MATH4630 Assignment 2 -Solved

Question 1: Posten (1962) performed an experiment with 2 factors: velocity (2 levels: V1 
and V2) and lubricants (three types: L1, L2, and L3). The ultimate torque x1 and the ultimate 
strain x2 of homogeneous pieces of bar steel are measured at each treatment combinations. 
The data are given below: 
V1 
V2 
x1 
x2 
x1 
x2 
L1 7.80 90.4 7.12 85.1 
7.10 88.9 7.06 89.0 
7.89 85.9 7.45 75.9 
7.82 88.8 7.45 77.9 
L2 9.00 82.5 8.19 66.0 
8.43 92.4 8.25 74.5 
7.65 82.4 7.45 83.1 
7.70 87.4 7.45 86.4 
L3 7.60 94.1 7.06 81.2 
7.00 86.6 7.04 79.9 
7.82 85.9 7.52 86.4 
7.80 88.8 7.70 76.4 
a. State clearly the model in terms of the overall mean, main e↵ects and interaction 
e↵ects. This should include all the necessary assumptions and constraints such that 
we can answer the rest of the questions. 
b. Obtain the all the necessary sum of squares. 
c. Is there any evidence that treatment e↵ect exists? 
d. Regardless of the answer in part (c), test for interaction e↵ect and then test for main 
e↵ect. 
Question 2: Using the data set given in Question 1 and ignoring the velocity factor. 
a. Is there any evidence that lubricant e↵ect exists? 
b. Obtain the 95% confifidence ellipsoid for the mean di↵erence between the L1 and L3. 
c. Is there evidence of heterogeneity in variance? 
1Question 3: To compare two types of coating for resistance to corrosion, 15 pieces of pipe 
were coated with each type of coating. Two pipes, one with each type of coating, were buried 
together and left for the same length of time at 14 loactions. Corrosion for the coating was 
measured by two variables: 
x1 = maximum depth of pit in thousandths of an inch 
x2 = number of pits 
The data are: 
Coating 1 Coating 2 
Location 
x1 
x2 
x1 
x2 
1 73 31 51 35 
2 43 19 41 14 
3 47 22 43 19 
4 53 26 41 29 
5 58 36 47 34 
6 47 30 32 26 
7 52 29 24 19 
8 38 36 43 37 
9 61 34 53 24 
10 56 33 52 27 
11 56 19 57 14 
12 34 19 44 19 
13 55 26 57 30 
14 65 15 40 7 
15 75 18 68 13 
Do the two coatings di↵er signifificantly in their e↵ect on corrosion? Clearly state the needed 
assumptions for your analysis. 
Question 4: 
Let x
e1,...,x
en be a sample from N2(µ
e
, ⌃), where ⌃ is a diagonal matrix 
with σ
2
1 
and σ22 be the diagonal entries. Derive the likelihood ratio statistic for testing 
H0 : σ
2
1 
= σ22 = σ2. 
24630 Assignment 2 
Ravish Kamath: 213893664 
04 November, 2022 
Question 1 
Posten (1962) performed an experiment with 2 factors: velocity (2 levels: V1 and V2) and lubricants (three 
types: L1, L2, and L3). The ultimate torque x1 and the ultimate strain x2 of homogeneous pieces of bar steel 
are measured at each treatment combinations. The data are given below: 
## Lubricant Velocity torque strain 
## 1 L1 V1 7.80 90.4 
## 2 L1 V1 7.10 88.9 
## 3 L1 V1 7.89 85.9 
## 4 L1 V1 7.82 88.8 
## 5 L1 V2 7.12 85.1 
## 6 L1 V2 7.06 89.0 
## 7 L1 V2 7.45 75.9 
## 8 L1 V2 7.45 77.9 
## 9 L2 V1 9.00 82.5 
## 10 L2 V1 8.43 92.4 
## 11 L2 V1 7.65 82.4 
## 12 L2 V1 7.70 87.4 
## 13 L2 V2 8.19 66.0 
## 14 L2 V2 8.25 74.5 
## 15 L2 V2 7.45 83.1 
## 16 L2 V2 7.45 86.4 
## 17 L3 V1 7.60 94.1 
## 18 L3 V1 7.00 86.6 
## 19 L3 V1 7.82 85.9 
## 20 L3 V1 7.80 88.8 
## 21 L3 V2 7.06 81.2 
## 22 L3 V2 7.04 79.9 
## 23 L3 V2 7.52 86.4 
## 24 L3 V2 7.70 76.4 
(a) State clearly the model in terms of the overall mean, main effffects and interaction effffects. This should 
include all the necessary assumptions and constraints such that we can answer the rest of the questions. 
(b) Obtain all the necessary sum of squares. 
(c) Is there any evidence that treatment effffect exists? 
(d) Regardless of the answer in part (c), test for interaction effffect and then test for main effffect. 
1Solution 
4630 Assignment 2 Ravish Kamath 213893664 
Solution 
Part A 
E(Xlkr) = µ + ·l + —k + “lk + ‘lkr 
= µ + Lubricantl + V elocityk + (Lubricant ◊ V elocity)lk + ‘lkr 
where: 
l = L1, L2, L3 
k = V1, V2 
r = 1, ..., 4 
µ represents the overall mean of the model. 
Lubricantl represents the Lth fifixed lubricant effffect of factor 1 
V elocityk represents the kth fifixed velocity effffect of factor 2 
(Lubricant ◊ V elocity)lk represents the interaction effffect between lubricant and velocity at the lkth level. 
Assumptions: ‘lkr iid 
≥ N2(0, Σ) and qLl=3L1 = qV 
2 
k=
V 1 —k = qLl=3L1 “lk = qV 
2 
k=
V 1 “lk = ˜
0. 
Part B 
Running it through SAS, we get: 
SSE = 5 
3.
1532 ≠14.
9535 
≠14
.9535 535
.
9725 6 SSLubricant = 5 
1.
6927 
≠9.
6989 
≠9
.
6989 56
.
3233 6 SSV eloctiy = 5 
0.
6240 14
.
8834 
14
.
8834 354
.
9704 6 
SSInteraction = 5 
0
.0290 
≠0
.102 
≠
0
.102 4
.7233 
6 SST reatment = SSLubricant + SSV elocity + SSInteraction 
= 5 2
.
3457 5.
0825 
5
.0825 416
.
0170 6 
SST otal = 5 
5.
4989 
≠9.
8710 
≠9
.
8710 951
.
9895 6 
Part C 
Let a = 3, b = 2, p = 2, n = 4. 
SSE = matrix(c(3.1532, -14.9535, -14.9535, 535.9725), nrow = 2, ncol = 2, byrow = T) 
SSLub = matrix(c(1.6927, -9.6989, -9.6989, 56.3233), nrow = 2, ncol = 2, byrow = T) 
SSVel = matrix(c(0.6240, 14.8834, 14.8834, 354.9704), nrow = 2, ncol = 2, byrow = T) 
SSInt = matrix(c(0.0290, -0.102, -0.102, 4.7233), nrow = 2, ncol = 2, byrow = T) 
SStr = SSLub + SSVel + SSInt 
wilkslam = det(SSE)/det(SSE + SStr) 
wilkslam 
## [1] 0.2854371 
a = 3 
b = 2 
p = 2 
n = 4 
#chi-squared observed 
chi_obs = -(a*b*(n-1)- ((p+1) - (a*b - 1))/2)*log(wilkslam) 
chi_obs 
## [1] 23.82094 
2Solution 
4630 Assignment 2 Ravish Kamath 213893664 
#P-Value 
pchisq(chi_obs, df = 10, lower.tail = F) 
## [1] 0.00809013 
Since it is a really small p-value, implying that based offff the data, there is evidence that treatment effffect 
exists. 
3Solution 

Part D 
Figure 1: Lubrication Effffect 
Based offff SAS, testing for existence of lubricant effffect gives us a 0.1082 p-value. This is a large p-value 
hence this implies that lubrication effffect is not signifificant. 
Figure 2: Velocity Effffect 
Based of SAS, testing for existence of velocity effffect gives 0.0009 p-value, which is quite small. This implies 
that velocity effffect is signifificant. 
4Solution 
4630 Assignment 2 Ravish Kamath 213893664 
Figure 3: Interaction Effffect 
Based of SAS, testing for existence of interaction effffect gives 0.9881 p-value. This is a large value which 
means that interaction effffect is not signifificant. 
54630 Assignment 2 Ravish Kamath 213893664 
Question 2 
Using the data set given in Question 1 and ignoring the velocity factor. 
(a) Is there any evidence that lubricant effffect exists? 
(b) Obtain the 95% confifidence ellipsoid for the mean difffference between the L1 and L3. 
(c) Is there evidence of heterogeneity in variance? 
Solution 
Part A 
n1_indices = which(df$Lubricant == 'L1') 
n1 = dim(df[n1_indices,])[1] 
n2_indices = which(df$Lubricant == 'L2') 
n2 = dim(df[n2_indices,])[1] 
n3_indices = which(df$Lubricant == 'L3') 
n3 = dim(df[n3_indices,])[1] 
grp_obs = c(n1, n2 ,n3) 
n = sum(grp_obs) 
g = nlevels(factor(df$Lubricant)) 
p = 2 
#Finding the means for each group and overall mean 
L1mean = as.vector(colMeans(df[n1_indices[1]:tail(n1_indices, n = 1),3:4])) 
L2mean = as.vector(colMeans(df[n2_indices[1]:tail(n2_indices, n = 1),3:4])) 
L3mean = as.vector(colMeans(df[n3_indices[1]:tail(n3_indices, n = 1),3:4])) 
lub_group_mean = cbind(L1mean, L2mean, L3mean) 
mean = (n1*L1mean + n2*L2mean + n3*L3mean)/(n1 + n2 + n3) 
#Finding the sample covariance for each lubricant 
X1 = as.matrix(df[n1_indices[1]:tail(n1_indices, n = 1),3:4]) 
X2 = as.matrix(df[n2_indices[1]:tail(n2_indices, n = 1),3:4]) 
X3 = as.matrix(df[n3_indices[1]:tail(n3_indices, n = 1),3:4]) 
S1 = 1/(n1 - 1)*(t(X1)%*%X1 - n1*colMeans(X1)%*%t(colMeans(X1))) 
S2 = 1/(n2 - 1)*(t(X2)%*%X2 - n2*colMeans(X2)%*%t(colMeans(X2))) 
S3 = 1/(n3 - 1)*(t(X3)%*%X3 - n3*colMeans(X3)%*%t(colMeans(X3))) 
S = list(S1, S2, S3) 
#Calculate the Within Matrix 
W = matrix(0,p,p) 
Wnew = matrix(0,p,p) 
for (i in 1:g){ 
Wnew = as.matrix(lapply(S[i], '*', (grp_obs[i] -1))) 
Wnew = matrix(unlist(Wnew), ncol = 2, byrow = T) 
W = W + Wnew 
} 
print(W) 
## [,1] [,2] 
## [1,] 3.806237 -0.172125 
## [2,] -0.172125 895.666250 
6Solution 
4630 Assignment 2 Ravish Kamath 213893664 
#Calculating the Between Matrix 
B = matrix(0,p,p) 
Bnew = matrix(0,p,p) 
for (i in 1:g){ 
Bnew = grp_obs[i]*(lub_group_mean[,i] - mean)%*%t(lub_group_mean[,i] - mean) 
B = B + Bnew 
} 
print(B) 
## [,1] [,2] 
## [1,] 1.692658 -9.698917 
## [2,] -9.698917 56.323333 
#Wilks Lambda and finding p-value 
wilkslam = det(W)/det(W + B) 
wilkslam 
## [1] 0.6635755 
fobs = (n - g - 1)/(g - 1)*((1 - sqrt(wilkslam))/sqrt(wilkslam)) 
fobs 
## [1] 2.275942 
pvalue = pf(fobs, df1 = 2*(g-1),df2 = 2*(n - g - 1), lower.tail = F) 
pvalue 
## [1] 0.07793588 
P-value is 0.07. If we have our – = 0.05, then there is no evidence to reject H0. Hence this implies that 
based on the data, the lubricant effffect does not exist. 
Part B 
#two sample difference 
tau1_hat = L1mean - mean 
tau2_hat = L2mean - mean 
tau3_hat = L3mean - mean 
tau_vec = matrix(data = c(tau1_hat, tau2_hat, tau3_hat), nco = 3, byrow = F) 
A = c(1, 0, -1) 
tau_hat = tau_vec%*%A 
tau_hat 
## [,1] 
## [1,] 0.01875 
## [2,] 0.32500 
#Finding the pooled variance for L1 and L3 
Spooled_L1andL3 = ((n1 - 1)*S1+(n3 - 1)*S3)/(n1 + n3 - 2 ) 
#Variance for tau_hat 
var_tau_hat = (1/n1 + 1/n3)*Spooled_L1andL3 
var_tau_hat 
## torque strain 
## torque 0.03048281 0.0810067 
## strain 0.08100670 7.5951339 
7Solution 
4630 Assignment 2 Ravish Kamath 213893664 
#95% Ellipsoid 
plot(tau_hat[1], tau_hat[2] , type="p", xlim=c(-2, 2), 
ylim=c(-2, 2), xlab="L1mean", ylab="L3mean", 
main = '95% C.I. for mean difference between L1 and L3') 
tau1 = matrix(seq(-2, 2, 0.05), ncol=1, byrow=T) 
ntau1 = nrow(tau1) 
tau3 = matrix(seq(-2, 2, 0.05), ncol=1, byrow=T) 
ntau3 = nrow(tau3) 
for (i in 1:ntau1) { 
for (j in 1:ntau3) { 
tau = matrix(c(tau1[i, 1], tau3[j, 1]), ncol=1, byrow=T) 
Tsq_obs = t((tau_hat - tau))%*%solve(var_tau_hat)%*%(tau_hat-tau) 
Fcomp = c( ( (n1 + n3 - 2) - p + 1)/((n1 + n3 - 2)*p )* Tsq_obs) 
Fcrit = qf(0.05, p, n1 + n3 - p - 1) 
if (Fcomp < Fcrit) points(tau1[i, 1], tau3[j, 1], pch="*") 
} 
} 
points(tau_hat[1], tau_hat[2], col='red') 
Figure 4: 95 percent Ellipsoid 
8Solution 
4630 Assignment 2 Ravish Kamath 213893664 
Part C 
#Taking the determinants of the sample co variances 
detS1 = det(S1) 
detS2 = det(S2) 
detS3 = det(S3) 
detS = c(detS1, detS2, detS3) 
#Finding the Spooled and its determinant 
Spooled = W/(n1 + n2 + n3 - g) 
detSpooled = det(Spooled) 
detSpooled 
## [1] 7.73036 
#Doing the Box Test for equality of co variances 
grp_obs = c(n1, n2 ,n3) 
lambda = rep(1,1) 
lambda_new = rep(0,1) 
for (i in 1:g){ 
lambda_new = (detS[i]/detSpooled)ˆ((grp_obs[i]-1)/2) 
lambda = lambda*lambda_new 
} 
print(lambda) 
## [1] 0.1140832 
M = -2*log(lambda) 
M 
## [1] 4.341655 
u = (sum(1/(grp_obs - 1)) - 1/(sum(grp_obs - 1)))*((2*pˆ2+3*p-1)/(6*(p+1)*(g-1))) 
u 
## [1] 0.1375661 
C = (1 - u)*M 
pchisq(C, df = ((1+p)*p*(g-1))/2, lower.tail = F) 
## [1] 0.7112208 
Since the p-value is large, then this implies that based offff the data, there is no evidence of heterogeneity in 
the variance. 
94630 Assignment 2 Ravish Kamath 213893664 
Question 3 
To compare two types of coating for resistance to corrosion, 15 pieces of pipe were coated with each type of 
coating. Two pipes, one with each type of coating, were buried together and left for the same length of time 
at 14 loactions. Corrosion for the coating was measured by two variables: 
x1 = maximum depth of pit in thousandiths of an inch 
x2 = number of pits 
The data are: 
## Coating x1 x2 
## 1 1 73 31 
## 2 1 43 19 
## 3 1 47 22 
## 4 1 53 26 
## 5 1 58 36 
## 6 1 47 30 
## 7 1 52 29 
## 8 1 38 36 
## 9 1 61 34 
## 10 1 56 33 
## 11 1 56 19 
## 12 1 34 19 
## 13 1 55 26 
## 14 1 65 15 
## 15 1 75 18 
## 16 2 51 35 
## 17 2 41 14 
## 18 2 43 19 
## 19 2 41 29 
## 20 2 47 34 
## 21 2 32 26 
## 22 2 24 19 
## 23 2 43 37 
## 24 2 53 24 
## 25 2 52 27 
## 26 2 57 14 
## 27 2 44 19 
## 28 2 57 30 
## 29 2 40 7 
## 30 2 68 13 
Do the two coatings diffffer signifificantly in their effffect on corrosion? Clearly state the needed assumptions for 
your analysis. 
Solution 
Assumptions: 
(1) Xl,1, ..., Xl,15 is a random sample size from a population with µl, where l = Coating1, Coating2. The 
random samples from difffferent populations and independent. 
(2) All population have a common variance matrix Σ. 
(3) Each population is multivariate normal. 
Furthermore, let H0 : µ1 = µ2 and Ha : µ1 
”
= 
µ2 
10Solution 
4630 Assignment 2 Ravish Kamath 213893664 
n1_indices = which(df$Coating == 1) 
n1 = dim(df[n1_indices,])[1] 
n2_indices = which(df$Coating == 2) 
n2 = dim(df[n2_indices,])[1] 
grp_obs = c(n1, n2) 
n = grp_obs[1] 
p = 2 
g = nlevels(factor(df$Coating)) 
#Difference of Mean 
C1 = df[n1_indices, 2:3] 
C2 = df[n2_indices, 2:3] 
d = C1 - C2 
d_bar = colMeans(d) 
d_bar 
## x1 x2 
## 8.000000 3.066667 
# sample Variance covariance matrix 
S = cov(d) 
S 
## x1 x2 
## x1 121.57143 17.07143 
## x2 17.07143 21.78095 
HotellingT = n*t(d_bar)%*%solve(S)%*%d_bar 
HotellingT 
## [,1] 
## [1,] 10.8189 
F_Obs = (n-p)/((n-1)*p)*HotellingT 
F_Obs 
## [,1] 
## [1,] 5.02306 
pvalue = pf(F_Obs, df1 = p, df2 = 15 - p, lower.tail = F) 
pvalue 
## [,1] 
## [1,] 0.02419613 
Since p-value is small, we can reject H0, which implies that based offff the data, the two coating do diffffer 
signifificantly in their effffect on corrosion. 
114630 Assignment 2 Ravish Kamath 213893664 
Question 4 
Let 
˜
x
1, ..., 
˜
x
n be a sample from N2(
˜ 
µ, 
Σ) where Σ 
is a diagonal matrix with 
‡
2
1 
and ‡
2
2 
be the diagonal entries. 
Derive the likelihood ratio statistics for testing 
H
0 : ‡
2
1 
= ‡
2
2 
= ‡2. 
Solution 
12