White''s general heteroscedasticity test, Advanced Statistics

The Null Hypothesis - H0:  γ1 = γ2 = ...  =  0  i.e.  there is no heteroscedasticity in the model

The Alternative Hypothesis - H1:  at least one of the γi's are not equal to zero i.e. the squared residuals are related to one of the independent variables.

Reject H0 if nR2   >   2094_Tests for Heteroscedasticity.png

Regression Analysis: sqresi versus totexp, age, ...

* sqnk is highly correlated with other X variables

* sqnk has been removed from the equation.

 

The regression equation is

sqresi = 0.00086 - 0.000117 totexp + 0.000765 age + 0.00007 nk

         + 0.000000 sqtotexp - 0.000009 sqage + 0.000000 totexpage

         + 0.000026 totexpnk - 0.000077 agenk

 

Predictor         Coef     SE Coef      T      P     VIF

Constant      0.000857    0.007288   0.12  0.906

totexp     -0.00011676  0.00004906  -2.38  0.017  49.148

age          0.0007649   0.0003256   2.35  0.019  73.466

nk            0.000072    0.002941   0.02  0.980  24.250

sqtotexp    0.00000019  0.00000010   2.00  0.045  13.958

sqage      -0.00000879  0.00000394  -2.23  0.026  62.515

totexpage   0.00000021  0.00000097   0.21  0.831  37.830

totexpnk    0.00002598  0.00001464   1.77  0.076  18.920

agenk      -0.00007694  0.00007905  -0.97  0.331  32.566

S = 0.0112807   R-Sq = 1.2%   R-Sq(adj) = 0.6%

 

Analysis of Variance

Source            DF         SS         MS     F      P

Regression         8  0.0022313  0.0002789  2.19  0.026

Residual Error  1493  0.1899897  0.0001273

  Lack of Fit    639  0.0804237  0.0001259  0.98  0.601

  Pure Error     854  0.1095659  0.0001283

Total           1501  0.1922209

 

 332 rows with no replicates

Source     DF     Seq SS

totexp      1  0.0006642

age         1  0.0000000

nk          1  0.0000026

sqtotexp    1  0.0005240

sqage       1  0.0005895

totexpage   1  0.0000013

totexpnk    1  0.0003292

agenk       1  0.0001206

 

MTB > let k4=1502*0.012

MTB > print k4

Data Display

K4    18.0240

Inverse Cumulative Distribution Function

Chi-Square with 8 DF

P( X <= x )        x

       0.95  15.5073

Since nrsq = (1502*0.012) 18.024 > 15.5073 = 2094_Tests for Heteroscedasticity.png, there is sufficient evidence to reject H0 which suggests that there is heteroscedasticity in the model from White's general heteroscedasticity test at the 5% significance level.  Both Breusch Pagan test and White's general heteroscedasticity test seem to indicate that totexp is the culprit as the T value is significant and the P-value is 0.000.

Posted Date: 3/5/2013 6:04:19 AM | Location : United States







Related Discussions:- White''s general heteroscedasticity test, Assignment Help, Ask Question on White''s general heteroscedasticity test, Get Answer, Expert's Help, White''s general heteroscedasticity test Discussions

Write discussion on White''s general heteroscedasticity test
Your posts are moderated
Related Questions
Protocol is the formal document outlining the proposed process for carrying out the clinical trial. The basic features of the document are to study the objectives, patient selecti

we are testing : Ho: µ=40 versus Ha: µ>40 (a= 0.01) Suppose that the test statistic is z0=2.75 based on a sample size of n=25. Assume that data are normal with mean mu and standa

There is high level of fluctuation in a zigzag pattern in the time series for RESI1 which indicates that there is possibly negative autocorrelation present. Column C11 show

4-13. Students in a management science class have just received their grades on the first test. The instructor has provided information about the first test grades in some previou

Yate s' continuity correction : When the testing for independence in contingency table, a continuous probability distribution, known as chi-squared distribution, is used as the app

Literature controls : The patients with the disease of interest who have received, in the past, one of two treatments under the investigation, and for whom the results have been pu

Catastrophe theory : A theory of how little is the continuous changes in the independent variables which can have unexpected, discontinuous effects on the dependent variables. Exam

The alternative process to make use of the chi-squared statistic for assessing the independence of the two variables forming a two-by-two contingency table particularly when expect

Non linear mapping (NLM ) is a technique for obtaining a low-dimensional representation of the set of multivariate data, which operates by minimizing a function of the differences

This is extension of the EM algorithm which typically converges more slowly than EM in terms of the iterations but can be much faster in the whole computer time. The general idea o