1. (6.16)Let contrast C = [~-I000~ } .We test the hypothesis0-1s=H Ha :C,u#O.1,1749.533,1509.133,1723.3) and.87289.71 94110.21, 76137.10 80777.35, 91917.09 90171.58; 90171.58 103766.36= n(CX)'(CS-1C')-l(CX) = 253.7966and theHo: C,u = 0By Splus, we can calculate that ,u' = (1906.-,~105616.30 94613.5394613.53 101510.1287289.71 76137.1094110.21 80777.35Then we can calculate the statistic T 2critical value (n-1)(q-1)Fq-ln-q+l(a) =9.53891< 253.7966. So we can reject the null(n-q+1) .vhypothesis, i.e., the four measures are not equal. Let a' = (0.5,0.5,-0.5,-0.5) , we canconstruct the 95% simultaneous confidence interval for a contract in the mean levelsrepresenting a comparison of the dynamic measurements with the static measurements,that isi.e. [124.3569,298.8431]0.02491758-0.019631590.02493980-0.026405630.020384020.024917580.020384020.016558420.01963159and Sl ='Ul =2. (6.17)(a) We use the logarithmic transformation for the data. First we calculate the mean andcovariance matrix for the two samples.[4.900659] r 4.6219283.940286 L[4.725444) l O.011072004 0.008019142 O.OO~15Y64~-.4.477574 and S2 = 0.008019142 0.006416726 0.0060052713.703186 0.008159648 0.006005271 0.006772758We can find that the covariance matrices are not quite different, so we can assume thatLI = L2 .We test the hypothesisHo : II I = 112 H H : , .r r " ~ ..~'U2 ='u, * .ll".. 1 1 (n1 + n2 -2) p( 8cntlca va ueF p.n1+n2-p-1 a) = .833461 < 85.34112. So we reject the nulln +n -p -11 2hypothesis, i.e., the two population mean vectors are not equal.(b) The hypothesis in part (a) is rejected. The linear combination of mean componentsmost responsible for rejecting H o is-1 --aoc Spool(X1-X2)=( c) The simultaneous confidence intervals for l' (}/I -}/2) is,- -~ 1 11 (XI -X2):tc 1 (-+-)Spool1nl n2We can choose l' = (1,0,0), (0,1,0), (0,0,1) , so the simultaneous confidence intervals for thecomponent mean are respectively[0.05776762, 0.2926638][0.05239685,0.2363126]~~ [0.1290622,0.3451377]The corresponding Bonferroni intervals are respectively[0.07702893, 0.2734025][0.06747781,0.2212316][0.1467803,0.3274197]We can find that Bonferroniin this situation.3. (6.22)For the modelx ij = .u + Ti + elj ,with constraint: Lf=l ni Ti = 0, we want to test the hypothesisH o : Ti = T2 = T3 = 0 H H a :not all ofthem equal to 0We ca~ construct the one-way MANOV A table as the followi~g -Source of Variation Sum of Squares De~ee gf FreedomTreatment (Species) 7877.333 -2288.933 9 -I = 2-2288.933 1151.853658.86 479.02]479.02 1697.228536.193 -1809.913-1809.913 2849.073n-g=147Residualn-l=149TotalFrom the output of S-plus,Df Pillai Trace approx. F num df den df P-valueSpecies 2 1.14 97.4238 4 294 OResiduals 147We can find that we should reject the null hypothesis, that is, the means of the threespecies are not equal to each other. We can calculate the 95% simultaneous confidenceintervals for all Dair of difference isAbout the assumption ~1 = ~2 = ~3 ' we calculate s~ /si7 .We can find that theassumption seems not supported by our calculation, because ~f the ratio are greaterthan 4 or less than 0.25./* S-plus program for 6.16. */stiff <- read.table("h:\\stat524\\stiffness.data")ic <- cbind(rep(l, 3), diag(rep(-l, 3)mu <- apply(stiff, 2, mean) ;S <- var(stiff);n <- dim{stiff) [1] ;q <- dim{stiff) [2] ;T2 <- n * t(C %*% mu) %*% solve(C %*% S %*% t(C))fcrit <- (n-l)*(q-l)/(n-q+l)*qf(O.95, q-l, n-q+l);%*% c %*% mu;t(a) %*% s %*% a / n)t(a) %*% s %*% a / n) ;a <- c(0.5, 0.5, -0.5, -0.5) ;CI.low <- t(a) %*% mu -sqrt(fcrit *cr.up <- t(a) %*% mu + sqrt(fcrit */* S-plus program for 6.17. * /carapace <- read.table('h:\\stat524\\carapace.data', header=T)female <- log{carapace[1:24, 2:4])female.mu <- apply {female, 2, mean)female.S <- var{female)n1 <- dim{ female) [1]male <- log(carapace[25:48, 2:4])male.mu <- apply (male, 2, mean)male.S <- var(male)n2 <- dim(male) [1]p <- dim( female) [2 ]S.pool <- {{nl-l)*female.S+{n2-1)*male.S)/{nl+n2-2)T2 <- t{female.mu-male.mu)%*%solve{{1/nl+l/n2)*S.pool)%*%{female.mu-male.mu)fcrit <- {nl+n2-2)*p/{nl+n2-p-l)*qf{O.95, p, nl+n2-p-l)alpha <- solve(S.pool)%*%(female.mu-male.mu)a <- diag(rep(l, 3)for(i in 1:3){diff <- t(a[ ti]) %*% (female.mu-male.mu)limit <- sqrt(fcrit*(1/nl+l/n2)*t(a[ ti])%*%S.pool%*%aprint(exp(diff-limit»print(exp(diff+limit»ibon.limit <- qt(1-0.05/6, nl+n2-2)*sqrt((1/nl+l/n2)*t(a[i])%*%S.pool%*%a[,i])print(exp(diff-bon.limit)print(exp(diff+bon.limit)/* Splus program for 6.22. */iris <- read.table(lh:\\stat524\\iris.data', heade=T)attach(iris)iris.manova <- man ova (cbind(Petalwidth, Sepalwidth)-Species, data=iris)iris. summary <- summary(iris.manova)samplel <- cbind(iris$Petalwidth[iris$No==l],iris$Sepalwidth[iris$No==l])sample2 <- cbind(iris$Petalwidth[iris$No==2],iris$Sepalwidth[iris$No==2])sample3 <- cbind(iris$Petalwidth[iris$No==3],iris$Sepalwidth[iris$No==3])mul <- apply (samplel, 2, mean)mu2 <- apply(sample2, 2, mean)mu3 <- apply(sample3, 2, mean)w <- iris.summary$SS$Residualsbon <- qt(1-0.05/12,