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
β π₯π΄ = 278 β π₯π΅ = 282 β π₯πΆ = 280
Hypotheses
π»0 = ππ΄ = ππ΅ = ππΆ (Likely)
π»π΄ = π΄π‘ππππ π‘ πππ ππππ πππππππ ππππ ππ‘βπππ .
π₯Μ π΄ =
β π₯π΄
π
=
278
5
= 55.6
π₯Μ π΅ =
β π₯π΅
π
=
282
5
= 56.4
π₯Μ πΆ =
β π₯πΆ
π
=
280
5
= 56
Conclusion: π»0 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
β π₯π΄ = 189 β π₯π΅ = 283 β π₯πΆ = 280
Solution
Hypotheses
π»0 = ππ΄ = ππ΅ = ππΆ (UnLikely)
π»π΄ = π΄π‘ππππ π‘ πππ ππππ πππππππ ππππ ππ‘βπππ .
π₯Μ π΄ =
β π₯π΄
π
=
189
5
= 37.8
π₯Μ π΅ =
β π₯π΅
π
=
283
5
= 56.6
π₯Μ πΆ =
β π₯πΆ
π
=
280
5
= 56
Conclusion: π»0 is rejected because π₯Μ π΄ 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
π»0 = ππ΄ = ππ΅ = ππΆ
π»π΄ = π΄π‘ππππ π‘ πππ ππππ πππππππ ππππ ππ‘βπππ .
2- Level of significance Ξ± = 0.05
3- Computing estimates of π2
: π
Μπππ‘π€πππ
2
and π
Μπ€ππ‘βππ
2