Description

Assignment 2
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:

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% confidence ellipsoid for the mean di↵erence between the L1 and L3.
c. Is there evidence of heterogeneity in variance?
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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 thousandths of an inch x2 = number of pits
The data are:
Location Coating 1 Coating 2
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 significantly in their e↵ect on corrosion? Clearly state the needed assumptions for your analysis.
Question 4: Let x1,…,xn be a sample from N2(µ,⌃), where ⌃ is a diagonal matrix with be the diagonal entries. Derive the likelihood ratio statistic for testinge
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4630 Assignment 2
Ravish Kamath: 213893664
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 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 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.
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 fixed lubricant e ect of factor 1 V elocityk represents the kth fixed velocity e ect of factor 2
(Lubricant ◊ V elocity)lk represents the interaction e ect between lubricant and velocity at the lkth level.
Assumptions: ‘lkr iid≥ N2(0, ) and qLl=3L1 = qkV=2V 1—k = qLl=3L1“lk = qVk=2V 1“lk = ˜0.
Part B
Running it through SAS, we get:
SSE SSLubricant SSV eloctiy
SSInteraction SSTreatment = SSLubricant + SSV elocity + SSInteraction
SSTotal
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
#P-Value
pchisq(chi_obs, df = 10, lower.tail = F)
## [1] 0.00809013
Since it is a really small p-value, implying that based o the data, there is evidence that treatment e ect exists.
Part D

Figure 1: Lubrication E ect
Based o SAS, testing for existence of lubricant e ect gives us a 0.1082 p-value. This is a large p-value hence this implies that lubrication e ect is not significant.

Figure 2: Velocity E ect
Based of SAS, testing for existence of velocity e ect gives 0.0009 p-value, which is quite small. This implies that velocity e ect is significant.

Figure 3: Interaction E ect
Based of SAS, testing for existence of interaction e ect gives 0.9881 p-value. This is a large value which means that interaction e ect is not significant.

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% confidence ellipsoid for the mean di erence 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
#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 e ect 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
#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 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 o the data, there is no evidence of heterogeneity in the variance.

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 di er significantly in their e ect 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 di erent populations and independent.
(2) All population have a common variance matrix .
(3) Each population is multivariate normal.
Furthermore, let H0 : µ1 = µ2 and Ha : µ1 =” µ2
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 o the data, the two coating do di er significantly in their e ect on corrosion.
Question 4
Let ˜x1,…,˜xn be a sample from N2(µ,˜ ) where His a diagonal matrix with0 : ‡12 = ‡22 = ‡2. ‡12 and ‡22 be the diagonal entries.
Derive the likelihood ratio statistics for testing
Solution