Definition

The randomized complete block design is used when there is one factor of interest and the experimenter wants to control a single source of extraneous variation. When there are two possible sources of extraneous variation, a Latin square design is the appropriate design for the experiment.

In general, a Latin square design can be used to compare t treatment means in the presence of two extraneous sources of variability, which we block off into t rows and t columns. The t treatments are then randomly assigned to the rows and columns so that each treatment appears in every row and every column of the design.

Model

The model for a response in a Latin square design can be written in the form

where the terms of the model are defined as follows:

A 4 x 4 Latin square design :

Advantages and Disadvantages

Illustration

A nonprofit consumer product testing organization is in the process of evaluating five major brands of room air cleaners. In order to make the ratings as realistic as possible, the organization’s engineers decided to evaluate the air cleaners outside the testing laboratory in residential homes. To control for variations due to the differing air quality in the homes and due to time of the year variation in external air pollution, the engineers decided to use a cleaner of each brand in each of five homes and to run the tests at five different months. The factors to be considered in the study are

1. Brand of air cleaner: B1, B2, B3, B4, B5

2. Residential home: H1, H2, H3, H4, H5

3. Month: M1, M2, M3, M4, M5

This example illustrates a situation in which the experimental units (rooms in home) are affected by two sources of extraneous variation, the home and the month of the year.

Hypothesis Testing

Hypothesis:

1. Treatments Effect:

   \(H_{0}:\tau_{1}=\tau_{2}=...=\tau_{t}\) (treatment has no effect on response)

   \(H_{1}:\) Not all \(\tau_{k}\)’s are equal

2. Row Effect:

   \(H_{0}:\beta_{1}=\beta_{2}=...=\beta_{t}\) (row has no effect on response)

   \(H_{1}:\) Not all \(\beta_{i}\)’s are equal

3. Column Effect:

   \(H_{0}:\gamma_{1}=\gamma_{2}=...=\gamma_{t}\) (column has no effect on response)

   \(H_{1}:\) Not all \(\gamma_{j}\)’s are equal

Analysis of variance table for a t x t Latin square design:

with,

Example

The consumer product rating organization decided to design the study of home air cleaners as a Latin square design using five homes and five months as blocking variables. The response variable is the CADR value obtained from a room air cleaner in a given home during a given month.Each brand of cleaner is observed in all five homes during all five months. The data from this study are given in Table below. Industry standards are that a CADR value above 300 is considered excellent and a CADR value below 100 is considered poor. Use these data to answer the following questions.

1. Write an appropriate statistical model for this experimental situation.

2. Conduct an analysis of variance to compare the mean CADR for the five brands of air cleaners. Use alpha = 0.05.

Answer:

1. The experiment was conducted as a Latin square design with t = 5 rows (months), t = 5 columns (homes), and t = 5 treatments (brands of air cleaners/row). An appropriate statistical model for this study is

   

2. Analysis of variance to compare the mean CADR for the five brands of air cleaners.

I.Hypotheses:

\(H_{0}:\tau_{1}=\tau_{2}=...=\tau_{t}\) (treatment has no effect on response)

\(H_{1}:\) Not all \(\tau_{k}\)’s are equal

II.Significance Level: 5%

III.Test Statistics

Complete ANOVA Table:

Complete ANOVA Table: (Using R)

#Data
library(readxl)
example1<-read_excel("D:/MATERI KULIAH S2 IPB/ASPRAK 2/LATIN.xlsx")
example1

## # A tibble: 25 x 4
##    month home  brand respon
##    <chr> <chr> <chr>  <dbl>
##  1 M1    Home1 B1       162
##  2 M2    Home1 B2       115
##  3 M3    Home1 B3       149
##  4 M4    Home1 B4       229
##  5 M5    Home1 B5       328
##  6 M1    Home2 B2        89
##  7 M2    Home2 B3       192
##  8 M3    Home2 B4       273
##  9 M4    Home2 B5       273
## 10 M5    Home2 B1       205
## # ... with 15 more rows

# change block variables into factors
example1$home <- as.factor(example1$home)
example1$month <- as.factor(example1$month)
example1$brand <- as.factor(example1$brand)

#anova table
anova1 = aov(respon~ home + month + brand, example1)
summary(anova1)

##             Df Sum Sq Mean Sq F value Pr(>F)  
## home         4   2178     544   0.189 0.9398  
## month        4  44513   11128   3.854 0.0308 *
## brand        4  58034   14509   5.025 0.0130 *
## Residuals   12  34647    2887                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

IV.Rejection Region:

Reject H0 if \(F_{test}>F_{4;12;0.05}=3.26\) .

V.Conclusion

Because F_test=5.02>F_table=3.26, H0 is rejected. Therefore, we conclude that there is significant evidence of a difference in the mean CADR values for the five brands of air cleaners.

Excercise

A petroleum company was interested in comparing the miles per gallon achieved by four different gasoline blends (A, B, C, and D). Because there can be considerable variability due to differences in driving characteristics and car models, these two extraneous sources of variability were included as ‘’blocking’’ variables in the study.The researcher selected four different brands of cars and four different drivers. The drivers and brands of cars were assigned to blends in the manner displayed in the following table. The mileage (in mpg) obtained over each test run was recorded as follows

1. Write a model for this experimental setting.

2. Conduct an analysis of variance to compare the mean of miles per gallon achieved by four different gasoline blends (A, B, C, and D). Use alpha=0.05.