apply the chi-square test for independence in a matrix - Maple Help

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Student[Statistics][ChiSquareIndependenceTest] - apply the chi-square test for independence in a matrix

Calling Sequence

ChiSquareIndependenceTest(X, level_option)

Parameters

X

-

Matrix of categorized data

level_option

-

(optional) equation of the form level=float.

Description

• 

The ChiSquareIndependenceTest function computes the chi-square test for independence in a Matrix. This tests whether two factors in a population are independent of one another. The rows represent the levels of one factor; the columns represent the levels of another. The entries in the Matrix are interpreted as counts of observations with the given combination of levels.

• 

The first parameter X is a Matrix of categorized data samples.

• 

level=float

  

This option is used to specify the level of analysis (minimum criteria for the observed data to be considered well-fit to the expected data). By default, this value is 0.05.

Examples

withStudent[Statistics]:

Specify the matrices of categorized data values.

X:=Matrix32,12,14,22,6,9:

Y:=Matrix2,4,4,9,7,12:

Perform the independence test on the first sample.

ChiSquareIndependenceTestX,level=0.05

Chi-Square Test for Independence
--------------------------------
Null Hypothesis:
Two attributes within a population are independent of one another
Alt. Hypothesis:
Two attributes within a population are not independent of one another

Dimensions:              3
Total Elements:          95
Distribution:            ChiSquare(2)
Computed statistic:      10.7122
Computed pvalue:         0.00471928
Critical value:          5.99146454710798

Result: [Rejected]
This statistical test provides evidence that the null hypothesis is false

hypothesis=false,criticalvalue=5.99146454710798,distribution=ChiSquare2,pvalue=0.00471928013704082,statistic=10.71219801

(1)

Perform the independence test on the second sample.

ChiSquareIndependenceTestY,level=0.05

Chi-Square Test for Independence
--------------------------------
Null Hypothesis:
Two attributes within a population are independent of one another
Alt. Hypothesis:
Two attributes within a population are not independent of one another

Dimensions:              3
Total Elements:          38
Distribution:            ChiSquare(2)
Computed statistic:      0.128915
Computed pvalue:         0.937576
Critical value:          5.99146454710798

Result: [Accepted]
This statistical test does not provide enough evidence to conclude that the null hypothesis is false

hypothesis=true,criticalvalue=5.99146454710798,distribution=ChiSquare2,pvalue=0.937575872647938,statistic=0.1289151874

(2)

See Also

Statistics[ChiSquareIndependenceTest], Student, Student/Statistics/ChiSquareIndependenceTest/overview, Student[Statistics], Student[Statistics][HypothesisTest]

References

  

Kanju, Gopal K. 100 Statistical Tests. London: SAGE Publications Ltd., 1994.

  

Sheskin, David J. Handbook of Parametric and Nonparametric Statistical Procedures. London: CRC Press, 1997.


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