Hi Amy-
To diagnose whether a particular x variable is the source of
heteroskedasticity, you would plot the sqrt(abs(
studentized residuals)) from the full regression against the variable in
question.
(Plotting the sqrt(abs(rstudent(lm.out))) against the x variable gives you
a more robust picture of whether spread is changing in x, but it is not as
good for diagnosing the actual relationship between the variance and x as
the plot of the abs(rstudent(lm.out)) against x.))
In this case, though, Kevin has specified the exact relationship between
the variance of the error and an independent variable, so you're all set
to run the regression.
Hope this helps!
Alison
On Sat, 4 Dec 2004, Amy Louise Catalinac wrote:
Hi everyone,
I am unsure about something in 3c. The question says we are to conduct a
weighted least squares regression that would be appropriate if the variance of
the error equals sigma-squared times x variable (assets). From our S-L plot in
(b) above, we know obviously that that the magnitude of errors is proportional
to our independent variable assets. According to Alison's sheet (p6) and
Kevin's lecture slides (403-404), to first ascertain the type of transformation
necessary we should plot the abs(student(res)) against our x variable.
My question is pretty simple: is this the studentized residuals from the whole
regression (lm(interlocks ~ sqrt(assets) + sector + nation), or just the
regression of lm(interlocks ~ sqrt(assets) that we plot against our x variable?
The exmaple in lecture and section don't make this completely clear to me.
The reason I ask is because after we use wls to downweight the observations of
X, and check whether it worked or not by using another S-L plot.... which shows
that it did work (the line is flat), I also want to check by graphing the
residuals from the weighted regresion against the original x value sqrt(assets)
- here again, is it the residuals from the whole regression, or just from the
regession of Y on our x variable?
Sorry to clog everyone's mailboxes if this is an annoying or obvious question
(roomies are away this weekend so I'm kind of on my own!!)
Thanks a lot,
Amy
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