Design of Experiments - Latin Square Designs

Design of Experiments Latin Sqaures

• An industrial engineer is investigating the effect of four assembly methods (A, B, C, D) on the assembly time for a color television component. Four operators are selected for the study. Furthermore, the engineer knows that each assem- bly method produces such fatigue that the time required for the last assembly may be greater than the time required for the first, regardless of the method. That is, a trend develops in the required assembly time. To account for this source of variability, the engineer uses the Latin square design shown below. Analyze the data from this experiment ( \(\alpha = 0.05\)) and draw appropriate conclusions.

knitr::kable(tibble::tribble(
         ~V1,          ~V2,    ~V3,     ~V4, ~V5,
   "1",       "C=10", "D=14",  "A=7","B=8",
  "2",       "B=7", "C=18",  "D=11","A=8",
  "3",       "A=5", "B=10",  "C=11","D=9",
     "4",       "D=10", "A=10",  "B=12","C=14"
  ), col.names = c("Order of Assembly", "Operator 1","Operator 2",  "Operator 3", "Operator 4"))

Order of Assembly	Operator 1	Operator 2	Operator 3	Operator 4
1	C=10	D=14	A=7	B=8
2	B=7	C=18	D=11	A=8
3	A=5	B=10	C=11	D=9
4	D=10	A=10	B=12	C=14

First we will try to solve the problem implementing the formula.

order=rep(1:4,each=4)
operator=rep(1:4,4)
method=c("C","D","A","B", "B","C","D","A","A","B","C","D",
         "D","A","B","C")
time=c(10,14,7,8,7,18,11,8,5,10,11,9,10,10,12,14)
data=data.frame(order,operator,method,time)

head(data)

  order operator method time
1     1        1      C   10
2     1        2      D   14
3     1        3      A    7
4     1        4      B    8
5     2        1      B    7
6     2        2      C   18

str(data)

'data.frame':   16 obs. of  4 variables:
 $ order   : int  1 1 1 1 2 2 2 2 3 3 ...
 $ operator: int  1 2 3 4 1 2 3 4 1 2 ...
 $ method  : chr  "C" "D" "A" "B" ...
 $ time    : num  10 14 7 8 7 18 11 8 5 10 ...

data[,c(1,2)]=lapply(data[,c(1,2)],factor)
str(data)

'data.frame':   16 obs. of  4 variables:
 $ order   : Factor w/ 4 levels "1","2","3","4": 1 1 1 1 2 2 2 2 3 3 ...
 $ operator: Factor w/ 4 levels "1","2","3","4": 1 2 3 4 1 2 3 4 1 2 ...
 $ method  : chr  "C" "D" "A" "B" ...
 $ time    : num  10 14 7 8 7 18 11 8 5 10 ...

dim(data)

[1] 16  4

mean_order=aggregate(data$time,list(data$order),mean)

library(dplyr)

Attaching package: 'dplyr'

The following objects are masked from 'package:stats':

    filter, lag

The following objects are masked from 'package:base':

    intersect, setdiff, setequal, union

mean_order=data %>% group_by(order) %>% summarise_at(vars(time), list(mean = mean,SD=sd))
mean_order

# A tibble: 4 × 3
  order  mean    SD
  <fct> <dbl> <dbl>
1 1      9.75  3.10
2 2     11     4.97
3 3      8.75  2.63
4 4     11.5   1.91

boxplot(time~order,data = data,main="Box Plot of Time aganist Order", xlab="Order",ylab="Time")

mean_operator=data %>% group_by(operator) %>% summarise_at(vars(time), list(mean = mean,SD=sd))
mean_operator

# A tibble: 4 × 3
  operator  mean    SD
  <fct>    <dbl> <dbl>
1 1         8     2.45
2 2        13     3.83
3 3        10.2   2.22
4 4         9.75  2.87

boxplot(time~operator,data = data,main="Box Plot of Time aganist Operator", xlab="Operator",ylab="Time")

mean_method=data %>% group_by(method) %>% summarise_at(vars(time), list(mean = mean,SD=sd))
mean_method

# A tibble: 4 × 3
  method  mean    SD
  <chr>  <dbl> <dbl>
1 A       7.5   2.08
2 B       9.25  2.22
3 C      13.2   3.59
4 D      11     2.16

boxplot(time~method,data = data,main="Box Plot of Time aganist Method", xlab="Method",ylab="Time")

data[,c(1,2,3)]=lapply(data[,c(1,2,3)],factor)
str(data)

'data.frame':   16 obs. of  4 variables:
 $ order   : Factor w/ 4 levels "1","2","3","4": 1 1 1 1 2 2 2 2 3 3 ...
 $ operator: Factor w/ 4 levels "1","2","3","4": 1 2 3 4 1 2 3 4 1 2 ...
 $ method  : Factor w/ 4 levels "A","B","C","D": 3 4 1 2 2 3 4 1 1 2 ...
 $ time    : num  10 14 7 8 7 18 11 8 5 10 ...

p=4
SS_method <- sum(aggregate(time ~ method, data = data, sum)$time^2)/p-sum(data$time)^2/(p^2)
SS_method

[1] 72.5

SS_order=sum((aggregate(time~order,data,sum)$time)^2)/p-(sum(data$time)^2)/(p^2)
SS_order

[1] 18.5

SS_operator=sum((aggregate(time~operator,data,sum)$time)^2)/p-(sum(data$time)^2)/(p^2)
SS_operator

[1] 51.5

SS_total=sum(data$time^2)-(sum(data$time))^2/(p^2)
SS_total

[1] 153

SS_error=SS_total-SS_order-SS_operator-SS_method
SS_error

[1] 10.5

Now we shall compute various mean sum of squares

MS_order=SS_order/(p-1)
MS_order

[1] 6.166667

MS_operator=SS_operator/(p-1)
MS_operator

[1] 17.16667

MS_method=SS_method/(p-1)
MS_method

[1] 24.16667

MS_error=SS_error/(p^2-3*(p-1)-1)
MS_error

[1] 1.75

F_method=MS_method/MS_error
F_method

[1] 13.80952

F_order=MS_order/MS_error
F_order

[1] 3.52381

F_operator=MS_operator/MS_error
F_operator

[1] 9.809524

pvalue_method=1-pf(F_method,p-1,p^2-3*(p-1)-1)
pvalue_method

[1] 0.00421304

pvalue_order=1-pf(F_order,p-1,p^2-3*(p-1)-1)
pvalue_order

[1] 0.08851868

pvalue_operator=1-pf(F_operator,p-1,p^2-3*(p-1)-1)
pvalue_operator

[1] 0.009925869

ANOVA Table for Latin Square

Source of Variation	Sum of Squares	Df	MS	F ratio	p-value
Method	72.5	3	24.1666667	13.8095238	0.004213
Order	18.5	3	6.1666667	3.5238095	0.0885187
Operator	51.5	3	17.1666667	9.8095238	0.0099259
Error	10.5	6	1.75
Total	153	15

Direct Execution of Latin Sqaure in R

matrix(data$time,p,p)

     [,1] [,2] [,3] [,4]
[1,]   10    7    5   10
[2,]   14   18   10   10
[3,]    7   11   11   12
[4,]    8    8    9   14

matrix(data$method,p,p)

     [,1] [,2] [,3] [,4]
[1,] "C"  "B"  "A"  "D" 
[2,] "D"  "C"  "B"  "A" 
[3,] "A"  "D"  "C"  "B" 
[4,] "B"  "A"  "D"  "C" 

model=lm(time~order+operator+method,data=data)
anova(model)

Analysis of Variance Table

Response: time
          Df Sum Sq Mean Sq F value   Pr(>F)   
order      3   18.5  6.1667  3.5238 0.088519 . 
operator   3   51.5 17.1667  9.8095 0.009926 **
method     3   72.5 24.1667 13.8095 0.004213 **
Residuals  6   10.5  1.7500                    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1