我觉得我错过了一些明显的东西，但我们开始吧。我有两次（或更多）治疗的一式三份测量的自相关数据。像这样的东西：

t <- 3:20 #times in my real dataset are possibly not always equidistant
a <- structure(c(0.652492388457625, 0.905172522010166, 1.23437705454616, 
                 1.48003667490842, 1.77876898946135, 1.99175317367897, 2.31666502140984, 
                 2.43520651415548, 2.67903421794922, 2.84115747823017, 2.89693734873647, 
                 2.91199679761145, 2.85645436179354, 2.99371033437697, 2.99965220711105, 
                 2.84984814715963, 2.64275376547326, 2.64060469520379, 0.481029734912324, 
                 0.8466803252367, 1.31126162780809, 1.56745630574946, 1.74865844658142, 
                 1.80367117155375, 2.06688393210808, 2.24500095501872, 2.52978288460243, 
                 2.69073206006205, 2.89657418056785, 2.93759772556246, 2.99305951550274, 
                 2.89146932307489, 2.88890777189028, 2.7974672802907, 2.70933381639295, 
                 2.66799551352975, 0.624178180970784, 0.867127935268765, 1.09752295578438, 
                 1.35037796202753, 1.60094288950107, 1.97949255710341, 2.15496378191076, 
                 2.42556913246041, 2.54331160179646, 2.67440414122285, 2.84249532365163, 
                 2.95278639560433, 3.06192227561515, 3.03297885461444, 3.04101341059534, 
                 3.01736966686846, 2.80061410999215, 2.69852643323913), 
               .Dim = c(18L, 3L), .Dimnames = list(NULL, c("a1", "a2", "a3")))
b <- structure(c(0.516527990622755, 0.84883434472028, 1.04202664437099, 
                 1.3100841689546, 1.48050413266838, 1.7824492800856, 1.96557179831706, 
                 2.17419105778186, 2.2453178060978, 2.35460428313729, 2.49308342865959, 
                 2.62343038370418, 2.70831189685371, 2.79459971623943, 2.94938536147398, 
                 3.04822554887815, 3.00287042052314, 2.91673487674283, 0.589490441973075, 
                 0.751768045201717, 0.917973959434798, 1.17617337222852, 1.39497560590896, 
                 1.65920945485901, 1.87749014780468, 2.11880355292648, 2.372755207219, 
                 2.46211141942227, 2.59688733749884, 2.72270421752644, 2.79848710425447, 
                 2.81134394947587, 2.75390203306788, 2.78499114431362, 2.86001341271914, 
                 2.95652300178809, 0.558662398944567, 0.834996005844121, 0.988238211915554, 
                 1.27569591423003, 1.38577342414377, 1.62664982549252, 1.83299700801392, 
                 2.04943560731628, 2.22950648854987, 2.38533269800646, 2.49845003387994, 
                 2.60036098089373, 2.61941602504858, 2.71298500309883, 2.78126388719353, 
                 3.04792375845498, 3.02691814463875, 3.06667590650438), 
               .Dim = c(18L, 3L), .Dimnames = list(NULL, c("b1", "b2", "b3")))

matplot(t,a,pch=1,xlab="",ylab="",col="blue")
matlines(t,a,col="blue", lty=2)

matpoints(t,b,pch=16,col="red")
matlines(t,b,col="red", lty=2)

我想知道治疗不同的时间段。我想避免拟合任何类型的模型。（我的科学数据有模型，但已知它们只是我的某些数据范围的近似值，我担心模型错误可能会掩盖差异。）我的想法是计算平均值并构建置信度像这样的间隔（使用正态性假设）：

a_means <- apply(a,1,mean)
a_sds <- apply(a,1,sd)
a_lwr <- a_means-qt(0.975,3)*a_sds/sqrt(3)
a_upr <- a_means+qt(0.975,3)*a_sds/sqrt(3)

b_means <- apply(b,1,mean)
b_sds <- apply(b,1,sd)
b_lwr <- b_means-qt(0.975,3)*b_sds/sqrt(3)
b_upr <- b_means+qt(0.975,3)*b_sds/sqrt(3)

DF <- data.frame(treat=factor(rep(1:2, each=length(t))), 
                 time=rep(t, 2),
                 mean=c(a_means,b_means),
                 lwr=c(a_lwr,b_lwr),
                 upr=c(a_upr,b_upr))

library(ggplot2)
p <- ggplot(DF, aes(x=time, y=mean, ymin=lwr, ymax=upr)) +
  geom_ribbon(aes(fill=treat), alpha=0.3) +
  geom_line(aes(color=treat))
print(p)

我构建置信区间的方式显然没有考虑自相关。

• 有没有办法构建某种“自相关置信区间”？
• 我可以使用“不相关的置信区间”吗？与自相关置信区间相比，我能否以某种方式估计它是否太窄或太宽？
• 有没有更好的方法来解决我的问题？