Fractional Factorial Design, Blocking & Confounding Variable
Isaiah C. Mireles
06/03/25
Q1)
First we must define :
Block Effect : the average effect of blocks
Two Level Design ( \(2^k\) ) : we have k-factors at 2 levels
Measuring Block-effect :
\[ \text{Block-eff} = \bar{y}_{\text{1Block1}} - \bar{y}_{\text{2Block2}} \]
So, TRUE!
Reference :
- EXAMPLE 7.2 (pg. 310-311)
Q2)
\[ 2^k \text{design} \text{ with } 2^p \text{ blocks for } k > p \\ k = 7\\ p = 3 \]
\[ \sum_{r=1}^{7} \binom{7}{r} = 127 \text{ Total effects} \]
Main-Eff : \(\{A, B, C, D, E, F, G\}\)
Interaction-Eff :
21 two-factor interactions
35 three-factor
35 four-factor
21 five-factor
7 six-factor
1 seven-factor
\[ N= \text{runs }=2^7=128 \\ n=N/8=\text{Replicates = 16} \]
FALSE , The number of effects confounded with blocks is \(2^3-1 = 7\).
Q3)
A) a \(\implies\) acde \(*\) a = cde
Odd, so different Block
B) acd \(\implies\) acde \(*\) acd = e
Odd, so different Block
C) bcd \(\implies\) acde \(*\) bcd = acb
Odd, so different Block
D) be \(\implies\) acde \(*\) be = acd
Odd, so different Block
E) abe \(\implies\) acde \(*\) abe = cd
Even, so same Block
E is the answer
Q4)
False, that is exactly what makes it confounded–its the fact it cannot be separated. You cannot separate the variation due to the interaction from the variation due to blocks.
Q5)
## run_number A B C D Run_Label y
## 1 1 -1 -1 -1 -1 (1) 23
## 2 2 1 -1 -1 -1 a 15
## 3 3 -1 1 -1 -1 b 16
## 4 4 1 1 -1 -1 ab 18
## 5 5 -1 -1 1 -1 c 25
## 6 6 1 -1 1 -1 ac 16
## 7 7 -1 1 1 -1 bc 17
## 8 8 1 1 1 -1 abc 26
## 9 9 -1 -1 -1 1 d 28
## 10 10 1 -1 -1 1 ad 16
## 11 11 -1 1 -1 1 bd 18
## 12 12 1 1 -1 1 abd 21
## 13 13 -1 -1 1 1 cd 36
## 14 14 1 -1 1 1 acd 24
## 15 15 -1 1 1 1 bcd 33
## 16 16 1 1 1 1 abcd 34Okay, this is a \(2^4\) factorial-design. We are going to base how we do this on lecture 7 [Slide 22-31].
In this problem we are asked to confound the highest interaction term (ABCD); that is, make it so we cannot separate the effect of ABCD and our blocks.
Further, how do we determine which Run_Label goes to which block?
- We take the row, product. Therefore defining the contrast for the confounding interaction.
## Confounded Block :## run_number A B C D y Run_Label Block_membership
## 16 16 1 1 1 1 34 abcd 1## Block 1:## run_number A B C D y Run_Label Block_membership
## 1 2 1 -1 -1 -1 15 a -1
## 2 3 -1 1 -1 -1 16 b -1
## 3 5 -1 -1 1 -1 25 c -1
## 4 8 1 1 1 -1 26 abc -1
## 5 9 -1 -1 -1 1 28 d -1
## 6 12 1 1 -1 1 21 abd -1
## 7 14 1 -1 1 1 24 acd -1
## 8 15 -1 1 1 1 33 bcd -1## Block 2:## run_number A B C D y Run_Label Block_membership
## 1 1 -1 -1 -1 -1 23 (1) 1
## 2 4 1 1 -1 -1 18 ab 1
## 3 6 1 -1 1 -1 16 ac 1
## 4 7 -1 1 1 -1 17 bc 1
## 5 10 1 -1 -1 1 16 ad 1
## 6 11 -1 1 -1 1 18 bd 1
## 7 13 -1 -1 1 1 36 cd 1
## 8 16 1 1 1 1 34 abcd 1Generally, we compute the effect via :
\[ \text{Est. Eff = }\frac{2}{n * 2^k} * \text{Contrast}_{\text{Eff}} \\ \text{n = # Replicates; k = # factors; Contrast = Block} \]
# Est Block Eff + ABCD Eff
cat("Est Block Eff + ABCD Eff : ")## Est Block Eff + ABCD Eff :mean(Block1$y) - mean(Block2$y)## [1] 1.25# Number of contrasts to compute :
n_contrast <- choose(n = 4, 1:4)
names(n_contrast) <- c("1 factor", "2 factors", "3 factors", "4 factors")
n_contrast## 1 factor 2 factors 3 factors 4 factors
## 4 6 4 1# Generate Classifications :
dat <- dat %>%
# 2 factor Interaction : 6
mutate(AB = A * B) %>%
mutate(AC = A * C) %>%
mutate(AD = A * D) %>%
mutate(BC = B * C) %>%
mutate(BD = B * D) %>%
mutate(CD = C * D) %>%
# 3 factor Interaction : 4
mutate(ABC = A * B * C) %>%
mutate(ABD = A * B * D) %>%
mutate(ACD = A * C * D) %>%
mutate(BCD = B * C * D) %>%
# 4 factor Interaction : 1
mutate(ABCD = A * B * C * D) %>%
# Select Col in relevant order :
select(run_number:D, AB:ABCD, y, Run_Label, Block_membership)
# Calculate Contrast :
# Function to compute contrast :
compute_contrast <- function(var) {
contrast <- sum(dat$y[dat[[var]] == 1]) - sum(dat$y[dat[[var]] == -1])
return(contrast)
}
# Apply to main effects
contrasts <- c(colnames(dat[2:16]))
summary_table <- data.frame(
Contrast_calc = sapply(contrasts, compute_contrast)
)
summary_table <- summary_table %>% mutate(Est_eff = 2*Contrast_calc/2^4)
summary_table## Contrast_calc Est_eff
## A -26 -3.25
## B 0 0.00
## C 56 7.00
## D 54 6.75
## AB 56 7.00
## AC 4 0.50
## AD -14 -1.75
## BC 18 2.25
## BD 4 0.50
## CD 32 4.00
## ABC 6 0.75
## ABD 0 0.00
## ACD -8 -1.00
## BCD 6 0.75
## ABCD -10 -1.25Except note that ABCD is actually the block effect + ABCD since it is confounded
library(FrF2)## Loading required package: DoE.base## Loading required package: grid## Loading required package: conf.design## Registered S3 method overwritten by 'DoE.base':
## method from
## factorize.factor conf.design##
## Attaching package: 'DoE.base'## The following objects are masked from 'package:stats':
##
## aov, lm## The following object is masked from 'package:graphics':
##
## plot.design## The following object is masked from 'package:base':
##
## lengths# Vector of effect estimates (excluding intercept)
effects <- summary_table$Est_eff
# Label each effect
names(effects) <- contrasts
# Half-normal plot
halfnormal(effects, alpha = 0.05, main = "Half-Normal Plot of Effects")
As we can see from the half-normal plot the most significant features are :
\[ \{AB, C, D, CD, A\} \]
Previously (ex. 6.36) we got :
if you notice, the results of the test hasn’t changed. Indicating that
blocking the variables hasnt made a significant difference in the
overall performance of our model. Therefore it appears that blocking
appears to add more complexity than is required than is required to
describe what is going on, on average.
Reference :