Lecture 15. Hypothesis testing in the linear model

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1 14. Lecture 15. Hypothesis testing in the linear model Lecture 15. Hypothesis testing in the linear model 1 (1 1)

2 Preliminary lemma 15. Hypothesis testing in the linear model Preliminary lemma Lemma 15.1 Suppose Z N n (0, σ 2 I n ) and A 1 and A 2 and symmetric, idempotent n n matrices with A 1 A 2 = 0. Then Z T A 1 Z and Z T A 2 Z are independent. Proof: Let W i = A i Z, i = 1, 2 and W 2n 1 = ( W1 By Proposition 11.1(i), W N 2n (( 0 0 ) ( ) A1 = AZ, where A =. W 2 2n n A 2 ) ( )), σ 2 A1 0 check. 0 A 2 So W 1 and W 2 are independent, which implies W 1 T W 1 = Z T A 1 Z and W 2 T W 2 = Z T A 2 Z are independent.. Lecture 15. Hypothesis testing in the linear model 2 (1 1)

3 Hypothesis testing 15. Hypothesis testing in the linear model Hypothesis testing ( ) β0 Suppose X = ( X 0 X 1 ) and β =, where n p n p 0 n (p p 0) β 1 rank(x ) = p, rank(x 0 ) = p 0. We want to test H 0 : β 1 = 0 against H 1 : β 1 0. Under H 0, Y = X 0 β 0 + ε. Under H 0, MLEs of β 0 and σ 2 are ˆβ 0 = (X 0 T X 0 ) 1 X 0 T Y ˆσ 2 = RSS 0 n and these are independent, by Theorem So fitted values under H 0 are where P 0 = X 0 (X 0 T X 0 ) 1 X 0 T. = 1 n (Y X 0ˆβ 0 ) T (Y X 0ˆβ 0 ) Ŷ = X 0 (X 0 T X 0 ) 1 X 0 T Y = P 0 Y, Lecture 15. Hypothesis testing in the linear model 3 (1 1)

4 Geometric interpretation 15. Hypothesis testing in the linear model Geometric interpretation Lecture 15. Hypothesis testing in the linear model 4 (1 1)

5 15. Hypothesis testing in the linear model Generalised likelihood ratio test Generalised likelihood ratio test The generalised likelihood ratio test of H 0 against H 1 is ( ) n ( 1 2πˆσ 2 exp 1 2ˆσ (Y X ˆβ) T (Y X ˆβ) ) Λ Y (H 0, H 1 ) = ( ) 2 n ( ) 1 exp 1 (Y X ˆβ0 ) T (Y X ˆβ0 ) 2ˆσ 2 = ( ˆσ 2 ˆσ 2 2πˆσ 2 ) n 2 = ( ) n RSS0 RSS 2 = (1 + RSS 0 RSS RSS ) n 2 We reject H 0 when 2 log Λ is large, equivalently when (RSS0 RSS) RSS Using the results in Lecture 8, under H 0 ( 2 log Λ = n log 1 + RSS ) 0 RSS RSS is large. is approximately a χ 2 p 1 p 0 rv. But we can get an exact null distribution. Lecture 15. Hypothesis testing in the linear model 5 (1 1)

6 15. Hypothesis testing in the linear model Null distribution of test statistic Null distribution of test statistic We have RSS = Y T (I n P)Y (see proof of Theorem 13.3 (ii)), and so RSS 0 RSS = Y T (I n P 0 )Y Y T (I n P)Y = Y T (P P 0 )Y. Now I n P and P P 0 are symmetric and idempotent, and therefore rank(i n P) = n p, and rank(p P 0 ) = tr(p P 0 ) = tr(p) tr(p 0 ) = rank(p) rank(p 0 ) = p p 0. Also (I n P)(P P 0 ) = (I n P)P (I n P)P 0 = 0. Finally, Y T (I n P)Y = (Y X 0 β 0 ) T (I n P)(Y X 0 β 0 ) since (I n P)X 0 = 0, Y T (P P 0 )Y = (Y X 0 β 0 ) T (P P 0 )(Y X 0 β 0 ) since (P P 0 )X 0 = 0, Lecture 15. Hypothesis testing in the linear model 6 (1 1)

7 15. Hypothesis testing in the linear model Null distribution of test statistic Applying Lemmas 13.2 (Z T A i Z σ 2 χ 2 r ) and 15.1 to Z = Y X 0 β 0, A 1 = I n P, A 2 = P P 0 to get that under H 0, RSS = Y T (I n P)Y χ 2 n p and these rvs are independent. So under H 0, RSS 0 RSS = Y T (P P 0 )Y χ 2 p p 0 F = YT (P P 0 )Y/(p p 0 ) Y T (I n P)Y/(n p) = (RSS 0 RSS)/(p p 0 ) F p p0,n p. RSS/(n p) Hence we reject H 0 if F > F p p0,n p(α). RSS 0 - RSS is the reduction in the sum of squares due to fitting β 1. Lecture 15. Hypothesis testing in the linear model 7 (1 1)

8 15. Hypothesis testing in the linear model Arrangement as an analysis of variance table Arrangement as an analysis of variance table Source of degrees of sum of squares mean square F statistic variation freedom (df) Fitted model p p 0 RSS 0 - RSS (RSS 0 RSS) (p p 0) (RSS 0 RSS)/(p p 0) RSS/(n p) Residual n p RSS RSS (n p) n p 0 RSS 0 The ratio (RSS0 RSS) is sometimes known as the proportion of variance RSS 0 explained by β 1, and denoted R 2. Lecture 15. Hypothesis testing in the linear model 8 (1 1)

9 Simple linear regression We assume that 15. Hypothesis testing in the linear model Simple linear regression Y i = a + b(x i x) + ε i, i = 1,..., n, where x = x i /n, and ε i, i = 1,..., n are iid N(0, σ 2 ). Suppose we want to test the hypothesis H 0 : b = 0, i.e. no linear relationship. From Lecture 14 we have seen how to construct a confidence interval, and so could simply see if it included 0. Alternatively, under H 0, the model is Y i N(a, σ 2 ), and so â = Y, and the fitted values are Ŷi = Y. The observed RSS 0 is therefore RSS 0 = i (y i y) 2 = S yy. The fitted sum of squares is therefore RSS 0 RSS = ( (y i y) 2 (y i y ˆb(x i x)) 2) = ˆb 2 (x i x) 2 = ˆb 2 S xx. i Lecture 15. Hypothesis testing in the linear model 9 (1 1)

10 15. Hypothesis testing in the linear model Simple linear regression Source of d.f. sum of squares mean square F statistic variation Fitted model 1 RSS 0 RSS = ˆb 2 S xx ˆb 2 S xx F = ˆb 2 S xx / σ 2 Residual n 2 RSS = i (y i ŷ) 2 σ 2 n 1 RSS 0 = i (y i y) 2 Note that the proportion of variance explained is ˆb 2 S xx /S yy = where r is Pearson s Product Moment Correlation coefficient r = S xy / S xx S yy. From lecture 14, slide 5, we see that under H 0, s.e.(ˆb) = σ/ S xx. So ˆb = ˆb Sxx s.e.(ˆb) σ = t. S2 xy S xx S yy = r 2, ˆb s.e.(ˆb) t n 2, where Checking whether t > t n 2 ( α 2 ) is precisely the same as checking whether t 2 = F > F 1,n 2 (α), since a F 1,n 2 variable is tn 2 2. Hence the same conclusion is reached, whether based on a t-distribution or the F statistic derived from an analysis-of-variance table. Lecture 15. Hypothesis testing in the linear model 10 (1 1)

11 15. Hypothesis testing in the linear model Simple linear regression Example 12.1 continued As R code > fit=lm(time~ oxy.s ) > summary.aov(fit) Df Sum Sq Mean Sq F value Pr(>F) oxy.s e-06 *** Residuals Signif. codes: 0 *** ** 0.01 * Note that the F statistic, 41.98, is , the square of the t statistic on Slide 5 in Lecture 14. Lecture 15. Hypothesis testing in the linear model 11 (1 1)

12 15. Hypothesis testing in the linear model One way analysis of variance with equal numbers in each group One way analysis of variance with equal numbers in each group Assume J measurements taken in each of I groups, and that Y i,j = µ i + ε i,j, where ε i,j are independent N(0, σ 2 ) random variables, and the µ i s are unknown constants. Fitting this model gives RSS = I J i=1 j=1 (Y i,j ˆµ i ) 2 = I J i=1 j=1 (Y i,j Y i. ) 2 on n I degrees of freedom. Suppose we want to test the hypothesis H 0 : µ i = µ, i.e. no difference between groups. Under H 0, the model is Y i,j N(µ, σ 2 ), and so ˆµ = Y.., and the fitted values are Ŷi,j = Y... The observed RSS 0 is therefore RSS 0 = i (y i,j y.. ) 2. j Lecture 15. Hypothesis testing in the linear model 12 (1 1)

13 15. Hypothesis testing in the linear model One way analysis of variance with equal numbers in each group The fitted sum of squares is therefore RSS 0 RSS = ( (yi,j y.. ) 2 (y i,j y i. ) 2) = J (y i. y.. ) 2. i j i Source of d.f. sum of squares mean square F statistic variation Fitted model I 1 J i (y i. y.. ) 2 J i (y i. y.. )2 (I 1) F = J i (y i. y.. )2 (I 1) σ 2 Residual n I i j (y i,j y i. ) 2 σ 2 n 1 i j (y i,j y.. ) 2 Lecture 15. Hypothesis testing in the linear model 13 (1 1)

14 15. Hypothesis testing in the linear model One way analysis of variance with equal numbers in each group Example 13.1 As R code > summary.aov(fit) Df Sum Sq Mean Sq F value Pr(>F) x Residuals The p-value is 0.35, and so there is no evidence for a difference between the instruments. Lecture 15. Hypothesis testing in the linear model 14 (1 1)