----------------------------
First problem
----------------------------
> attach(ex06.3)
> ex06.3
> f1=lm(dosage~drug)
> boxplot(dosage~drug)
> par(mfrow=c(2,2))
> plot(f1)

(a) The residuals vs. predicted show variance increasing with fitted values, i.e. a megaphone opening to the right.  The side-by-side boxplots also show increasing variance.  However the normal probability plot looks fine.

> library(MASS)
> par(mfrow=c(1,1))
> boxcox(dosage~drug)

(b) lambda=0 is suggested, the natural log transformation.  

> ldose=log(dosage)
> f2=lm(ldose~drug)
> anova(f2)
Analysis of Variance Table

Response: ldose
          Df Sum Sq Mean Sq F value    Pr(>F)    
drug       3 2.4925 0.83084  8.6037 0.0001944 ***
Residuals 36 3.4764 0.09657                      
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> lines(pairwise(f2,drug))
            
4 -0.232 |  
3 -0.192 |  
2  0.027 | |
1  0.397   |

(c) We reject H0: mu1=mu2=mu3=mu4 at the 5% level as p=0.0002.  There are differences in dose across drug levels.  Drugs 3 and 4 are significantly different (and better -- lower dosages are more effective) from drug 1.  There is not a significant difference between 3 and 4 though.

> par(mfrow=c(2,2))
> plot(f2)

(d) Yes, constant variance and normality of the residuals seem fine for the log-transformed response.

> qqnorm(dosage[drug==1])
> qqnorm(dosage[drug==2])
> qqnorm(dosage[drug==3])
> qqnorm(dosage[drug==4])

> brown.forsythe.test(dosage~drug)

        One-way analysis of means, Brown-Forsythe (unequal variances)

data:  dosage and drug
F = 9.8651, num df = 3.000, denom df = 31.188, p-value = 9.931e-05

> oneway.test(dosage~drug) 

        One-way analysis of means (not assuming equal variances)

data:  dosage and drug
F = 8.0021, num df = 3.000, denom df = 19.672, p-value = 0.001105

> source("http://people.stat.sc.edu/hansont/stat506/gh.R")
> gh(dosage,drug) 
$result1
        n  Mean  Variance
Group1 10 13.86 13.320444
Group2 10  9.73 11.549000
Group3 10  7.70  5.297778
Group4 10  7.41  5.714333

$Games.Howell
            t       df           p
1:2 2.6188885 17.90913 0.075114238
1:3 4.5145203 15.18118 0.002022012
1:4 4.6750451 15.52163 0.001401697
2:3 1.5640031 15.82156 0.425571417
2:4 1.7657355 16.15464 0.324312165
3:4 0.2763521 17.97428 0.992366287


(e) The four NPP's all look fairly straight; normality within each group is reasonable.  The p-values for both Welch and Brown-Forsythe indicate we reject H0: mu1=mu2=mu3=mu4 on the original scale.  The lines plot using Games-Howell is

4 3 2 1
    ---
-----   

This is the same as what we got using Tukey on the log-transformed data.

----------------------------
Second problem
----------------------------
> attach(ex06.4)
> names(ex06.4)
[1] "company"  "breakage"
> boxcox(breakage~company)

(a) The maximum is closes to zero, so we consider the log-transformation. 

> compare.to.best(f1,company,lowisbest=T)
          difference allowance
* 1 - 4    1.4163417 0.4384483
* 2 - 4    0.7650023 0.4384483
* 3 - 4    0.5298419 0.4384483
best is 4  0.0000000        NA

(b) Best (lowest breakage) is company 4, which is significantly better than the other three.
 
> par(mfrow=c(2,2))
> plot(f1)

(c) I think the residuals vs. predicted look fine; no evidence of non-constant variance.  The normal probability plot shows a bit of curvature, but nothing that makes me worried.

----------------------------
Ex. 6.5
----------------------------
(a) Right-opening megaphone show means variance is increasing with the response.  Tells us little about normality and nothing about independence.
(b) The NPP shows high curvature; these residuals are not normal.  Tells us nothing about constant variance or independence.  
(c) Residuals vs. time order shows "clumps" of negative, then positive, then negative residuals as time increases: these are not independent.  Doesn't tell us about constant variance or normality.
(d) Residuals vs. predicted here shows constant variance reasonable.  Tells us little about normality and nothing about independence.
 
----------------------------
Ex. 7.2
----------------------------
> power.anova.test(means=c(10,11,11),ns=c(6,6,6),sigma2=4,alpha=.05)
[1] 0.1170648

The power is only 0.12 under these values.  

> sample.size.anova(.9,.05,means=c(10,11,11),sigma2=4)
$nis
[1] 77 77 77

$power
[1] 0.9002725

To get a power of at least 0.90 we need a sample size of n=77 in each group, giving a power of 0.90.

----------------------------
Pr. 7.2
----------------------------

> sample.size.anova(.95,.01,means=c(45,32,60),sigma2=35)
$nis
[1] 4 4 4

$power
[1] 0.9808418

We need n=4 in each group to achieve a power over 0.95.  The actual power is 0.98.