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April 2005 Q9

M

Michael

Member
I've been working through the exam questions in the revision notes, booklet 7 (GLMs) and got stuck on question 9, part iii, calculating the deviance residual.

I checked the ASET for the 2005 exam, and found that the answer given there is different. I like the ASET answer because I can actually understand what's going on, but the revision booklet answer matches the answer in the examiner's report... Which answer is right?
 
Sorry, not sure as I don't have the ASET 2005, could you briefly describe the method used there?

Slightly off topic, I found that the Acted notes had little material on calculating Pearson or deviance residuals... & I was shocked to see it come up so often in past papers!
 
MissAussie said:
Sorry, not sure as I don't have the ASET 2005, could you briefly describe the method used there?

Slightly off topic, I found that the Acted notes had little material on calculating Pearson or deviance residuals... & I was shocked to see it come up so often in past papers!
Click to expand...

The answer in the ASET is just the usual business of taking the sign of (yi - y^i) and multiplying by the square root of the contribution of yi to the scaled deviance. I had no idea what was going on in the examiners report though, there's just some random equation which gives a different answer.

Know what you mean about the notes too, I was surprised to see these sorts of questions come up so often.
 
Hello

The answer in the revision booklets and the examiners' report is correct.

In part (ii)(c) of the question, we are asked to work out a scaled deviance formula. In doing so, a number of cancellations of terms take place. Whilst many of these terms cancel at a total level they wouldn't cancel if we were to write out each term separately (i = 1, i = 2, etc).

In part (iii), we have to work out the deviance residual for i = 1. In order to do this, we need to go back to the "non-cancelled" version of the scaled deviance. This is ALWAYS the case when doing the deviance residual.

So let's break down the workings in the revision booklets for part (iii).

Log LS = -n - sum (log yi).

For i = 1, this gives us -1 - log (y1).

Log LM = -e(-alpha) sum (yi) - malpha - e(-beta) sum (yi) - (n-m) beta

(The first sum is i = 1 to m and the second sum is i = m+1 to n.)

alpha = log (1/m * sum (yi)) and beta = log(1/(n-m) * sum (yi)).

(The alpha sum is i = 1 to m and the beta sum is i = m+1 to n.)

From the question, alpha = log(14.2) and beta = log(18.7). These are constants.

For i = 1, Log LM = -e(-log 14.2) * y1 - 1 log 14.2 = -y1/14.2 - log 14.2.

This gives a scaled deviance of:

2[log LS - LogLM] = 2[-1 - log (y1) - (-y1/14.2 - log 14.2)] = 0.4006

since y1 = 7 is given in the question.

The deviance = scaled deviance for the exponential model since phi = 1.

So the deviance residual = sign(7 - 14.2) * root 0.4006 = negative 0.633

All the logs in my workings are natural logs.

Anna
 
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