Question

Breusch-Pagan / Cook-Weisberg test for heteroscedasticity

Ho:	Constant variance
Variable	fitted values of Dropout.
chi2(1)	8.14
Prob > F	0.0043

Is heteroscedasticity present here? why

Answer 1

Heteroscedasticity is present, since the p-value of the statistic is less than 0.01, meaning that the test statistic is significant at even 1% significance level.

Heteroscedasticity exists when the variance of the residual varies throughout the independent variable. To test for heteroscedasticity, among several tests, Breuch-Pagan test assumes that the variance varies linearly with the independent variable, as below.

For a regression , we estimate the regression, and obtain the squared residuals 's. Then, we set the auxiliary regression as , ie we regress squared residual on the independent variables. After this, we obtain the R-squared to test the joint significance that and at least one coefficient is statistically significant. The null implies that , ie the error variance is constant. In this case, the joint significance test is done through chi-square test, as for the R-squared of the auxiliary regression be , we have , for p be the number of parameters to be estimated in the auxiliary model, and p-1 is the df (note that the number independent variables are usually same for the auxiliary regression, but may sometimes be less than the original model).

In this case, , and there is supposedly 1 df since the original model is bi-variate model. But, the p-value is lower than even 0.01 meaning that the statistic is significantly different from zero for even 1% significance. Hence, we reject the null that (constant error variance ), and conclude that the heteroscedasticity is present in the model.

Note : It is mentioned that the variable is fitted values of dropout. Now, if it is the independent variable, there is no issue, but if Dropout is the dependent variable, the test is concerned with (a form of) White's test for heteroscedasticity. In this test, the residual squares and fitted values of the dependent variable is obtained, and then we regress , and for the null , the F or chi-square statistic is checked. If F or chi-square values are significant (as in the question too), then heteroscedasticity is present.