The Breusch-Pagan Test (Numerical Example)

To perform the Breusch-Pagan test for the detection of heteroscedasticity, use the data from the following file Table_11.3.

Step 1:

The estimated regression is $\hat{Y}_i = 9.2903 + 0.6378X_i$

Step 2:

The residuals obtained from this regression are:

$\hat{u}_i$	$\hat{u}_i^2$	$p_i$
-5.31307	28.22873	0.358665
-8.06876	65.10494	0.827201
6.49801	42.22407	0.536485
0.55339	0.30624	0.003891
-6.82445	46.57318	0.591743
1.36447	1.86177	0.023655
5.79770	33.61333	0.427079
-3.58015	12.81744	0.162854
0.98662	0.97342	0.012368
8.30908	69.04085	0.877209
-2.25769	5.09715	0.064763
-1.33584	1.78446	0.022673
8.04201	64.67391	0.821724
10.47524	109.73066	1.3942
6.23093	38.82451	0.493291
-9.09153	82.65588	1.050197
-12.79183	163.63099	2.079039
-16.84722	283.82879	3.606231
-17.35860	301.32104	3.828481
2.71955	7.39595	0.09397
2.39709	5.74604	0.073007
0.77494	0.60052	0.00763
9.45248	89.34930	1.135241
4.88571	23.87014	0.303286
4.53063	20.52658	0.260804
-0.03614	0.00131	1.66E-05
-0.30322	0.09194	0.001168
9.50786	90.39944	1.148584
-18.98076	360.26909	4.577455
20.26355	410.61159	5.217089

The estimated $\tilde{\sigma}^2$ is $\frac{\sum u_i^2}{n} = \frac{2361.15325}{30} = 78.7051$.

Compute a new variable $p_i = \frac{\hat{u}_i^2}{\hat{\sigma^2}}$

Step 3:

Assuming $p_i$ is linearly related to $X_i(=Z_i)$ and run the regression of $p_i=\alpha_1+\alpha_2Z_{2i}+v_i$.

The regression Results are: $\hat{p}_i=-0.74261 + 0.010063X_i$

Step 4:

Obtain the Explained Sum of Squares (ESS) = 10.42802.

Step 5:

Compute: $\Theta = \frac{1}{2} ESS = \frac{10.42802}{2}= 5.2140$.

The Breusch-Pagan test follows Chi-Square Distribution. The $\chi^2_{tab}$ value at a 5% level of significance and with ($k-1$) one degrees of freedom is 3.8414. The $\chi_{cal}^2$ is greater than $\chi_{tab}^2$, therefore, results are statistically significant. There is evidence of heteroscedasticity at a 5% level of significance.

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