Park Glejser Test: Numerical Example

To detect the presence of heteroscedasticity using the Park Glejser test, consider the following data.

Year	1992	1993	1994	1995	1996	1997	1998
Yt	37	48	45	36	25	55	63
Xt	4.5	6.5	3.5	3	2.5	8.5	7.5

The step by step procedure of conducting Park Glejser test:

Step 1: Obtain estimate the regression equation

$$\hat{Y}_i = 19.8822 + 4.7173X_i$$

Obtain the residuals from this estimated regression equation:

Residuals

-4.1103

-2.5450

8.6071

1.9657

-6.6756

-4.9797

7.7377

Take the absolute values of these residuals and consider it as your dependent variables to perform the different functional forms suggested by Glejser.

Step 2: Regress the absolute values of $\hat{u}_i$ on the $X$ variable that is thought to be closely associated with $\sigma_i^2$. We will use the following function forms.

	Functional Form	Results
1)	$|\hat{u}_t| = \beta_1 + \beta_2 X_i +v_i$	$|\hat{u}_i| = 5.2666-0.00681X_i,\quad R^2=0.00004$ $t_{cal} = -0.014$
2)	$|\hat{u}_t| = \beta_1 + \beta_2 \sqrt{X_i} +v_i$	$|\hat{u}_i| = 5.445-0.0962X_i,\quad R^2=0.000389$ $t_{cal} = -0.04414$
3)	$|\hat{u}_t| = \beta_1 + \beta_2 \frac{1}{X_i} +v_i$	$||\hat{u}_i| = 4.9124+1.3571X_i,\quad R^2=0.00332$ $t_{cal} = -0.12914$
4)	$|\hat{u}_t| = \beta_1 + \beta_2 \frac{1}{\sqrt{X_i}} +v_i$	$\hat{u}_i| = 4.7375+1.0428X_i,\quad R^2=0.00209$ $t_{cal} = 0.10252$

Since none of the residual regression is significant, therefore, the hypothesis of heteroscedasticity is rejected. Therefore, we can say that there is no relationship between the absolute value of the residuals ($u_i$) and the explanatory variable $X$.

How to perform White General Heteroscedasticity?