1 Problem 1: Blocking a 2^3 factorial design

Three factors are studied:

Each factor has two levels. One replicate of the 2^3 factorial design has eight treatment combinations.

1.1 1(a). Two blocks

For two blocks, the block degrees of freedom are:

\[ 2-1=1 \]

Therefore, one factorial effect must be confounded with blocks.

We choose the highest-order interaction, ABC, so the main effects and two-factor interactions remain estimable.

The block defining relation is:

\[ I = \text{Block}\times ABC \]

The block assignment is determined by the sign of ABC.

1.2 1(b). Assign treatment combinations to two blocks

treatments <- data.frame(
  Treatment = c("(1)", "a", "b", "ab", "c", "ac", "bc", "abc"),
  A = c(-1, 1, -1, 1, -1, 1, -1, 1),
  B = c(-1, -1, 1, 1, -1, -1, 1, 1),
  C = c(-1, -1, -1, -1, 1, 1, 1, 1)
)

treatments$ABC <- with(treatments, A * B * C)

treatments$Block <- ifelse(
  treatments$ABC == -1,
  "Block 1",
  "Block 2"
)

knitr::kable(
  treatments,
  caption = "Two-block layout; ABC is confounded with blocks"
)
Two-block layout; ABC is confounded with blocks
Treatment A B C ABC Block
(1) -1 -1 -1 -1 Block 1
a 1 -1 -1 1 Block 2
b -1 1 -1 1 Block 2
ab 1 1 -1 -1 Block 1
c -1 -1 1 1 Block 2
ac 1 -1 1 -1 Block 1
bc -1 1 1 -1 Block 1
abc 1 1 1 1 Block 2

The resulting blocks are:

  • Block 1 (ABC = -1): (1), ab, ac, bc
  • Block 2 (ABC = +1): a, b, c, abc

The block labels can be reversed without changing the confounding structure.

2 Problem 2: Blocking the design into four blocks

2.1 2(a). Effects confounded with blocks

Four blocks require:

\[ 4-1=3 \]

block degrees of freedom. Therefore, three independent factorial effects are confounded with blocks.

To preserve the main effects, choose AB and AC as block generators.

Their generalized interaction is:

\[ (AB)(AC)=A^2BC=BC \]

Therefore, the three effects confounded with blocks are:

\[ \boxed{AB,\ AC,\ BC} \]

2.2 2(b). Assign treatment combinations to four blocks

The four blocks are determined by the signs of AB and AC.

treatments$AB <- with(treatments, A * B)
treatments$AC <- with(treatments, A * C)
treatments$BC <- with(treatments, B * C)

treatments$Block4 <- paste0(
  "Block (AB=",
  ifelse(treatments$AB == 1, "+", "-"),
  ", AC=",
  ifelse(treatments$AC == 1, "+", "-"),
  ")"
)

four_block_layout <- treatments[
  order(treatments$Block4, treatments$Treatment),
  c("Treatment", "A", "B", "C", "AB", "AC", "BC", "Block4")
]

knitr::kable(
  four_block_layout,
  caption = "Four-block layout; AB, AC, and BC are confounded with blocks"
)
Four-block layout; AB, AC, and BC are confounded with blocks
Treatment A B C AB AC BC Block4
2 a 1 -1 -1 -1 -1 1 Block (AB=-, AC=-)
7 bc -1 1 1 -1 -1 1 Block (AB=-, AC=-)
6 ac 1 -1 1 -1 1 -1 Block (AB=-, AC=+)
3 b -1 1 -1 -1 1 -1 Block (AB=-, AC=+)
4 ab 1 1 -1 1 -1 -1 Block (AB=+, AC=-)
5 c -1 -1 1 1 -1 -1 Block (AB=+, AC=-)
1 (1) -1 -1 -1 1 1 1 Block (AB=+, AC=+)
8 abc 1 1 1 1 1 1 Block (AB=+, AC=+)

Each block contains two treatment combinations. The three block-confounded effects cannot be estimated separately from block effects.

3 Problem 3: Half-normal plot and collected data

The observations supplied in the assignment correspond to the two-block layout from Problem 1(b), where ABC is confounded with blocks.

3.1 3.1 Enter the observed data

dat <- data.frame(
  Treatment = c("(1)", "a", "b", "ab", "c", "ac", "bc", "abc"),
  A = c(-1, 1, -1, 1, -1, 1, -1, 1),
  B = c(-1, -1, 1, 1, -1, -1, 1, 1),
  C = c(-1, -1, -1, -1, 1, 1, 1, 1),
  ToolLife = c(22, 32, 35, 55, 44, 40, 60, 39)
)

dat$ABC <- with(dat, A * B * C)

dat$Block <- ifelse(
  dat$ABC == -1,
  "Block 1",
  "Block 2"
)

knitr::kable(
  dat,
  caption = "Observed tool-life data and block assignment"
)
Observed tool-life data and block assignment
Treatment A B C ToolLife ABC Block
(1) -1 -1 -1 22 -1 Block 1
a 1 -1 -1 32 1 Block 2
b -1 1 -1 35 1 Block 2
ab 1 1 -1 55 -1 Block 1
c -1 -1 1 44 1 Block 2
ac 1 -1 1 40 -1 Block 1
bc -1 1 1 60 -1 Block 1
abc 1 1 1 39 1 Block 2

3.2 3.2 Calculate factorial effects

For a 2^3 design with one replicate, each factorial effect is its contrast divided by:

\[ 2^{3-1}=4 \]

effect_names <- c("A", "B", "C", "AB", "AC", "BC", "ABC")

effect_values <- sapply(effect_names, function(term) {

  factors <- strsplit(term, "")[[1]]

  sign <- Reduce(
    `*`,
    lapply(factors, function(x) dat[[x]])
  )

  sum(sign * dat$ToolLife) / 4
})

effects <- data.frame(
  Effect = effect_names,
  Estimate = as.numeric(effect_values),
  Absolute_Effect = abs(as.numeric(effect_values))
)

knitr::kable(
  effects,
  digits = 4,
  caption = "Estimated factorial effects"
)
Estimated factorial effects
Effect Estimate Absolute_Effect
A 1.25 1.25
B 12.75 12.75
C 9.75 9.75
AB -1.75 1.75
AC -13.75 13.75
BC -5.25 5.25
ABC -6.75 6.75

The ABC effect is confounded with the two-block comparison. It is not independently estimable as a treatment effect.

3.3 3.3 Generate the half-normal plot

The DoE.base package provides halfnormal() for an unreplicated factorial model.

if (!requireNamespace("DoE.base", quietly = TRUE)) {
  install.packages("DoE.base")
}

library(DoE.base)

model_full <- lm(
  ToolLife ~ A * B * C,
  data = dat
)

halfnormal(model_full)

3.4 3.4 Identify effects that appear important

effect_ranking <- effects[
  order(-effects$Absolute_Effect),
]

knitr::kable(
  effect_ranking,
  digits = 4,
  caption = "Effects ranked by absolute magnitude"
)
Effects ranked by absolute magnitude
Effect Estimate Absolute_Effect
5 AC -13.75 13.75
2 B 12.75 12.75
3 C 9.75 9.75
7 ABC -6.75 6.75
6 BC -5.25 5.25
4 AB -1.75 1.75
1 A 1.25 1.25

The half-normal plot is used to judge which effects depart from the straight-line pattern formed by smaller effects.

With only one replicate, there is no independent pure-error estimate. The half-normal plot is therefore a screening tool rather than a conventional replicated ANOVA test.

3.5 3.5 Examine the block contrast

block_means <- aggregate(
  ToolLife ~ Block,
  data = dat,
  mean
)

knitr::kable(
  block_means,
  digits = 4,
  caption = "Mean tool life by block"
)
Mean tool life by block
Block ToolLife
Block 1 44.25
Block 2 37.50 
block_difference <- diff(block_means$ToolLife)

cat(
  "Difference between block means:",
  round(block_difference, 4),
  "hours\n"
)
## Difference between block means: -6.75 hours
cat(
  "Absolute ABC factorial contrast:",
  round(abs(effect_values["ABC"]), 4),
  "\n"
)
## Absolute ABC factorial contrast: 6.75

3.6 3.6 Conclusions

Question 3(a): Which effects appear significant?

The largest estimated treatment effects are:

  • AC = -13.75
  • B = 12.75
  • C = 9.75

These are the leading candidates for active effects based on magnitude. The ABC contrast represents the block effect and is confounded with the blocks.

Question 3(b): Does the block appear significant?

The two block means differ by 6.75 hours.

However, the block effect is exactly confounded with ABC in this design. Therefore, the data cannot distinguish a block effect from an ABC interaction effect.

The observed block contrast is 6.75 hours, but it cannot be tested independently using this single replicate.

4 Overall summary

  1. With two blocks, one effect is confounded with blocks: ABC.
  2. With four blocks, three effects are confounded with blocks: AB, AC, and BC.
  3. The largest observed treatment-effect estimates are AC, B, and C.
  4. The ABC contrast is confounded with blocks, so its effect cannot be separated from blocking.