Statistical modelling
In
this example the deviance statistic can be used to determine whether the model
that specifies the separate variance components fits the data better
than the fixed model that contains only the residual variance component.
The deviance in this output, namely 1817.10, is lower than the model without random effects (namely 1855.50 seen by rerunning this model). The difference between the two values: 1855.80
- 1817.10 = 38.70 with 688 − 686 = 2 degrees of freedom. This
approximates to a Chi-square distribution (Chi-square (df = 2)
= 38.70).
This is significant (P<0.001) and shows that the mixed
model provides a better fit than the fixed effects model without the ram
and ewe variance components.
|
|
**Estimated Variance Components **
Random
term |
Component |
S.e. |
RAM_ID |
0.067 |
0.089 |
EWE_ID |
1.457 |
0.283 |
***
Residual variance model ***
Parameter
|
Estimate
|
S.e.
|
Sigma2
|
3.427
|
0.266
|
**Approximate stratum variances ***
|
|
Effective
d.f.
|
RAM_ID |
4.733
|
57.66
|
EWE_ID |
6.490
|
297.74
|
*units* |
3.427
|
332.60
|
* Matrix of
coefficients of components for each stratum
RAM_ID |
10.31
|
0.42
|
1.00
|
EWE_ID |
0.00
|
2.10
|
1.00
|
*units* |
0.00
|
0.00
|
1.00
|
*** Deviance: -2*Log-Likelihood ***
Deviance |
d.f. |
1817.10 |
685 |
***
Wald tests for fixed effects ***
Fixed
term |
Wald
statistic |
d.f.
|
Wald/d.f.
|
Chi-sq
prob
|
*
Sequentially adding terms to fixed model |
YEAR |
230.32
|
5
|
46.06
|
<0.001
|
SEX |
9.66
|
1
|
9.66
|
0.002
|
AGEWEAN |
63.84
|
1
|
63.84
|
<0.001
|
DL |
30.44
|
1
|
30.44
|
<0.001
|
DQ |
78.41
|
1
|
78.41
|
<0.001
|
RAM_BRD |
6.64
|
1
|
6.64
|
0.010
|
EWE_BRD |
2.91
|
1
|
2.91
|
0.088
|
|
|