Design and Analysis of Experiments

Problem 1
Problem 2

Problem 1

Let us first read the data as follows,

cloth = read.csv("C:/Users/Lenovo/OneDrive/Desktop/Cloth.csv", header = TRUE)
cloth

Here we will take bolts as blocks and chemicals as treatments. Now, the ANOVA table for RCBD is as follows,

r = c(t(as.matrix(cloth[,c(2:6)]))) # response data 
r

##  [1] 73 68 74 71 67 73 67 75 72 70 75 68 78 73 68 73 71 75 75 69

f = colnames(cloth[,c(2:6)])   # treatment levels 
k = 5                  # number of control blocks 
n = 4                    # number of treatment levels 
blk = gl(k, 1, n*k, factor(f))   # blocking factor 
blk

##  [1] B1 B2 B3 B4 B5 B1 B2 B3 B4 B5 B1 B2 B3 B4 B5 B1 B2 B3 B4 B5
## Levels: B1 B2 B3 B4 B5

tm = gl(n, k, k*n)            # matching treatment  
tm

##  [1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4
## Levels: 1 2 3 4

av_rcbd = aov(r ~ tm + blk)
summary(av_rcbd)

##             Df Sum Sq Mean Sq F value   Pr(>F)    
## tm           3  12.95    4.32   2.376    0.121    
## blk          4 157.00   39.25  21.606 2.06e-05 ***
## Residuals   12  21.80    1.82                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Now, observe that, p-value for treatments = 0.121 > 0.05, so we accept null hypothesis at level 0.05. So, all treatments have same effects at level 0.05.

Now, the ANOVA table for completely randomized design (RCD) is as follows,

av_rcd = aov(r ~ tm)
summary(av_rcd)

##             Df Sum Sq Mean Sq F value Pr(>F)
## tm           3  12.95   4.317   0.386  0.764
## Residuals   16 178.80  11.175

Now, observe that, p-value for treatments = 0.764 > 0.05, so we accept null hypothesis at level 0.05. So, all treatments have same effects at level 0.05.

Now, the efficiency for any design is defined as, \[ efficiency = \frac{1}{variance} = \frac{df}{SSE}, \text{ df = degrees of freedom} \]

Now the efficiency of RCBD and RCD are as follows,

eff_rcbd = 12/21.8
eff_rcbd

## [1] 0.5504587

eff_rcd = 16/178.8
eff_rcd

## [1] 0.08948546

As, RCBD has higher efficiency, so it’s better than RCD.

Problem 2

Let us first read the data as follows,

grain = read.csv("C:/Users/Lenovo/OneDrive/Desktop/Grain.csv", header = TRUE)
grain

Here we will take Furnance as blocks and stirring rate as treatments. Now, the RCBD is as follows,

r2 = c(t(as.matrix(grain[,c(2:5)]))) # response data 
r2

##  [1]  8  4  5  6 14  5  6  9 14  6  9  2 17  9  3  6

f2 = colnames(grain[,c(2:5)])   # block levels 
k2 = 4                  # number of control blocks 
n2 = 4                    # number of treatment levels 
blk2 = gl(k2, 1, n2*k2, factor(f2))   # blocking factor 
blk2

##  [1] F1 F2 F3 F4 F1 F2 F3 F4 F1 F2 F3 F4 F1 F2 F3 F4
## Levels: F1 F2 F3 F4

tm2 = gl(n2, k2, k2*n2)            # matching treatment  
tm2

##  [1] 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4
## Levels: 1 2 3 4

av2_rcbd = aov(r2 ~ tm2 + blk2)
summary(av2_rcbd)

##             Df Sum Sq Mean Sq F value Pr(>F)  
## tm2          3  22.19    7.40   0.853 0.4995  
## blk2         3 165.19   55.06   6.348 0.0133 *
## Residuals    9  78.06    8.67                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Now, observe that, p-value for treatments = 0.4995 > 0.05, so we accept null hypothesis at level 0.05. So, all treatments have same effects at level 0.05.

Now, the completely randomized design (RCD) is as follows,

av2_rcd = aov(r2 ~ tm2)
summary(av2_rcd)

##             Df Sum Sq Mean Sq F value Pr(>F)
## tm2          3  22.19   7.396   0.365   0.78
## Residuals   12 243.25  20.271

Now, observe that, p-value for treatments = 0.78 > 0.05, so we accept null hypothesis at level 0.05. So, all treatments have same effects at level 0.05.