davidhaitaopan 发表于 2010-3-8 16:36
给个例子，希望有帮助

First, I perform the Tukey’s Test ,also called Tukey’s Honestly Significant Difference Test( abbreviated as HSD).
We already knew that the HSD test focuses on the experiment wise error rate, a, and assumes that the number of replicates at each level are equal. To use the function, an aov object is required as the first argument.

> #eg 3-1
> X=c(575,542,530,539,570,
+     565,593,590,579,610,
+     600,651,610,637,629,
+     725,700,715,685,710)
> A=factor(rep(1:4,each=5))
> #construct a data.frame
> shuju=data.frame(X,A)
> aov.shuju=aov(X~A,data=shuju)
> shuju.tukey=TukeyHSD(aov.shuju,ordered=T)
> shuju.tukey
  Tukey multiple comparisons of means
    95% family-wise confidence level
    factor levels have been ordered

Fit: aov(formula = X ~ A, data = shuju)

$A
     diff        lwr       upr     p adj
2-1  36.2   3.145624   69.25438   0.0294279
3-1  74.2   41.145624  107.25438  0.0000455
4-1  155.8  122.745624  188.85438  0.0000000
3-2  38.0   4.945624   71.05438    0.0215995
4-2  119.6   86.545624  152.65438  0.0000001
4-3  81.6   48.545624   114.65438  0.0000146


We can see immediately that all pairs of means are significantly different. Therefore, each power setting results in a mean etch rate that differs from the mean etch rate at any other power setting.
We also can plot the confidence intervals, which is more explicitly.


> plot(shuju.tukey)



Conclusions are the same as before.

Now we perform the LSD test to our above example. The LSD test essentially involves performing a series of pair wise t test. It is important to note that with LSD, we have to specify the individual error rate, not the experimentwise or family error rate.
Assume here that we wish to find the LSD between level 1 and 3.

> n1=sum(aov.shuju$model$A=='1')
> n1
[1] 5
> n3=sum(aov.shuju$model$A=='3')
> n3
[1] 5
> s=sqrt(sum((aov.shuju$residuals)^2)/aov.shuju$df.residual)
> s
[1] 18.26746
> s^2
[1] 333.7
> tcrit=qt(0.025,aov.shuju$df.residual,lower.tail=F)
> tcrit
[1] 2.119905
> LSD=tcrit*s*sqrt(1/n1+1/n3)
> LSD
[1] 24.49202
> #and
> y1.bar=sum(shuju$X[A=='1'])/5
> y1.bar
[1] 551.2
> y3.bar=sum(shuju$X[A=='3'])/5
> y3.bar
[1] 625.4
> y1.bar-y3.bar
[1] -74.2
>

Because   , so this implies that level 1 and level 3 ‘s means are significantly different. You can follow the exactly the same routine to compare the rest pair of means.