Question 3.7

mt1 <- c(3129, 3000, 2865, 2890)
mt2 <- c(3200, 3300, 2975, 3150)
mt3 <- c(2800, 2900, 2985, 3050)
mt4 <- c(2600,2700,2600,2765)
combinedGroups <- data.frame(cbind(mt1,mt2,mt3,mt4))
stackedGroups <- stack(combinedGroups)
dat <- aov(values~ind, data = stackedGroups)
#summary(dat)

c.

\[ LSD = t_{\frac{\alpha}{2},df} \sqrt{\frac{2MSE}{n}} \\ = t_{\frac{0.05}{2},12} \sqrt{\frac{12826*2}{4}} \\ = 2.1788 \sqrt{\frac{12826*2}{4}} \\ = 2.1788 \sqrt{6413} \\ = 2.1788 * 80.081 \\ = 174.4804828 \]

## [1] 185.25
## [1] 37.25
## [1] 304.75
## [1] 222.5
## [1] 490
## [1] 267.5

Except for the pairing Mixing Technique 1 and 3 all other pairs differ significantly in mean.

d. and e. Plots

comments:

The normal probability plot shows us that the data is distributed normally because the residual points form somewhat of a line.

The residual vs fitted graph shows us that the assumption of equal variance still holds..

f.

Question 3.10

b. Fisher’s LSD

\[ LSD = t_{\frac{\alpha}{2},df} \sqrt{\frac{2MSE}{n}} \\ = t_{\frac{0.05}{2},20} \sqrt{\frac{8.06*2}{5}} \\ = 2.1788 \sqrt{\frac{8.06*2}{5}} \\ = 2.1788 \sqrt{3.224} \\ = 2.1788 * 1.796 \\ = 3.913 \]

## [1] 5.6
## [1] 7.8
## [1] 11.8
## [1] 1
## [1] 2.2
## [1] 6.2
## [1] 4.6
## [1] 4
## [1] 6.8
## [1] 10.8

The pairing of 15 Percent Cotton Weight and 35 Percent Cotton weight as well as 20 Percent Cotton Weight and 25 Percent cotton weight do not differ significantly in mean. All other pairings do differ significantly in mean.

c. Analyze Residuals

plot(dat2)

Comment:

The assumption for normality and constant variance both hold.

Question 3.44

library(pwr)
pwr.anova.test(k = 4, n = NULL, f = sqrt((5)^2/25), sig.level= 0.05, power = 0.90)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 4.658119
##               f = 1
##       sig.level = 0.05
##           power = 0.9
## 
## NOTE: n is number in each group

There would need to be 5 observations taken from each group.

Question 3.45

  1. The effect would decrease and the number of observations would increase to 7

  2. The effect would decrease and the number of observations would increase to 8

  3. The higher sigma increases the more observations are needed to meet the same power of the test.

  4. You can use this approach to find a range for sigma.