Я использую модель LME, определенную как:
mod4.lme <- lme(pRNFL ~ Init.Age + Status + I(Time^2), random= ~1|Patient/EyeID,data = long1, na.action = na.omit)
Вывод:
> summary(mod4.lme)
Linear mixed-effects model fit by REML
Data: long1
AIC BIC logLik
2055.295 2089.432 -1018.647
Random effects:
Formula: ~1 | Patient
(Intercept)
StdDev: 7.949465
Formula: ~1 | EyeID %in% Patient
(Intercept) Residual
StdDev: 12.10405 2.279917
Fixed effects: pRNFL ~ Init.Age + Status + I(Time^2)
Value Std.Error DF t-value p-value
(Intercept) 97.27827 6.156093 212 15.801950 0.0000
Init.Age 0.02114 0.131122 57 0.161261 0.8725
StatusA -27.32643 3.762155 212 -7.263504 0.0000
StatusF -23.31652 3.984353 212 -5.852023 0.0000
StatusN -0.28814 3.744980 57 -0.076940 0.9389
I(Time^2) -0.06498 0.030223 212 -2.149921 0.0327
Correlation:
(Intr) Int.Ag StatsA StatsF StatsN
Init.Age -0.921
StatusA -0.317 0.076
StatusF -0.314 0.088 0.834
StatusN -0.049 -0.216 0.390 0.365
I(Time^2) -0.006 -0.004 0.001 -0.038 -0.007
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-2.3565641 -0.4765840 0.0100608 0.4670792 2.7775392
Number of Observations: 334
Number of Groups:
Patient EyeID %in% Patient
60 119
Я хотел получить сравнение между моими факторами «Статус» (названными A, N,F и H).Итак, я сделал модель Emmeans, используя этот код:
emmeans(mod4.lme, pairwise ~ Status, adjust="bonferroni")
Вывод для этого:
> emmeans(mod4.lme, pairwise ~ Status, adjust="bonferroni")
$emmeans
Status emmean SE df lower.CL upper.CL
H 98.13515 2.402248 57 93.32473 102.94557
A 70.80872 2.930072 57 64.94135 76.67609
F 74.81863 3.215350 57 68.38000 81.25726
N 97.84701 2.829706 57 92.18062 103.51340
Degrees-of-freedom method: containment
Confidence level used: 0.95
$contrasts
contrast estimate SE df t.ratio p.value
H - A 27.3264289 3.762155 212 7.264 <.0001
H - F 23.3165220 3.984353 212 5.852 <.0001
H - N 0.2881375 3.744980 57 0.077 1.0000
A - F -4.0099069 2.242793 212 -1.788 0.4513
A - N -27.0382913 4.145370 57 -6.523 <.0001
F - N -23.0283844 4.359019 57 -5.283 <.0001