R in Action 学习笔记 － 第九章－Analysis of Variance

One-way ANOVA

> install.packages("multcomp")

> library(multcomp)

> attach(cholesterol)

条件：因为 one-way ANOVA 的因变量需要满足正态分布，并且各组等方差

In a one-way ANOVA, the dependent variable is assumed to be normally distributed, and have equal variance in each group 

Step-1：用QQPlot检查是否符合正态分布条件

use a Q-Q plot to assess the normality assumption 

> install.packages("car")

> library(car)

> qqPlot(lm(response~trt,data=cholesterol),simulate=TRUE,main="Q-Q Plot",labels=FALSE)

#data fall within the 95 percent confidence envelope, 

#suggesting that the normality assumption has been met fairly well 

Step-2: 用 ANOVA

> fit <- aov(response~trt)

> summary(fit)
            Df Sum Sq Mean Sq F value   Pr(>F)    
trt          4 1351.4   337.8   32.43 9.82e-13 ***
Residuals   45  468.8    10.4                     
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

结论：

The ANOVA F test for treatment (trt) is significant (p < .0001), providing evidence that the five treatments aren't all equally effective

可视化

> library(gplots)

> plotmeans(response~trt,xlab="Treatment",ylab="Response",main="Mean Plot with 95% CI")

Step-3: 用其他test方法可得到相同结论

Bartlett’s test 

> bartlett.test()

Fligner–Killeen test  

> fligner.test()

Brown–Forsythe test 

> hov()

> bartlett.test(response~trt,data=cholesterol)


Bartlett test of homogeneity of variances


data:  response by trt
Bartlett's K-squared = 0.57975, df = 4, p-value = 0.9653#indicates that the variances in the five groups don’t differ significantly 

Step-4：看具体两两组间的差异

tell you which treatments differ from one another 

> TukeyHSD(fit)#TukeyHSD() function provides a test of all pairwise differences between group means 
  Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = response ~ trt)

$trt
                  diff        lwr       upr     p adj
2times-1time   3.44300 -0.6582817  7.544282 0.1380949#for1timeand2timesaren’t significantly different from each other (p = 0.138) 
4times-1time   6.59281  2.4915283 10.694092 0.0003542#the difference between 1time and 4times is significantly different (p < .001) 
drugD-1time    9.57920  5.4779183 13.680482 0.0000003
drugE-1time   15.16555 11.0642683 19.266832 0.0000000
4times-2times  3.14981 -0.9514717  7.251092 0.2050382
drugD-2times   6.13620  2.0349183 10.237482 0.0009611
drugE-2times  11.72255  7.6212683 15.823832 0.0000000
drugD-4times   2.98639 -1.1148917  7.087672 0.2512446
drugE-4times   8.57274  4.4714583 12.674022 0.0000037
drugE-drugD    5.58635  1.4850683  9.687632 0.0030633

可视化

> par(las=2)＃rotatesthe axis labels 

> par(mar=c(5,8,4,2))＃increases the left margin area so that the labels fit 

> plot(TukeyHSD(fit))

可视化#reproduces the Tukey HSD test, along with a different graphical representation of the results 

> par(mar=c(5,4,6,2))#increased the top margin to fit the letter array 

> tuk <- glht(fit,linfct=mcp(trt="Tukey"))

#glht() provides a much more comprehensive set of methods for multiple mean comparisons 

#you can use for both linear models and generalized linear models 

> plot(cld(tuk,level=.05),col="lightgrey")#significance level to use 0.05, or 95percent confidence in this case

参考：

http://www.inside-r.org/packages/cran/multcomp/docs/glht

http://www.stat.wmich.edu/wang/664/egs/Rmice.html

图表说明：具有相同字母的两组间没有显著的不同

Groups that have the same letter don’t have significantly different means 

taking the cholesterol-lowering drug in5 mg dosesfour times a day was better than taking a 20 mg dose once per day 

The competitordrugD wasn’t  superior to this four-times-per-day regimen 

competitor drugE was superior to both drugD and all three dosage strategies for our focus drug 

条件：因为 ANOVA 对outlier比较敏感

analysis of variance methodologies can be sensitive to the presence of outliers 

Step-5: 所以要检查是否没有outlier

> outlierTest(fit)

No Studentized residuals with Bonferonni p < 0.05
Largest |rstudent|:
   rstudent unadjusted p-value Bonferonni p
19 2.251149           0.029422           NA #there’s no indication of outliers in the cholesterol data (NA occurs when p > 1) 

最后：

Taking the Q-Q plot, Bartlett’s test, and outlier test together, the data appear to fit the ANOVA model quite well.  

总结：

1. 先用 qqPlot 看变量是否符合正态分布，car 包

2. 用 aov 再 summary 看多组间是否具有显著差异性，multcomp包

3. 用 ghlt 的 Tukey 看两两间的差异性

4. 用 outlierTest 看是否满足没有 outlier