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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
140427245
249309137
3583310659
4474414035
534368444
μi45.63798.644
μ = (45.6 + 37 + 98.6 + 44) / 4 = 56.3
The square of errors within groups (Square Sum within group):
SSw = ΣiΣj(yij - μi)2 = 3576.4
The square of errors between groups (Square Sum between group):
SSb = Σii - μ)2 = 2427.56
Given the degrees of freedom, the expected average is:
MSw = SSw / a(n-1) = 238.43
MSb = SSb / a-1 = 606.89
Value of Fcrit :
F0 = MSb/MSw = 2.545

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 = 2.545 < 5.86, then we assume that the introduced factor did not have 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 30.39 30.384 30.369 30.366 30.395 30.362 30.328 30.319 30.39 30.325
M2 30.394 30.314 30.365 30.311 30.318 30.366 30.327 30.339 30.301 30.316
M3 30.392 30.369 30.339 30.343 30.355 30.368 30.356 30.394 30.346 30.314
M4 30.304 30.342 30.31 30.344 30.309 30.367 30.341 30.301 30.381 30.34
M5 30.624 30.596 30.461 30.433 30.5 30.335 30.51 30.636 30.328 30.462
Operator 2
M1 30.383 30.45 30.362 30.337 30.385 30.47 30.313 30.52 30.367 30.55
M2 30.707 30.766 30.731 30.608 30.541 30.358 30.429 30.408 30.641 30.432
M3 30.326 30.334 30.337 30.343 30.466 30.5 30.433 30.402 30.443 30.409
M4 31.169 31.1 30.82 30.432 30.969 31.188 31.218 31.019 31.144 30.45
M5 30.464 30.681 30.673 30.489 30.601 30.501 30.56 30.642 30.579 30.318

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
3047.547 1518.7791528.768 607.765 608.972 607.569 612.848 610.393
SSA = 0.998
SSB = 0.951
SSAB = 1.204
SSE = 1.314

MSA = 0.998
MSB = 0.238
MSAB = 0.301
MSE = 0.329

FA = 3.033
FB = 0.723
FAB = 0.915

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 No 3.033 > 2.77
The impact of the employee's qualifications on the result Yes 0.723 < 2.01
The mutual influence of the employee's qualifications and the choice of the machine on the result Yes 0.915 < 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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