本文是实验设计与分析（第6版，Montgomery著，傅珏生译) 第3章单因子实验：方差分析3.11思考题3.6 R语言解题。主要涉及单因子方差分析，正态性假设检验，残差与拟合值的关系图，TukeyHSD法。
(a) Is there evidence to indicate that dosage level affects bioactivity? Use α = 0.05.
X<-c(24,28,37,30,37,44,31,35,42,47,52,38)
A<-factor(rep(1:3, each=4))
miscellany<-data.frame(X,A)
aov.mis<-aov(X~A, data=miscellany)
> summary(aov.mis) Df Sum Sq Mean Sq F value Pr(>F)
A 2 450.7 225.33 7.036 0.0145 *
Residuals 9 288.2 32.03
Signif. codes:
0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
There appears to be a different in the dosages.
(b) If it is appropriate to do so, make comparisons between the pairs of means. What conclusions can you draw?
> TukeyHSD(aov.mis)
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = X ~ A, data = miscellany)
$A
diff lwr upr p adj
2-1 7 -4.172869 18.17287 0.2402975
3-1 15 3.827131 26.17287 0.0114434
3-2 8 -3.172869 19.17287 0.1680265
Because there appears to be a difference in the dosages, the comparison of means is appropriate.
opar <- par(mfrow=c(2,2),cex=.8)
plot(aov.mis)
par(opar)
