White General Heteroscedasticity Test (Numerical Example)

One important assumption of Regression is that the variance of the Error Term is constant across observations. If the error has a constant variance, then the errors are called homoscedastic, otherwise heteroscedastic. In the case of heteroscedastic errors (non-constant variance), the standard estimation methods become inefficient. Typically, to assess the assumption of homoscedasticity, residuals are plotted.

White general Heteroscedasticity test

We will consider the following data, to test the presence of heteroscedasticity using White General Heteroscedasticity test.

IncomeEducationJob Experience
529
9.7418
28.4821
8.8812
21814
26.61016
25.41216
23.1129
22.51218
19.5125
21.7127
24.8139
30.11412
24.81417
28.51519
26156
38.91617
22.1161
33.11710
48.32117

White General Heteroscedasticity Test

To perform the White General Heteroscedasticity test, the general procedure is

Step 1: Run a regression and obtain $\hat{u}_i$ of this regression equation.

The regression model is: $income = \beta_1+\beta_2\, educ + \beta_3\, jobexp + u_i$

The Regression results are: $Income_i=-7.09686 + 1.93339 educ_{i} + 0.649365 jobexp_{i}$

Step 2: Run the following auxiliary regression

$$\hat{u}_i^2=\alpha_1+\alpha_2X_{2i}+\alpha_3 X_{3i}+\alpha_4 X_{2i}^2+\alpha_5X_{3i}^2+\alpha_6X_{2i}X_{3i}+vi $$

that is, regress the squared residuals on a constant, all the explanatory variables, the squared explanatory variables, and their respective cross-product.

Here in auxiliary regression education, $Y$ is income, $X_2$ is educ, and $X_3$ is jobexp.

The results from auxiliary regression are:

$$Y=42.6145  -0.10872\,X_{2i} – 5.8402\, X_{3i} -0.15273\, X_{2i}^2 + 0.200715\, X_{3i}^2 + 0.226517\,X_{2i}X_{3i}$$

Step 3: Formulate the null and alternative hypotheses

$H_0: \alpha_1=\alpha_2=\cdots=\alpha_p=0$

$H_1$: at least one of the $\alpha$s is different from zero

Step 4: Reject the null and conclude that there is significant evidence of heteroscedasticity when the statistic is bigger than the critical value.

The statistic with computed value is:

$$n \cdot R^2 \, \Rightarrow = 20\times 0.4488 = 8.977$$

The statistics follow asymptotically $\chi^2_{df}$, where $df=k-1$. The Critical value is $\chi^2_5$ at a 5% level of significance is  11.07. 

Since the calculated value is smaller than the tabulated value, therefore, the null hypothesis is accepted. Therefore, based on the White general heteroscedasticity test, there is no heteroscedasticity.

Download the data file: White’s test Related Data

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