Question:
The p-value of the Breusch-Pagan test is 0.0005. Based on this data and the information in the tables, there is evidence of:
A. serial correlation only.
B. serial correlation and heteroskedasticity.
C. heteroskedasticity only.
Why is there evidence of heteroskedasticity?
Is it because when we refer to the chi-square table, the critical value at 3 degrees of freedom and at α = 0.005 is 12.838? And since the test statistic is 17.7, this exceeds the critical value and so we reject the null hypothesis (that there is no heteroskedasticity)?
We need to know a significance level for the test. Whic I don’t see in tehe question.
Lets assume sigificance is 5% = 0.05
Method 1 - easiest
p vlaue from BP test = 0.0005 < signifcance level =0.05
There reject null of homoskedacity and accept al hetroskedacity
Method 2 - involves table
From Chi squared table k =3 signifcance = 0.5
Critcial Value = 7.81
BP test sttat > 17.7 > 7.81 therefore reject null of BP test )as above)
I think there must be more data as we need threasholds for serial correlation too though we DW stat at 1.8 this is close to 2.
Hi MikeyF, alpha = 0.005 is just an assumption I was making. To be honest, I was not very sure what alpha to use
I understand your method, but doesn’t that involve assuming an alpha? And the answer could, depending on the specific question, change depending on the assumed alpha? Hence, isn’t it better to NOT assume an alpha?
Well yes as p is so low. BUt that was the idea you were rejecting earlier.
That is what I wrote as the first option.
But if you remember from Level 1
p < significnce level in order to reject null. we have no significance level.
But as p is is low we can just assume it is low enough,
