wordpress图片地址优化优化设计答案六年级上册

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                

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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)