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
