library(emmeans)
library(lme4)

# generate some sample data
# condition (Placebo, Treatment)
# type (some factor, e.g. two different medications)
# value (outcome of interest)
# subject

data <- data.frame(condition = as.factor(rep(c(0,1), times = 10)), value = rnorm(20), type = rep(c("A","B"), each = 10), subject = as.factor(rep(rep(1:5, each = 2), times = 2)))

# caluculate a mixed model
fit_data <- lmer(value ~ condition*type + (1|subject), data = data)

# emmeans post-hoc test
emmeans(fit_data, pairwise ~ condition | type)

# paired t-test
t.test(data[data$condition==0 & data$type=="A","value"], data[data$condition==1 & data$type=="A","value"], paired = TRUE)
t.test(data[data$condition==0 & data$type=="B","value"], data[data$condition==1 & data$type=="B","value"], paired = TRUE)

set.seed(12345)
data <- data.frame(condition = as.factor(rep(c(0,1), times = 10)),
    value = rnorm(20), type = rep(c("A","B"), each = 10),
    subject = as.factor(rep(rep(1:5, each = 2), times = 2)))

> t.test(data[data$condition==0 & data$type=="A","value"],
+        data[data$condition==1 & data$type=="A","value"], paired = TRUE)

    Paired t-test

data:  data[data$condition == 0 & data$type == "A", "value"] and
       data[data$condition == 1 & data$type == "A", "value"]
t = 1.9384, df = 4, p-value = 0.1246
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.3618957  2.0361137
sample estimates:
mean of the differences
               0.837109

> library(lme4)
> mod1 = lmer(value ~ condition*type + (1|subject), data = data)
> pairs(emmeans(mod1, ~ condition | type))

type = A:
 contrast estimate    SE df t.ratio p.value
 0 - 1       0.837 0.475 12  1.764  0.1032

type = B:
 contrast estimate    SE df t.ratio p.value
 0 - 1      -0.677 0.475 12 -1.426  0.1795

> mod2 = lmer(value ~ condition + (1|subject), data = data,
+             subset = (type == "A"))

> pairs(emmeans(mod2, ~ condition))
 contrast estimate    SE df t.ratio p.value
 0 - 1       0.837 0.432  4 1.938   0.1246