Applied Multivariable Statistic - Homework 10 Practice Problems | STAT 873, Assignments of Statistics

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HW Chapter 10

Problem

a) T^2 =55.71. p-value=0.0346. Below is my code.

prob2<-c(11, 33, 20, 28, 22, 20, 24, 30, 18, 27, 28, 26, 23, 22, 27, 26, 15, 31, 27, 18, 22, 16, 22, 30, 18, 21, 23, 18, 16, 15, 21, 24, 15, 17, 29, 18, 10, 12, 24, 20) prob2.2<-matrix(data = prob2, nrow=8, ncol=5) prob2. #The c(data) function creates a column vector of data. N<-nrow(prob2.2) p<-ncol(prob2.2) meanx1<-mean(prob2.2[,1]) meanx2<-mean(prob2.2[,2]) meanx3<-mean(prob2.2[,3]) meanx4<-mean(prob2.2[,4]) meanx5<-mean(prob2.2[,5]) mu.hat<-rbind(meanx1, meanx2, meanx3, meanx4, meanx5) mu.hat sigma.hat<-var(prob2.2) sigma.hat #remember that R is case sensitive! H<-matrix(data = c(1,1,1,1, -1,0,0,0, 0,-1,0,0, 0,0,-1,0, 0,0,0,-1),4,5) H q<-nrow(H) h<-matrix(0,q,1) h #PART A) #construct the Hotelling's T^2 for the hypothesis Ho:Hmu=h, Ha:Hmu<>h T.sq<-Nt(H%%mu.hat-h)%%solve(H%%sigma.hat%%t(H))%%(H%%mu.hat-h) p.val<-1-pf((N-q)/(q(N-1))*T.sq, q, N-q) cat("T.sq is", round(T.sq,2), "and the p-value is", round(p.val,4)) #critical value: qf(.95, q, N-q)

g) T^2 =47.93, p-value=0.

h)

Time

-

0

Mean response 0 5 10 15 20 25 Time

0

Control Treatment

T^2 =4.53, p-value=0.71 – test for interaction

i) Control vs. treatment main effect

T^2 =0.06, p-value=0.

Problem

a) I used the tests available in PROC DISCRIM in SAS

data set1; input gender$ height CC MUAC; datalines; F 80 58.4 14 F 75 59.2 15 F 78 60.3 15 F 75 57.4 13 F 79 59.5 14 F 78 58.1 14. F 75 58 12. F 64 55.5 11 F 80 59.2 12. M 78 60.6 16. M 76 58.1 12. M 92 63.2 14. M 81 59.0 14. M 81 60.8 15. M 84 59.5 14. ; run ; title2 'Proc factor output'; proc discrim data=set1 pool=test; class gender; var height CC MUAC; run ; The DISCRIM Procedure Test of Homogeneity of Within Covariance Matrices Notation: K = Number of Groups P = Number of Variables N = Total Number of Observations - Number of Groups N(i) = Number of Observations in the i'th Group - 1 __ N(i)/ || |Within SS Matrix(i)| V = ----------------------------------- N/ |Pooled SS Matrix| _ _ 2 | 1 1 | 2P + 3P - 1 RHO = 1.0 - | SUM ----- - --- | ------------- |_ N(i) N | 6(P+1)(K-1) DF = .5(K-1)P(P+1) _ _ | PN/2 | | N V | Under the null hypothesis: -2 RHO ln | ------------------ | | __ PN(i)/2 | | || N(i) _| is distributed approximately as Chi-Square(DF). Chi-Square DF Pr > ChiSq 3.321019 6 0.

Since the Chi-Square value is not significant at the 0.1 level, a pooled covariance matrix will be used in the discriminant function.

Problem #4 – I wrote a R program using ex10_1.R as a template. The test statistic is 28.49 and the

p-value is 0.000029 (using the distributional approximations given in class – different from Johnson’s

PROC IML program)

Problem #5 – I used PROC FACTOR in SAS. The test statistic is 12.20 and the p-value is 0.0067.

Below is the SAS code:

data set1; input gender$ height CC MUAC; datalines; F 80.00 58.40 14. F 75.00 59.20 15. F 78.00 60.30 15. F 75.00 57.40 13. F 79.00 59.50 14. F 78.00 58.10 14. F 75.00 58.00 12. F 64.00 55.50 11. F 80.00 59.20 12. ; run ; title2 'Proc factor output'; proc factor data=set1 method=ml; var height CC MUAC; run ;