STA 437: Applied Multivariate Statistics

Transcription

1 Al Nosedal. University of Toronto. Winter 2015

2 1 Chapter 5. Tests on One or Two Mean Vectors

3 If you can t explain it simply, you don t understand it well enough Albert Einstein.

4 Definition Chapter 5. Tests on One or Two Mean Vectors A (k k symmetric matrix) is positive definite if for all vectors x 0. x Ax > 0

5 Cauchy-Schwarz Inequality Let b and d be any two p 1 vectors. Then (b d) 2 (b b)(d d) with equality if and only if b = cd (or d = cb).

6 Extended Cauchy-Schwarz Inequality Let b and d be any two vectors, and let B be a positive definite matrix. Then (b d) 2 (b Bb)(d B 1 d) with equality if and only if b = cb 1 d (or d = cbb) for some constant c.

7 Maximization Lemma Let B be positive definite and d be a given vector. Then, for an arbitrary nonzero vector x, max x 0 (x d) 2 x Bx = d B 1 d with the maximum attained when x = cb 1 d for any constant c 0.

8 Hotelling s T 2 -Test We assume that a random sample y 1, y 2,..., y n is available from N p (µ, Σ), where y i contains the p measurements on the ith sampling unit. We estimate µ by ȳ and Σ by S. In order to test H 0 : µ = µ 0 versus H 1 : µ µ 0, we use the test statistic T 2 = n(ȳ µ 0 ) S 1 (ȳ µ 0 ). The distribution is indexed by two parameters, the dimension p and degrees of freedom ν = n 1. We reject H 0 if T 2 > T 2 α,p,n 1 and accept otherwise. Critical values of the T 2 -distribution are found in Table A.7.

9 Development of T 2 The development of this multivariate significance test proceeds as follows: a) We define a new variable: W n 1 = a 1 X 1 + a 2 X a p X p = X n p a p 1 where X j is an n-element column vector giving each of the n subjects score on dependent measure j; X = [X 1, X 2,..., X p ] is an n p data matrix whose ith row gives subject i s scores on each of the outcome variables; a is a p-element column vector giving the weights by which the dependent measures are to be multiplied before being added together.

10 Development of T 2 b) Our null hypothesis is that µ 1 = µ 10, µ 2 = µ 20,..., µ p = µ p0 are all true. If one or more of these equalities is false, the null hypothesis is false. This hypothesis can be expressed in matrix form as µ = µ 1 µ 2. µ p and it implies that µ w = a µ 0. = µ 10 µ 20. µ p0 = µ 0

11 Development of T 2 c) The variance of a linear combination of variables can readily be expressed as a linear combination of the variances and covariances of the original variables S 2 W = a Sa where S is the covariance matrix of the outcome variables. Thus the univariate t computed on the combined variable W is given by Squaring it yields t(a) = a X a µ 0 a Sa/n t 2 (a) = n a ( X µ 0 )( X µ 0 ) a a Sa

12 Development of T 2 Note that t 2 (a) depends on a, thus we will maximize t 2 (a). Using our maximization lemma where a = S 1 ( x µ 0 ) t 2 (a ) = T 2 = n( x µ 0 ) S 1 ( x µ 0 )

13 Example. Evaluating T 2 Let the data matrix for a random sample of size n = 3 from a bivariate Normal population be X = Evaluate the observed T 2 for µ 0 = [9, 5].

14 Solution Chapter 5. Tests on One or Two Mean Vectors x = ( ( )/3 ( )/3 ) = ( 8 6 ) s 11 = (6 8)2 +(10 8) 2 +(8 8) 2 2 = 4 s 12 = (6 8)(9 6)+(10 8)(6 6)+(8 8)(3 6) 2 = 3 s 22 = (9 6)2 +(6 6) 2 +(3 6) 2 2 = 9

15 Solution(cont.) ( ) 4 3 S = 3 9 S 1 = 1 ( ) ( ) 9 3 1/3 1/9 = /9 4/27

16 Solution (cont.) T 2 = n(ȳ µ 0 ) S 1 (ȳ µ 0 ). T 2 = 7 9

17 Example. Testing a multivariate mean vector Perspiration from 20 healthy females was analyzed. Three components, X 1 = sweat rate, X 2 = sodium content, and X 3 = potassium content, were measured, and the results, which we call the sweat data, are given in T5-1.DAT. Test the hypothesis H 0 : µ = [4, 50, 10] against H 1 : µ [4, 50, 10] at the level of significance α = 0.05.

18 Solution Chapter 5. Tests on One or Two Mean Vectors S = S 1 = x = T 2 = n(ȳ µ 0 ) S 1 (ȳ µ 0 ) =

19 Solution Chapter 5. Tests on One or Two Mean Vectors From Table A.7, we obtain the critical value T 0.05,3,19 = Comparing the observed T 2 = 9.74 with the critical value we see that T 2 = 9.74 < , and consequently, we can t reject H 0 at the 5% level of significance. Another way of finding the critical value for T 2. (n 1)p (n p) F p,n p(0.05) = (19)(3) F 3,17(0.05) = (19)(3) (3.20) =

20 ### Example 5.2 Sweat data data<-read.table(file="t5-1.dat") x.bar<-apply(data,2,fun=mean) x.bar mu.0<-c(4,50,10) difference<-x.bar-mu.0 difference<-matrix(difference,ncol=1) S.inv<-solve(cov(data))

21 n<-dim(data)[1] T.2<-n*t(difference)%*%S.inv%*%difference T.2 ## Critical value T.alpha<-(19*3/17)*qf(0.95,3,17) T.alpha

22 Example In Table 3.4 we have n = 10 observations on p = 3 variables. Desirable levels for y 1 and y 2 are 15.0 and 6.0, respectively, and the expected level of y 3 is We can, therefore, test the hypothesis H 0 : µ = [15, 6.0, 2.85] against H 1 : µ [15, 6.0, 2.85] at the level of significance α = 0.05.

23 Solution Chapter 5. Tests on One or Two Mean Vectors From Table A.7, we obtain the critical value T 0.05,3,9 = Comparing the observed T 2 = with the critical value we see that T 2 = > , and consequently, we reject H 0 at the 5% level of significance. Another way of finding the critical value for T 2. (n 1)p (n p) F p,n p(0.05) = (9)(3) F 3,7(0.05) = (9)(3) (4.35) =

24 ## Calcium data data<-read.table(file="t3_4_calcium.dat") data<-data[,-1] x.bar<-apply(data,2,fun=mean) x.bar mu.0<-c(15,6,2.85) difference<-x.bar-mu.0 difference<-matrix(difference,ncol=1)

25 S.inv<-solve(cov(data)) n<-dim(data)[1] T.2<-n*t(difference)%*%S.inv%*%difference T.2 ## Critical value crit.val<-(9*3)/(7)*qf(0.95,3,7) crit.val

26 Univariate Two-sample t-test In the one-variable case we obtain a random sample y 11, y 12,..., y 1n1 from N(µ 1, σ1 2 ) and a second random sample y 21, y 22,..., y 2n2 from N(µ 2, σ2 2 ). We assume that the two samples are independent and that σ1 2 = σ2 2 = σ2, say, with σ 2 unknown. From the two samples we calculate the pooled variance s 2 p = (n 1 1)s (n 2 1)s 2 2 n 1 + n 2 2 where n 1 + n 2 2 is the sum of the weights n 1 1 and n 2 1 in the numerator.,

27 Univariate Two-sample t-test To test H 0 : µ 1 = µ 2 vs H a : µ 1 µ 2, we use t = ȳ 1 ȳ 2 s p 1 n n 2 which has a t-distribution with n 1 + n 2 2 degrees of freedom when H 0 is true. We therefore reject H 0 if t t α/2,n 1 +n 2 2.

28 Multivariate Two-Sample T 2 -Test We wish to test H 0 : µ 1 = µ 2 vs H 1 : µ 1 µ 2. We obtain a random sample y 11, y 12,..., y 1n1 from N p (µ 1, Σ 1 ) and a second random sample y 21, y 22,..., y 2n2 from N p (µ 1, Σ 2 ). We assume that the two samples are independent and that Σ 1 = Σ 2 = Σ, say, with Σ unknown. where T 2 = n 1n 2 n 1 + n 2 (ȳ 1 ȳ 2 ) S p 1 (ȳ 1 ȳ 2 ) S p = 1 n 1 + n 2 2 [(n 1 1)S 1 + (n 2 1)S 2 ] We reject H 0 if T 2 T 2 α,p,n 1 +n 2 2. Critical values of T 2 are found in Table A.7.

29 Example Chapter 5. Tests on One or Two Mean Vectors Four psychological tests were given to 32 men and 32 women. The data are recorded in Table 5.1. The variables are y 1 = pictorial inconsistencies, y 2 = paper from board, y 3 = tool recognition, y 4 = vocabulary. The mean vectors are ŷ 1 = ŷ 2 =

30 Example (cont.) The covariance matrices of the two samples are S 1 = S 2 = Test the hypothesis H 0 : µ 1 = µ 2 versus H 1 : µ 1 µ 2 at the 0.01 significance level.

31 Univariate t-tests We give a procedure that could be used to check each variable following rejection of H 0 by a two-sample T 2 test: t j = ȳ 1j ȳ 2j [(n1 + n 2 )/n 1 n 2 ]s jj, j = 1, 2,..., p, where s jj is the jth diagonal element of S p. Reject H 0 : µ 1j = µ 2j if t j > t α/2,n 1 +n 2 2.

32 Examples Chapter 5. Tests on One or Two Mean Vectors Please, see tutorial 3.