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$
-5.3130728.228730.358665
-8.0687665.104940.827201
6.4980142.224070.536485
0.553390.306240.003891
-6.8244546.573180.591743
1.364471.861770.023655
5.7977033.613330.427079
-3.5801512.817440.162854
0.986620.973420.012368
8.3090869.040850.877209
-2.257695.097150.064763
-1.335841.784460.022673
8.0420164.673910.821724
10.47524109.730661.3942
6.2309338.824510.493291
-9.0915382.655881.050197
-12.79183163.630992.079039
-16.84722283.828793.606231
-17.35860301.321043.828481
2.719557.395950.09397
2.397095.746040.073007
0.774940.600520.00763
9.4524889.349301.135241
4.8857123.870140.303286
4.5306320.526580.260804
-0.036140.001311.66E-05
-0.303220.091940.001168
9.5078690.399441.148584
-18.98076360.269094.577455
20.26355410.611595.217089

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 degree 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.

See More about Breusch-Pagan Test

Bruesch-Pagan-Test-of-Heteroscedasticity

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