Hetrosadastisity

8/12/2019 Hetrosadastisity

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HetrosadastesityHomosadastesity is also one of the assumption ofLinear Regression Model.

homo means equal and scedasticity means

spread or variance. Homoscedasticity thus refersto as equal or same variances.===> E(ui) = ; remains constant

while i varies


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Consequence Hetrosadastesity

1. Due TO hetrosadastesity, variances of thecoefficients ( i)are larger, and consequently,their standard errors and confidence interval

are large, while t ratios are consequently smalland insignificant.

2) Estimated results are misleading.

3) OLS estimators are no longer efficient.


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Detection Hetrosadastesity

Tests use for detection of Hetrosadastesity:Park Test: Run a usual regression, like:lnY = 0 + 1lnXi + i Obtain residuals e i and make them squared, runregression of the following form:Lne2i = 0 + 1lnXi + i

If 1 happens to be statistically significant, it willindicate the existence of the problems ofheteroscedasticity. Lets do the Park test for evaluatingour Job satisfaction and organizational justice casefor checking existence of heteroscadasticity problem.


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Park Test: (Cont ) Convert data on all dependent and independentvariables JS, DJ,PJ, IJ, INJ and AEE into log using

TRANSFORMand COMPUTE VARIABLEcommandsin SPSS; let the newly log-variables have newnames LJS, LDJ,LPJ, LIJ, LINJ and LAEE.


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Park Test: (Cont ) After Converting data into log regress the model

as follows:

lnLJS = 0 + 1lnDJ + 2lnPJ + 3lnIJ + 4lnIN +5lnAEE + i

Obtain residuals using additional SPSS commands:ANALYZEREGRESSIONLINEARSAVERESIDUALSUNSTANDARDIZEDCONTINUEOK


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Park Test: (Cont ) The command in previous slide will estimate

residuals and put those in the last column of

the data file under name RES_1(this residualwill be in logarithmic form as all the variables inregression are in log from) . Make this variablesquare (as we need Lne 2i), using TRANSFORM

and COMPUTE commands.you can run regressionLne2i = 0 + 1lnDJ + 2lnPJ + 3lnIJ + 4lnIN +

5lnAEE + i


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Park Test: (Cont )

Model

Unstandardized Coefficients

Standardized

Coefficients t Sig.

B Std. Error Beta B Std. Error 1 (Constant) .240 .098 2.450 .015

LDJ -.157 .026 -.455 -6.124 .000

LPJ

-.008 .022 -.027 -.341 .733

LIJ .026 .024 .069 1.075 .283

LINJ -.056 .032 -.129 -1.748 .082

LAEE .021 .024 .046 .848 .397


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Park Test: (Cont )Result interpretation of Park Test:All the coefficients (LDJ, LPJ, LIJ, LINJ & LAEE) are

statistically insignificant except LDJ , suggestingno heterosedasticity problem.


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Goldfeld-Quant Test:The Goldfeld-Quant test suggests ordering or rank

observations according to the values of X i,

beginning with the lowest X i value. Then somecentral observations are omitted in a way thatthe remaining observations are divided into

two equal groups.


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Goldfeld- Quant Test: (Cont) These two data groups are used for running twoseparate regressions, and residual sum of squares

(RSS) are obtained; these RSSs (RSS 1 & RSS2) arethen used to compute Goldfeld-Quant F test,namely:

F = RSS2/dfRSS1/df

If the F is found significant (F -calculated > F-tabulated ) theproblem of heteroscedasticity is likely to exist.


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Goldfeld- Quant Test: (Cont) Lets run the stated test for Organizational justice andJob satisfaction case. In Parks test indicated that logof variable DJ was found the most collinear with thelog of the squared residuals; this suggested that wearrange data in ascending order using DJ variable asthe base, and then omit central 14 observations,which will leave 250 observation to be equally dividedin two parts of 125 observation each.

The SPSS command is: DATASORT CASESTake DJto the SORT -BY BOXASCENDING.


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Goldfeld- Quant Test: (Cont) Remove the 14 central observations, and savedata in two separate files, one having Group 1

data (the first 125 observations) and the secondhaving Group II data (having 125 laterobservations).


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Goldfeld- Quant Test: (Cont) Then running the required two regressions gives

the following TWO ANOVA tables:

ANOVAb

a Predictors: (Constant), AEE, Procedural justice, Interactive justice , INJ, Distributive justice

b Dependent Variable: Job satisfaction

Model Sum of

Squares df Mean

Square F Sig. 1 Regression 17.457 5 3.491 7.913 .000(a)

Residual 52.504 119 .441

Total 69.961 124


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Goldfeld- Quant Test: (Cont) ANOVAb

a Predictors: (Constant), AEE, Procedural justice, Interactive justice , INJ, Distributive justice

b Dependent Variable: Job satisfaction

Model Sum of

Squares df Mean

Square F Sig. 1 Regression 2.502 5 .500 3.112 .011(a)

Residual 19.139 119 .161

Total 21.641 124


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Goldfeld- Quant Test: (Cont) The residual sum of squares (RSS) of the two groupsare:

RSS1 = 52.504 with DF = 119RSSII = 19.139 with DF = 119

Calculating F, using the above valuesF = (RSSII/DF) / (RSS I/DF)

= (19.139/119) / 52.504/119= 0.3565F-calculated = 0.3565 < F -tabulated = 2.29 (at p = 0.05),

suggesting there exists no heteroscadasticity.


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Whites General Heteroscedasticity TestConsider the following three-variable regression

model:

Yi = 1 + 2X2i + 3X3i + u i

Step 1: Run the above regression and obtain

the residuals, u i .Step 2: Make Square of the Residual


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Whites General Heteroscedasticity Test (Cont) Step : 3 Run the following Auxiliary Regressionu2i = 1 + 2 X 2i + 3 X 3i + 4 X 22i + 5 X 23i + 6 X 2i X 3i + v i

Obtain the R2 from this (auxiliary) regression.

Step 4: Multiply R2 by n

Under the null hypothesis there is no hetrpsedasticity,when R2 is multipled by n, it approached tocalculated Chi-Square i.e.

n R2 ~ asy 2df


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Whites General Heteroscedasticity Test (Cont) Step : 5, Compare calculated Chi-square withtabulated value,

If 2df (Calculated) > 2df (Tabulated) indicatinghetrosedasticity problem. i.e.if calculated chi-sqr is greater than tabulated chi-sqr showing that ( 1 2 3 4 5 6 0), sothan u 2i 1 Showing variance does not remain constant, aproblem of hetrosedastisity


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Whites General Heteroscedasticity Test (Cont) Take the Example of Jobsatisfaction

In Spss Command

Analyze Regression Save Residual unstandardized continue Ok

Make this residual square (as we need e 2i), usingTRANSFORM and COMPUTE commands.


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Whites General Heteroscedasticity Test (Cont) Now run the following Auxiliary Regression

e 2i=a1+a2DJ+a3PJ+a4IJ+a5INJ+a6AEE+a7SDJ+a8SPJ+a9SIJ+a10SINJ+a11SAEE+a12DJPJ+a13DJINJ+a14DJAEE+a15IJINJ+a16IJAEE+a17INJAEE +e

using spss Commands .Transform .compute

varaibles , as we need(SDJ,SPJ,SIJ,SINJ,SAEE,DJPJ,DJIJ,DJINJ,DJAEE,PJIJ,PJINJ,PJAEE,IJINJ,IJAEE,INJAEE)


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Whites General Heteroscedasticity Test (Cont) Using SPSS Command

Analyze Regression Linaer Dependent

variable e 2i independent variables SDJ,SPJ,SIJ,SINJ,SAEE,DJPJ,DJIJ,DJINJ,DJAEE,PJIJ,PJI

NJ,PJAEE,IJINJ,IJAEE,INJAEE.OK


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Detection HetrosadastesityWhites General Heteroscedasticity Test (Cont)

Model Summary

a Predictors: (Constant), INJAEE, SIJ, SDJ, SPJ, SINJ, SAEE, IJAEE, DJIJ, PJAEE, DJAEE, PJIJ,DJINJ, IJINJ, DJPJ, PJINJ

df = all independent variables excluding constant(in Auxiliary Regression)

In our case df = 15

Model R R Sqr Adj R Sqr St.Error

1 .522(a) .272 .228 .50813


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Whites General Heteroscedasticity Test (Cont) 215 (Calculated) = R2 * n

We have n =264

So

215 (Calculated) = .272 * 264

= 71.808


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Whites General Heteroscedasticity Test (Cont) Compare 215 (Calculated) = 4.08 with Chi-Square

table at page 968 Gujrati

10% 5% 1% 215 (Tabulated) 28.41 31.41 37.57

215 (Calculated) = 71.808 > 215 (Tabulated) = 37.57Showing Hetrosedastisity problem


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Remedies of Hetrosadastesity

1. If we know , then we use the weighted leastsquares (WLS) estimation technique, i.e.,

Where i = standard deviation of the X i.2. Log -transformation:

It reduces the heteroscedasticity.

i

i

i

i

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i e X Y

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iii X LnY Ln 10


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Remedies of Hetrosadastesity

3. Other Transformation:(a)

After estimating the above model, both the sides arethen multiplied by X i.

(b)

Note: In case of transformed data, the diagnosticstatistics t- ratio and F- statistic

are valid only in large sample size.

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X X X

X X Y

0

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Y Y

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Hetrosadastisity

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