[R] Wald test and frailty models in coxph

[Julianno Sambatti]

Dear R members,

I am new in using frailty models in survival analyses and am getting  
some contrasting results when I compare the Wald and likelihood ratio  
tests provided by the r output.

I am testing the survivorship of different sunflower interspecific  
crosses using cytoplasm (Cyt), Pollen and the interaction Cyt*Pollen  
as fixed effects, and sub-block  as a random effect.  I stratified  
the analysis by developmental stage (G_stageSM) as an ordered factor  
(two classes). There is a lot of tied deaths in this dataset.

Below is the analysis summary.

coxph(formula = Surv(Death_day, Censor) ~ Pollen * Cyt + strata 
(G_stageSM) +
     frailty(Sub.block), data = SurvNMexpSM)

   n=1422 (1 observation deleted due to missingness)
                    coef    se(coef) se2   Chisq  DF   p
PollenHNA          -0.0966 0.177    0.177   0.30  1.0 5.9e-01
PollenHNP          -0.3160 0.122    0.122   6.65  1.0 9.9e-03
PollenPET          -0.0478 0.120    0.120   0.16  1.0 6.9e-01
CytXA              -0.2967 0.118    0.118   6.36  1.0 1.2e-02
frailty(Sub.block)                        507.64 38.4 0.0e+00
PollenHNA:CytXA     0.2732 0.205    0.205   1.77  1.0 1.8e-01
PollenHNP:CytXA     0.7020 0.169    0.169  17.27  1.0 3.3e-05
PollenPET:CytXA     0.0837 0.207    0.207   0.16  1.0 6.9e-01

                 exp(coef) exp(-coef) lower .95 upper .95
PollenHNA           0.908      1.101     0.641     1.285
PollenHNP           0.729      1.372     0.573     0.927
PollenPET           0.953      1.049     0.753     1.206
CytXA               0.743      1.345     0.590     0.936
PollenHNA:CytXA     1.314      0.761     0.879     1.966
PollenHNP:CytXA     2.018      0.496     1.449     2.810
PollenPET:CytXA     1.087      0.920     0.724     1.632

Iterations: 10 outer, 25 Newton-Raphson
      Variance of random effect= 0.81   I-likelihood = -6513.5
Degrees of freedom for terms=  3.0  1.0 38.4  3.0
Rsquare= 0.365   (max possible= 1 )
Likelihood ratio test= 647  on 45.4 df,   p=0
Wald test            = 20.6  on 45.4 df,   p=1


Although, the results seem to reflect what we observe, it called my  
attention that the Likelihood ratio test and Wald test p-values are  
exactly the opposite.

I performed the same analysis without frailty and obtained

Call:
coxph(formula = Surv(Death_day, Censor) ~ Pollen * Cyt + strata 
(G_stageSM),
     data = SurvNMexpSM)

   n=1422 (1 observation deleted due to missingness)
                    coef exp(coef) se(coef)       z      p
PollenHNA       -0.0193     0.981    0.170 -0.1139 0.9100
PollenHNP       -0.2582     0.772    0.119 -2.1642 0.0300
PollenPET       -0.0555     0.946    0.117 -0.4747 0.6400
CytXA           -0.2123     0.809    0.114 -1.8702 0.0610
PollenHNA:CytXA -0.0135     0.987    0.197 -0.0684 0.9500
PollenHNP:CytXA  0.4358     1.546    0.164  2.6600 0.0078
PollenPET:CytXA  0.0186     1.019    0.202  0.0924 0.9300

                 exp(coef) exp(-coef) lower .95 upper .95
PollenHNA           0.981      1.020     0.703     1.368
PollenHNP           0.772      1.295     0.611     0.976
PollenPET           0.946      1.057     0.752     1.190
CytXA               0.809      1.237     0.647     1.010
PollenHNA:CytXA     0.987      1.014     0.670     1.452
PollenHNP:CytXA     1.546      0.647     1.122     2.132
PollenPET:CytXA     1.019      0.982     0.686     1.513

Rsquare= 0.008   (max possible= 1 )
Likelihood ratio test= 11.3  on 7 df,   p=0.127
Wald test            = 11.3  on 7 df,   p=0.124
Score (logrank) test = 11.4  on 7 df,   p=0.123

Here, the wald and the Likelihood ratio tests seem to be telling the  
same thing

Does anyone have a clue on how to interpret these results?

Thanks

J Berg