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Analysis of variance

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ANOVA

ANOVA in statistics is a powerful tool for determining the influence of different groups of observations among themselves. The analysis of variance was introduced by Fisher, an English scientist who made a huge contribution to the development of science. ANOVA is an acronym for ANalysis Of VAriance.

Example

Suppose you want to conduct an empirical study of gasoline quality, for this you fill up the tank at one gas station and drive n kilometers, repeat such an experiment, say, five times, then conduct the same experiment, only at a different gas station. You have two sets of data - refueling A and refueling B. Certainly, the figures are scattered, but there is still some dependence, so that would determine whether refueling affects gasoline consumption (or the data are not related) You are using variance analysis.

The analysis of variance allows you to determine which of the factors affects more, intra-group or intergroup. In the example above, you will be able to determine how much the choice of gas station affects gasoline consumption. This is the essence of the dispersion analysis: to find out whether the selected factor is significant for the selected observations.

In a sense, the analysis of variance is similar to regression and correlation analyses, because it allows determine the influence of variables on each other.

Analysis

In theory, a simple model is built to analyze the variance, similar to the one studied in time series analysis.

Model

The model of the analysis of variance includes the average value, the effect of the experiment and a random error:

y = μ + τ + ε
τ - experiment effect, ε - random error

Single-factor

One-factor analysis of variance considers the influence of one criterion, it is done this way: we conduct two experiments, in one of them we include an additional factor and analyze whether this factor has made changes. As initial data, consider the results of a number of experiments:

NE1E2E3E4
1525111847
2433311248
360489246
442308433
5443910630
μi48.240.2102.440.8
μ = (48.2 + 40.2 + 102.4 + 40.8) / 4 = 57.9
The square of errors within groups (Square Sum within group):
SSw = ΣiΣj(yij - μi)2 = 1661.6
The square of errors between groups (Square Sum between group):
SSb = Σii - μ)2 = 2680.04
Given the degrees of freedom, the expected average is:
MSw = SSw / a(n-1) = 110.77
MSb = SSb / a-1 = 670.01
Value of Fcrit :
F0 = MSb/MSw = 6.049

Fischer's test: if the value of F0 turns out to be greater than the value of F λ,4,15, then the factor has an impact.

For n = 20 and a = 5, Fλ,n-a,a-1 = Fλ,15,4= 5.86
Since F0 = 6.049 > 5.86, then we assume that the introduced factor had an effecton the results of the experiment.

Two-factor

In two- factor analysis , three hypotheses are put forward for verification:

  • Factors A and B do not affect the result
  • Factor A does not affect the result
  • Factor B does not affect the result

To carry out a two-factor analysis, it is necessary to make groups of results: several measurements for all values of each of the factors, i.e.:

A1A2
B1X1a1,b1...XNa1,b1X1a1,b2...XNa1,b2
B2X1a1,b2...XNa1,b2X1a1,b2...XNa1,b2

Next, the average value for each factor value is calculated, i.e. the average for A1, the average for B1, etc. Then it is calculated the total average for all results. Let's set the number of criteria: k = 2 (the number of criteria A) and m = 2 (the number of criteria B).

T = ΣΣΣxijk
The sum of elements under the influence of factor A:
TAi = Σxi·k
The sum of elements under the influence of factor B:
TBj = Σx·jk
The sum of elements under the influence of factor AB:
TAiBj = Σxij·
SST = Σx2ijk - T2/N
SSA = ΣT2Ai/n·m - T2/N
SSB = ΣT2Bj/n·k - T2/N
SSAB = ΣΣT2AiBj/n - SSA - SSB - T2/N
SSE = ΣΣΣx2ijk - ΣΣT2AiBj/n

SST = SSA + SSB + SSAB + SSE

MSE = SSE/(n-1)·m·k
MSA = SSA/k-1
MSB = SSB/m-1
MSAB = SSAB/(m-1)·(k-1)
Test "Criterion A does notaffect the result", ν1= k-1:
FA = MSA/MSE
Test "Criterion B does notaffect the result", ν1= m-1:
FB = MSB/MSE
Test "Criteria A and B do notaffect the result", ν1 = (k-1)(m-1):
Fint = MSAB/MSE

For each F, if F > F α,ν12, then the hypothesis is rejected. ν2 = N-mk

Multifactorial

Multivariate analysis is similar to two-factor analysis - the same operations are performed, but the criteria are grouped and the influence of each of the factors is found iteratively.

With repeated measurements

The analysis of variance with repeated measurements indicates that several tests were performed for each criterion measurements of a random variable to obtain a more accurate result (since ANOVA) uses the intra-group sum of squares.

Application

Dispersion analysis is used in a wide variety of branches of science and production when it is necessary to study the dependence of the criteria on the difference in average values, while comparing not the average value, but the spread the results are around the mean, i.e. the variance.

Solving problems

As an example, let's give a problem from metrology. The plant houses five machines that produce shafts. It is necessary to determine whether the choice of a machine tool or the training of an employee affects the result of production. For analysis measurements are made for each machine and employee, the result is a table:

Operator 1
M1 30.51 30.521 30.604 30.507 30.366 30.634 30.386 30.677 30.515 30.462
M2 30.178 30.102 29.943 29.924 30.127 29.843 29.979 30.077 29.913 30.261
M3 30.324 30.303 30.318 30.39 30.327 30.38 30.357 30.354 30.361 30.321
M4 30.31 30.341 30.372 30.362 30.361 30.389 30.355 30.34 30.311 30.388
M5 30.369 30.324 30.386 30.397 30.332 30.306 30.393 30.317 30.341 30.315
Operator 2
M1 30.164 30.266 30.161 30.101 30.155 30.206 30.207 30.136 30.106 30.22
M2 30.452 30.378 30.582 30.5 30.544 30.557 30.41 30.547 30.313 30.551
M3 30.4 30.3 30.395 30.327 30.309 30.32 30.353 30.315 30.3 30.387
M4 30.618 30.615 30.313 30.492 30.518 30.651 30.559 30.501 30.415 30.516
M5 30.267 30.228 30.162 30.145 30.116 30.203 30.291 30.156 30.138 30.118

Let's use the method of two-factor analysis, factor A is the operator, factor B is the machine. Calculate the sums of squares, to do this, you need to calculate the average value for each of the groups:

TTA1TA2 TB1TB2TB3TB4TB5
3032.957 1515.9731516.984 606.904 605.181 606.841 608.727 605.304
SSA = 0.01
SSB = 0.418
SSAB = 1.872
SSE = 0.518

MSA = 0.01
MSB = 0.105
MSAB = 0.468
MSE = 0.13

FA = 0.077
FB = 0.808
FAB = 3.6

Critical values for the Fischer test:
Fcrit A = F0.1, 1, 90 = 2.77
Fcrit B = F0.1, 4, 90 = 2.01
Fcrit AB = F0.1, 4, 90 = 2.01

Results table:

The impact of the machine on the result Yes 0.077 < 2.77
The impact of the employee's qualifications on the result Yes 0.808 < 2.01
The mutual influence of the employee's qualifications and the choice of the machine on the result No 3.6 > 2.01

In excel/Open Calc

To solve the variance analysis in a spreadsheet, you will need the following formulas:

sumproduct Sum of products, used to find the sum of squares
finv Inverse value of the distribution F - Fisher criterion

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