= 57.2
∑(𝑥 − 𝑥̅𝐵)2
= 69.2
∑(𝑥 − 𝑥̅𝐶)2
= 370

6. For variance between groups, find sum of square between groups.
First to find 𝑥̅̅ =
∑ 𝑥𝐴 +∑𝑥𝐵 +∑ 𝑥𝐶
𝑛𝐴+𝑛𝐵+𝑛𝐶
=
278 + 282 + 280
5+5+5
= 56
Sum of square between groups
= 5{(𝑥̅𝐴 − 𝑥̅̅)2
+ (𝑥̅𝐵 − 𝑥̅̅)2
+ (𝑥̅𝐶 − 𝑥̅̅)2}
= 5{(55.6 − 56)2
+ (56.4 − 56)2
+ (56 − 56)2}
= 5(0.16 + 0.16 + 0)
= 1.6
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑤𝑒𝑛 𝑔𝑟𝑜𝑢𝑝𝑠 =
Sum of square between groups
𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑖𝑠 (𝑘 − 1)
=
1.6
3 − 1
= 0.8
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑤𝑖𝑡ℎ𝑖𝑛 𝑔𝑟𝑜𝑢𝑝𝑠 =
Sum of square within groups
𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑖𝑠 (𝑛 − 𝑘)
=
496.40
15 − 3
= 41.37
𝐹𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 =
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑤𝑒𝑛 𝑔𝑟𝑜𝑢𝑝𝑠
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑤𝑖𝑡ℎ𝑖𝑛 𝑔𝑟𝑜𝑢𝑝𝑠
=
0.8
41.37
= 0.019
𝐹𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 = 𝐹∝,(𝑘−1,𝑛−𝑘) = 𝐹0.05,(2,12) = 3.89
Conclusion: 𝐻0 is accepted, because
𝐹𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 < 𝐹𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒

7. Computing technique - II
Example-1
Consider the three machines in operation are performing with equal
efficiency. Random samples have been drawn from the machines, and
the deviations of samples from specifications have been recorded in
millimeters as shown below.
Machine-A Machine-B Machine-C
55 60 45
55 55 70
58 50 60
50 60 55
60 57 50
Solution
1- Hypotheses
𝐻0 = 𝜇𝐴 = 𝜇𝐵 = 𝜇𝐶
𝐻𝐴 = 𝐴𝑡𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑠 𝑓𝑟𝑜𝑚 𝑜𝑡ℎ𝑒𝑟𝑠.
2- Level of significance α = 0.05
3-Computation
Samples Machine-A Machine-
B
Machine-
C
𝑇𝑖. ∑ 𝑦𝑖𝑗
2
1 55(3025) 60(3600) 45(2025) 160 8650
2 55(3025) 55(3025) 70(4900) 180 10950
3 58(3364) 50(2500) 60(3600) 168 9464
4 50(2500) 60(3600) 55(3025) 165 9125
5 60(3600) 57(3249) 50(2500) 167 9349
𝑇.𝑗 278 282 280 G = 840 47538

8. 𝐶. 𝐹. =
𝐺2
𝑛𝑘
=
(840)2
3(5)
= 47040
𝑆𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑔𝑟𝑜𝑢𝑝𝑠
=
(278)2
+ (282)2
+ (280)2
5
− 𝐶. 𝐹
=
(278)2+(282)2+(280)2
5
− 47040
= 47041.6 − 47040 = 1.6
𝑇𝑜𝑡𝑎𝑙 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 = ∑ ∑ 𝑦𝑖𝑗
2
− 𝐶. 𝐹
𝑇𝑜𝑡𝑎𝑙 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 = 47538 − 47040
𝑇𝑜𝑡𝑎𝑙 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 = 498
Source of
variation
d.f. Sum of
squares
Mean
squares
F-ratio
Between
groups
2 1.6 0.8
𝐹 =
0.8
41.37
= 0.019
Within
groups
12 496.4 41.37
Total 14 498
𝐹𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 = 𝐹∝,(𝑘−1,𝑛−𝑘) = 𝐹0.05,(2,12) = 3.89
Conclusion: 𝐻0 is accepted, because
𝐹𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 < 𝐹𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒

9. Computing technique - I
Example-2
Consider the three machines in operation are performing with equal
efficiency. Random samples have been drawn from the machines, and
the deviations of samples from specifications have been recorded in
millimeters as shown below.
Machine-A Machine-B Machine-C
38 58 45
40 52 70
37 50 60
39 63 55
35 60 50
Solution
1- Hypotheses
𝐻0 = 𝜇𝐴 = 𝜇𝐵 = 𝜇𝐶
𝐻𝐴 = 𝐴𝑡𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑠 𝑓𝑟𝑜𝑚 𝑜𝑡ℎ𝑒𝑟𝑠.
2- Level of significance α = 0.05
3- Computing estimates of 𝜎2
: 𝜎
̂𝑏𝑒𝑡𝑤𝑒𝑒𝑛
2
and 𝜎
̂𝑤𝑖𝑡ℎ𝑖𝑛
2
Machine-A Machine-B Machine-C
38 58 45
40 52 70
37 50 60
39 63 55
35 60 50
∑ 𝑥𝐴 = 189 ∑ 𝑥𝐵 = 283 ∑ 𝑥𝐶 = 280

10. 𝑥̅𝐴 =
∑ 𝑥𝐴
𝑛
=
189
5
= 37.8
𝑥̅𝐵 =
∑ 𝑥𝐵
𝑛
=
283
5
= 56.6
𝑥̅𝐶 =
∑ 𝑥𝐶
𝑛
=
280
5
= 56
For variance within groups, find sum of squares within groups
Sum of square within groups = 14.8 + 119.2 + 370 = 504
For variance between groups, find sum of square between groups.
First to find 𝑥̅̅ =
∑ 𝑥𝐴 +∑𝑥𝐵 +∑ 𝑥𝐶
𝑛𝐴+𝑛𝐵+𝑛𝐶
=
189 + 283 + 280
5+5+5
= 50.13
Sum of square between groups
= 5{(𝑥̅𝐴 − 𝑥̅̅)2
+ (𝑥̅𝐵 − 𝑥̅̅)2
+ (𝑥̅𝐶 − 𝑥̅̅)2}
= 5{(37.8 − 50.13)2
+ (56.6 − 50.13)2
+ (56 − 50.13)2}
= 5(152.03 + 41.86 + 34.46)
= 1141.75
M.A (𝑥 − 𝑥̅𝐴)2 M.B (𝑥 − 𝑥̅𝐵)2 M.C (𝑥 − 𝑥̅𝐶)2
38 (38 − 37.8)2
= 0.04
58 (58 − 56.6)2
= 1.96
45 (45 − 56)2
= 121
40 4.84 52 21.16 70 196
37 0.64 50 43.96 60 16
39 1.44 63 40.96 55 01
35 7.84 60 11.56 50 36
∑(𝑥 − 𝑥̅𝐴)2
= 14.8
∑(𝑥 − 𝑥̅𝐵)2
= 119.2
∑(𝑥 − 𝑥̅𝐶)2
= 370

11. 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑤𝑒𝑛 𝑔𝑟𝑜𝑢𝑝𝑠 =
Sum of square between groups
𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑖𝑠 (𝑘 − 1)
=
1141.75
3 − 1
= 570.875
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑤𝑖𝑡ℎ𝑖𝑛 𝑔𝑟𝑜𝑢𝑝𝑠 =
Sum of square within groups
𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑖𝑠 (𝑛 − 𝑘)
=
504
15 − 3
= 42
𝐹𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 =
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑤𝑒𝑛 𝑔𝑟𝑜𝑢𝑝𝑠
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑤𝑖𝑡ℎ𝑖𝑛 𝑔𝑟𝑜𝑢𝑝𝑠
=
570.875
42
= 13.59
𝐹𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 = 𝐹∝,(𝑘−1,𝑛−𝑘) = 𝐹0.05,(2,12) = 3.89
Conclusion: 𝐻0 is rejected, because
𝐹𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 > 𝐹𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒.

12. Computing technique - II
Example-2
Consider the three machines in operation are performing with equal
efficiency. Random samples have been drawn from the machines, and
the deviations of samples from specifications have been recorded in
millimeters as shown below.
Machine-A Machine-B Machine-C
38 58 45
40 52 70
37 50 60
39 63 55
35 60 50
Solution
1- Hypotheses
𝐻0 = 𝜇𝐴 = 𝜇𝐵 = 𝜇𝐶
𝐻𝐴 = 𝐴𝑡𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑠 𝑓𝑟𝑜𝑚 𝑜𝑡ℎ𝑒𝑟𝑠.
2- Level of significance α = 0.05
3-Computation
Samples Machine-A Machine-
B
Machine-
C
𝑇𝑖. ∑ 𝑦𝑖𝑗
2
1 38(1444) 58(3364) 45(2025) 141 6833
2 40(1600) 52(2704) 70(4900) 162 9204
3 37(1369) 50(2500) 60(3600) 147 7469
4 39(1521) 63(3969) 55(3025) 157 8515
5 35(1225) 60(3600) 50(2500) 145 7325
𝑇.𝑗 189 283 280 G = 752 39346

13. 𝐶. 𝐹. =
𝐺2
𝑛𝑘
=
(752)2
3(5)
= 37700.27
𝑆𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑔𝑟𝑜𝑢𝑝𝑠
=
(189)2
+ (283)2
+ (280)2
5
− 𝐶. 𝐹
𝑆𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑔𝑟𝑜𝑢𝑝𝑠
=
(189)2+(283)2+(280)2
5
− 37700.27
= 38842 − 37700.27 = 1141.73
𝑇𝑜𝑡𝑎𝑙 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 = ∑ ∑ 𝑦𝑖𝑗
2
− 𝐶. 𝐹
𝑇𝑜𝑡𝑎𝑙 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 = 39346 − 37700.27
𝑇𝑜𝑡𝑎𝑙 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 = 1645.73
Source of
variation
d.f. Sum of
squares
Mean
squares
F-ratio
Between
groups
2 1141.73 570.87
𝐹 =
570.87
42
= 13.59
Within
groups
12 504 42
Total 14 1645.73
𝐹𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 = 𝐹∝,(𝑘−1,𝑛−𝑘) = 𝐹0.05,(2,12) = 3.89
Conclusion: 𝐻0 is rejected, because
𝐹𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 > 𝐹𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 .