The table gives us unit root test statistic of -19. Unit root test critical value at 5% level of significance is -2.89; heteroskedasticity test statistic is 2.017 and its critical value @ 5% level of significance is 1.96.
The question asks the model is best described as a) a unit root, b) reliable standard erros and c) heteroskedasticity in the error term variance. Answer is C.
I understand that the model exhibits heteroskedasticity since test statistic of 2.07 is bigger than critical value of 1.96, suggesting the value is significant. But how does the model pass the unit root test since the absolute value of -19 is bigger than that of 2.89.
So the H0: b1-1=0 and Ha:b1-1 is not equal to 0. Since 19 is bigger than 2.89, we reject the H0 and conclude that b1-1 is not equal to 0 so that the model has no unit root (since b1 is not equal to 1).
For heteroskedasticity, H0: heteroskedasticity is present and Ha: heteroskedasticity is not present; since we reject H0, then heteroskedasticity is present in the model!!
Looks like you got it! One of the easiest ways to make conclusions from a hypothesis test is to quickly jot down the null and alternative hypotheses to keep straight what you are failing to reject or rejecting.
Just remember that the null hypothesis of these two are opposite. For ARCH, the null is ARCH is not present. For Unit Root, the null is Unit Root is present.
In case of unit root, rejecting the null hypothesis in the example means the absence of Unit Root.
Oh…I guess It should be Ho: Heteroskedasticity=0(not present); Ha: Heteroskedasticity=/=0 (present). Since we reject Ho then we concolude heteroskedasticity is present! Correct?
