White General Heteroscedasticity Test (Numerical Example)
We will consider the following data, to test the presence of heteroscedasticity using the White General Heteroscedasticity test.
| income | educ | jobexp |
| 5 | 2 | 9 |
| 9.7 | 4 | 18 |
| 28.4 | 8 | 21 |
| 8.8 | 8 | 12 |
| 21 | 8 | 14 |
| 26.6 | 10 | 16 |
| 25.4 | 12 | 16 |
| 23.1 | 12 | 9 |
| 22.5 | 12 | 18 |
| 19.5 | 12 | 5 |
| 21.7 | 12 | 7 |
| 24.8 | 13 | 9 |
| 30.1 | 14 | 12 |
| 24.8 | 14 | 17 |
| 28.5 | 15 | 19 |
| 26 | 15 | 6 |
| 38.9 | 16 | 17 |
| 22.1 | 16 | 1 |
| 33.1 | 17 | 10 |
| 48.3 | 21 | 17 |
To perform the white 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 5% level of significance is 11.07.
Since the calculated value is smaller than the tabulated value, therefore, the null hypothesis is accepted. Therefore, on the basis of the White general heteroscedasticity test, there is no heteroscedasticity.
Download the data file: White’s test Related Data
