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st: RE: RE: Re: Loglinear quasi-symmetric agreement
From
"Scholes, Shaun" <[email protected]>
To
"[email protected]" <[email protected]>
Subject
st: RE: RE: Re: Loglinear quasi-symmetric agreement
Date
Thu, 7 Jun 2012 17:19:37 +0000
Actually, this appears to give different results:
version 9
use http://www.ats.ucla.edu/stat/stata/examples/icda/carcinoma, clear
xi: glm count i.px i.py i.symm, fam(poi) nolog
hope this helps
best wishes
Shaun
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of Scholes, Shaun
Sent: 07 June 2012 18:07
To: [email protected]
Subject: st: RE: Re: Loglinear quasi-symmetric agreement
Martyn, I can't help you with your question but it may be worth taking a close look at:
http://www.ats.ucla.edu/stat/stata/examples/icda/icdast9.htm
Best wishes
Shaun
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of Martyn Sherriff
Sent: 07 June 2012 16:24
To: [email protected]
Subject: st: Re: Loglinear quasi-symmetric agreement
I am trying to use loglinear models to assess agreement using the quasi-symmetry model and have used the data from Agresti (An Introduction to Categorical Analysis, p 245) to check my method.
+-------------------------+
| px py count qasym |
|-------------------------|
1. | 1 1 22 1 |
2. | 1 2 2 2 |
3. | 1 3 2 3 |
4. | 1 4 0 4 |
5. | 2 1 5 2 |
|-------------------------|
6. | 2 2 7 5 |
7. | 2 3 14 6 |
8. | 2 4 0 7 |
9. | 3 1 0 3 |
10. | 3 2 2 6 |
|-------------------------|
11. | 3 3 36 8 |
12. | 3 4 0 9 |
13. | 4 1 0 4 |
14. | 4 2 1 7 |
15. | 4 3 17 9 |
|-------------------------|
16. | 4 4 10 10 |
+-------------------------+
The simple symmetry model is satisfactory:
glm count i.px i.py, fam(poi) nolog
Generalized linear models No. of obs = 16
Optimization : ML Residual df = 9
Scale parameter = 1
Deviance = 117.9568605 (1/df) Deviance = 13.10632
Pearson = 120.2634516 (1/df) Pearson = 13.36261
Variance function: V(u) = u [Poisson]
Link function : g(u) = ln(u) [Log]
AIC = 10.79847
Log likelihood = -79.38776817 BIC = 93.00356
------------------------------------------------------------------------------
| OIM
count | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------
-------------+------
px |
2 | -4.07e-08 .2773501 -0.00 1.000 -.5435962 .5435962
3 | .3794896 .2545139 1.49 0.136 -.1193485 .8783277
4 | .0741079 .2723524 0.27 0.786 -.4596929 .6079088
|
py |
2 | -.8109302 .3469443 -2.34 0.019 -1.490929 -.1309318
3 | .9382696 .2270017 4.13 0.000 .4933544 1.383185
4 | -.9932518 .3701851 -2.68 0.007 -1.718801 -.2677022
|
_cons | 1.783249 .2588899 6.89 0.000 1.275834 2.290664
------------------------------------------------------------------------------
However when I attempt the quasi-symmetric model I get very large and equal standard errors which do not make sense to me:
. glm count i.px i.py i.qasym, fam(poi) nolog
note: 7.qasym omitted because of collinearity
note: 9.qasym omitted because of collinearity
note: 10.qasym omitted because of collinearity
Generalized linear models No. of obs = 16
Optimization : ML Residual df
= 3
Scale
parameter = 1
Deviance = .978304658 (1/df) Deviance = .3261016
Pearson = .621982784 (1/df) Pearson = .2073276
Variance function: V(u) = u [Poisson]
Link function : g(u) = ln(u) [Log]
AIC
= 4.237311
Log likelihood = -20.89849023 BIC = -7.339462
------------------------------------------------------------------------------
| OIM
count | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------
-------------+------
px |
2 | -10.64727 1131.109 -0.01 0.992 -2227.58 2206.286
3 | -9.987129 1131.109 -0.01 0.993 -2226.92 2206.945
4 | 8.229144 1131.109 0.01 0.994 -2208.703 2225.161
|
py |
2 | -11.32026 1131.109 -0.01 0.992 -2228.253 2205.613
3 | -8.486948 1131.109 -0.01 0.994 -2225.419 2208.445
4 | -9.017585 1131.109 -0.01 0.994 -2225.95 2207.915
|
qasym |
2 | 9.089909 1131.109 0.01 0.994 -2207.843 2226.023
3 | 5.887591 1131.109 0.01 0.996 -2211.045 2222.82
4 | -27.11437 2775.396 -0.01 0.992 -5466.791 5412.562
5 | 20.82242 2262.218 0.01 0.993 -4413.043 4454.688
6 | 18.70797 2262.217 0.01 0.993 -4415.157 4452.573
7 | 0 (omitted)
8 | 18.96654 2262.217 0.01 0.993 -4414.898 4452.831
9 | 0 (omitted)
10 | 0 (omitted)
|
_cons | 3.091024 .2132027 14.50 0.000 2.673155 3.508894
------------------------------------------------------------------------------
I would be grateful for any advice on what I am doing wrong. I am using Stata 12.
Thank you,
Martyn
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