Chi2testMixtures {AdaptGauss} | R Documentation |

## Pearson's chi-squared goodness of fit test

### Description

Chi2testMixtures is goodness of fit test which establishes whether an observed distribution (data) differs from a Gauss Mixture Model (GMM). Returns a P value of a special case of a chi-square test and visualizes data versus a given GMM.

### Usage

```
Chi2testMixtures(Data,Means,SDs,Weights,
IsLogDistribution,PlotIt,UpperLimit,VarName,NoRepetitions)
```

### Arguments

`Data` |
vector of data points (1:n) |

`Means` |
vector of Means of Gaussians (1:c) |

`SDs` |
vector of standard deviations, estimated Gaussian Kernels (1:c) |

`Weights` |
vector of relative number of points in Gaussians (prior probabilities) (1:c) |

`IsLogDistribution` |
Optional, if IsLogDistribution(i)==1, then mixture is lognormal, default vector of zeros of length 1:L |

`PlotIt` |
Optional, Default: FALSE, do a Plot of the compared cdfs and the KS-test distribution (Diff) |

`UpperLimit` |
Optional. test only for Data <= UpperLimit, Default = max(Data) i.e all Data. |

`VarName` |
If PlotIt=TRUE, the name of the inspected variable, default 'Data' |

`NoRepetitions` |
Optional, scalar, default =1000, Number of Repetitions for monte carlo sampling |

### Details

The null hypothesis is that the estimated data distribution does not differ significantly from the GMM. Let O_i be the observed features and E_i be the expected number E, than the test statistic is defined with the minimum chi-square estimate T=sum((O_i-E_i)^2/E_i)*1/m, where m the number of data points. The expected number Ei may be derived for each bin. If there is a significant difference between the O_i and the E_i, the Pvalue is small and the null hypothesis can be rejected.

Further details, see [Thrun & Ultsch, 2015].

### Value

List with

`Pvalue` |
Pvalue of a suiting chi-square , Pvalue ==0 if Pvalue <0.001 |

`BinCenters` |
bin centers |

`ObsNrInBin` |
No. of data in bin |

`ExpectedNrInBin` |
No. of data that should be in bin according to GMM |

`Chi2Value` |
the TestStatistic T i.e.: sum((ObsNrInBin(Ind)-ExpectedNrInBin(Ind))^2/ExpectedNrInBin(Ind)) with
Ind = find(ExpectedNrInBin>=10)
The value of |

### Note

The statistic assumption is that the the test statistic follows a chi square distribution. The number of degrees of freedom is equal to the number of datapoints n-1-3*c

### Author(s)

Rabea Griese, Michael Thrun

### References

Hartung, J., Elpelt, B., and Kloesener, K.H.: Statistik, 8. Aufl. Verlag Oldenburg (1991).

Thrun, M. C., Ultsch, A.: Models of Income Distributions for Knowledge Discovery, European Conference on Data Analysis, DOI 10.13140/RG.2.1.4463.0244, pp. 28-29, Colchester 2015.

*AdaptGauss*version 1.6 Index]