Chapter 7. Hypothesis Testing

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1 Chapter 7. Hypothesis Testing Joonpyo Kim June 24, 2017 Joonpyo Kim Ch7 June 24, / 63

2 Basic Concepts of Testing Suppose that our interest centers on a random variable X which has density function f(x : θ) where θ Ω. Suppose we think that θ Ω 0 or θ Ω 1 where Ω 0 and Ω 1 are disjoint subsets of Ω and Ω 0 Ω 1 = Ω. We label the hypotheses as H 0 : θ Ω 0 vs H 1 : θ Ω 1. Joonpyo Kim Ch7 June 24, / 63

3 Basic Concepts of Testing The decision rule to take H 0 or H 1 is based on a sample X 1,, X n from the distribution of X and hence, the decision could be wrong. Table displays the various situations. True State of Nature Decision H 0 is True H 1 is True Reject H 0 Type I error Correct decision Accept H 0 Correct decision Type II error Table: Testing Hypothesis and Decision Error Joonpyo Kim Ch7 June 24, / 63

4 Basic Concepts of Testing A test of H 0 versus H 1 is based on a subset C of data. This set C is called the rejection region and its corresponding decision rule is: Reject H 0 if (X 1,, X n ) C Retain H 0 if (X 1,, X n ) / C A type I error occurs if H 0 is rejected when it is true while a type II error occurs if H 0 is accepted when H 1 is true. The goal is to select a rejection region from all possible rejection regions which minimizes the probabilities of these errors, which often have a trade-off relationship. So we often consider type I error to be the worse of the two errors. Joonpyo Kim Ch7 June 24, / 63

5 Basic Concepts of Testing Definition. We say a rejection region C is of size α if where X = (X 1,, X n ) t. α = max θ Ω 0 P θ (X C) Over all rejection regions of size α, we want to consider rejection regions which have lower probabilities of type II error. Joonpyo Kim Ch7 June 24, / 63

6 Basic Concepts of Testing Definition. The probability 1 P (type II error) = P θ (X C) is called the power of the test at θ. It is the probability that the test detects the alternative θ when θ Ω 1. So minimizing the probability of type II error is equivalnet to maximizing power. Further, we define the power function of a rejection region to be γ C (θ) = P θ (X C), θ Ω 1. Joonpyo Kim Ch7 June 24, / 63

7 Basic Concepts of Testing Example. Let X 1,, X n be a random sample from N(µ, σ 2 ). Suppose that we want test the hypotheses H 0 : µ = 0 vs H 1 : µ > 0 with size α. (0 < α < 1) In here, we may use X. If X is large, it supports H 1, so if X is sufficiently large we reject H 0 and accpet H 1. Hence {X c} is a reasonable rejection region. Now we have P µ=0 (X c) = α and it implies that c = z α σ n. Joonpyo Kim Ch7 June 24, / 63

8 Basic Concepts of Testing Example. Now we assume the same situation as in previous one, except that in this example we consider H 0 : µ 0 vs H 1 : µ > 0. Also in this case the rejection region is {X c} and now we have Note that max µ 0 P µ(x c) = α. ( X µ max P µ(x c) = max P µ µ 0 µ 0 σ/ n c µ ) ( σ/ = P µ=0 Z c µ ) n σ/ = α n where Z N(0, 1). Joonpyo Kim Ch7 June 24, / 63

9 Basic Concepts of Testing Example. (Cont d) So we get c = z α σ n. Joonpyo Kim Ch7 June 24, / 63

10 Randomized Test Example. Now consider X 1,, X 8 P oisson(λ) and H 0 : λ = 5, H 1 : λ < 5. We again consider a rejection region {X c }, which is equivalent to {X 1 + X X 8 c}. Our goal is to find c which satisfies P λ=5 (X X 8 c) α where α = 0.05 is the size of the test. Joonpyo Kim Ch7 June 24, / 63

11 Randomized Test Example. (Cont d) Following table shows probabilities P (Z n) where Z P oisson(40). n P (Z n) If we set c = 29, then the size is , and if c = 30, then the size is Thus we can say that the rejection region is {X X 8 29}, but in this case size of the test is smaller then α, which may yield smaller power. Joonpyo Kim Ch7 June 24, / 63

12 Randomized Test Example. (Cont d) Let s consider following test rule. If X X 8 29, we reject H 0. If X X 8 > 30, we do not reject H 0. If X X 8 = 30, we reject H 0 with fixed probability p (0, 1) randomly. With this procedure we can make P λ=5 (reject H 0 ) = 0.05 exactly. Such p should satisfy ( )p = 0.05 and we can find p. Such test containing random step is called a randomized test. This example suggests that we need more general definition about size and power of the test. Joonpyo Kim Ch7 June 24, / 63

13 Randomized Test Definition. Let φ(x) be a rejection probability of H 0, i.e., Then Reject H 0 if φ(x) = 1 Reject H 0 randomly with probability φ(x) if φ(x) (0, 1) Accept H 0 if φ(x) = 0 is called a size of the test, and max E θ φ(x) θ Ω 0 γ φ (θ) = E θ φ(x), θ Ω is called a power function or test function of the test. The test with significance level α means that the size is less than or equal to α. Joonpyo Kim Ch7 June 24, / 63

14 Likelihood Ratio Test Definition. Let X = (X 1,, X n ) be a sample from f( : θ) and L(θ) be a likelihood function. A test with rejection region C α = { } { } supθ Ω L(θ) sup θ Ω0 L(θ) λ L(ˆθ Ω ) = λ = {2(l(ˆθ Ω ) l(ˆθ Ω 0 )) λ } L(ˆθ Ω 0 ) and satisfying max θ Ω0 P θ (X C α ) = α is called a likelihood ratio test. In here, ˆθ Ω is MLE on Ω, ˆθ Ω 0 is MLE on Ω 0, and l( ) is a log-likelihood. Joonpyo Kim Ch7 June 24, / 63

15 Likelihood Ratio Test Example. Let X 1,, X n N(µ, σ 2 ), < µ <, and 0 < σ 2 <. Consider the hypotheses H 0 : µ = µ 0, H 1 : µ µ 0. Our goal is to describe likelihood ratio test with level α. (0 < α < 1) For θ = (µ, σ 2 ) t, we can easily show that (exercise) ( ˆθ Ω = x, 1 n ( ˆθ Ω 0 = µ 0, 1 n ) t n (x i x) 2 i=1 ) t n (x i µ 0 ) 2 i=1 Joonpyo Kim Ch7 June 24, / 63

16 Likelihood Ratio Test Example. (Cont d) and so 2(l(ˆθ Ω ) l(ˆθ Ω 0 )) = n log Therefore the rejection region of the test is ( x µ 0 s/ n c 1 + (x µ 0) 2 ˆσ 2Ω where s is a sample standard deviation, and from sample distribution theory, we can easily obtain c = t α/2 (n 1). ). Joonpyo Kim Ch7 June 24, / 63

17 Likelihood Ratio Test Example. Let X 1,, X n N(µ, σ 2 ), < µ <, and 0 < σ 2 <. Consider the hypotheses H 0 : µ µ 0, H 1 : µ > µ 0. Our goal is to describe likelihood ratio test with level α. (0 < α < 1) For θ = (µ, σ 2 ) t, we can easily show that (exercise) ( ˆθ Ω = x, 1 n ( ˆθ Ω 0 = x µ 0, 1 n ) t n (x i x) 2 i=1 ) t n (x i ˆµ Ω 0 ) 2 i=1 Joonpyo Kim Ch7 June 24, / 63

18 Likelihood Ratio Test Example. (Cont d) and so 2(l(ˆθ Ω ) l(ˆθ Ω 0 )) = n log Therefore the rejection region of the test is ( x x µ 0 s/ n c 1 + (x x µ 0) 2 ˆσ 2Ω where s is a sample standard deviation. Note that x x µ 0 0. So if c 0, α = 1 and it is unreasonable. Thus we can assume that c > 0 and then rejection region is x µ 0 s/ n c. ). Joonpyo Kim Ch7 June 24, / 63

19 Likelihood Ratio Test Example. (Cont d) Meanwhile we know that ( ) ( X µ0 P θ S/ n c Z + (µ µ 0 )/(σ/ ) n) = P θ c V/(n 1) ( = P θ c V/(n 1) Z µ µ ) 0 σ/ n ( ) X µ0 P (µ0,σ 2 ) S/ n c for θ Ω 0 so we can obtain the test with level α if c = t α (n 1). Joonpyo Kim Ch7 June 24, / 63

20 Likelihood Ratio Test Example. Consider X 1,, X n Exp(θ) and hypotheses H 0 : θ = θ 0 vs H 1 : θ θ 0. We want to find the test under level α. (0 < α < 1) Log-likelihood is l(θ) = n log θ n x θ and hence ( 2(l(ˆθ Ω ) l(ˆθ Ω 0 x )) = 2n 1 log x ) c θ 0 θ 0 is rejection region. Joonpyo Kim Ch7 June 24, / 63

21 Likelihood Ratio Test Example. (Cont d) It is equivalent to x x c 1, c 2 θ 0 θ 0 c 1 log c 1 = c 2 log c 2 Figure: Graph of y = x 1 log x. Joonpyo Kim Ch7 June 24, / 63

22 Likelihood Ratio Test Example. (Cont d) So rejection region is X X c 1, c 2 θ 0 θ 0 c 1 log c 1 = c 2 log c 2 2nc2 2nc 1 pdf χ 2 (2n)(t)dt = 1 α because 2nX θ 0 χ 2 (2n). Joonpyo Kim Ch7 June 24, / 63

23 Likelihood Ratio Test Example. Let X 11,, X 1n1 N(µ 1, σ 2 ) and X 21,, X 2n2 N(µ 2, σ 2 ) be random samples. Consider hypotheses H 0 : µ 1 = µ 2 vs. H 1 : µ 1 µ 2. We would find a likelihood ratio test under level α. Let θ = (µ 1, µ 2, σ 2 ) t. Joonpyo Kim Ch7 June 24, / 63

24 Likelihood Ratio Test Example. (Cont d) Then log-likelihood is l(θ) = 1 2σ 2 2 n i i=1 j=1 (x ij µ i ) 2 n 1 + n 2 2 log(2πσ 2 ) and where ˆθ Ω 1 = X 1, X 2, n 1 + n 2 ˆθ Ω 0 1 = ˆµ 0, ˆµ 0, n 1 + n 2 2 n i (x ij x i ) 2 i=1 j=1 2 n i (x ij ˆµ 0 ) 2 i=1 j=1 ˆµ 0 = n 1X 1 + n 2 X 2 n 1 + n 2. t t Joonpyo Kim Ch7 June 24, / 63

25 Likelihood Ratio Test Example. (Cont d) From this and sample distribution theory, rejection region is X 1 X S n1 n2 p t α/2(n 1 + n 2 2). Joonpyo Kim Ch7 June 24, / 63

26 Likelihood Ratio Test Example. Similarly, we can test H 0 : σ 2 1 = σ 2 2 vs H 1 : σ 2 1 σ 2 2 under level α with X 11,, X 1n1 N(µ 1, σ 2 1 ) and X 21,, X 2n2 N(µ 2, σ2 2 ). It is left as exercise. Joonpyo Kim Ch7 June 24, / 63

27 Optimality of Testing In this section, optimal testing will be introduced. In testing, type I error probability is controlled via significance level, so type II error probability may determine the performance of test. Definition. Consider simple hypotheses H 0 : θ = θ 0 vs H 1 : θ = θ 1. Then test φ satisfying (i) E θ0 φ (X) α (ii) E θ1 φ (X) E θ1 φ(x) for any test φ such that Eθ0 φ(x) α is called Most Powerful test (MP test) at level α. Joonpyo Kim Ch7 June 24, / 63

28 Optimality of Testing Definition. Consider hypotheses H 0 : θ Ω 0 vs H 1 : θ Ω 1 where Ω = Ω 0 Ω 1 and Ω 0 Ω 1 = φ. Then test φ satisfying (i) sup θ Ω0 E θ φ (X) α (ii) E θ φ (X) E θ φ(x) θ Ω1 for any test φ such that sup θ Ω0 E θ φ(x) α is called Uniformly Most Powerful test (UMP test) at level α. Joonpyo Kim Ch7 June 24, / 63

29 Optimality of Testing Remark. Obviously, our interest is how to find MP and UMP test. Which test has maximum power among level α tests? Neyman-Pearson lemma gives the answer in simple vs simple case. Joonpyo Kim Ch7 June 24, / 63

30 Neyman-Pearson Lemma Theorem. Neyman-Pearson Lemma. Suppose that X p θ (x), where θ = θ 0 or θ 1. Consider testing problem of hypotheses H 0 : θ = θ 0 vs H 1 : θ = θ 1. Then ϕ satisfying 1 p θ1 (x)/p θ0 (x) > k (i) ϕ (x) = γ p θ1 (x)/p θ0 (x) = k 0 p θ1 (x)/p θ0 (x) < k (ii) E θ0 ϕ (X) = α Then ϕ is MP test at level α. Furthermore, for 0 < α < 1 such ϕ (i.e., such k > 0 and 0 < γ < 1) always exist. Joonpyo Kim Ch7 June 24, / 63

31 Neyman-Pearson Lemma Proof. Existence part will not be proved here. For the rest part, consider E θ1 (ϕ (X) ϕ(x)) ke θ0 (ϕ (X) ϕ(x)) for level α test ϕ. Define X 0 = {x : p θ0 (x) > 0} and X 1 = {x : p θ1 (x) > 0}. Note that E θ1 (ϕ (X) ϕ(x)) ke θ0 (ϕ (X) ϕ(x)) = (ϕ (x) ϕ(x))(p θ1 (x) kp θ0 (x))dx X 1 + k(ϕ (x) ϕ(x))p θ0 (x)dx X 0 \X 1 Joonpyo Kim Ch7 June 24, / 63

32 Neyman-Pearson Lemma Proof. (Cont d) so E θ1 (ϕ (X) ϕ(x)) ke θ0 (ϕ (X) ϕ(x)) 0. Also note that E θ0 (ϕ (X) ϕ(x)) = α E θ0 ϕ(x) 0. Therefore, E θ1 (ϕ (X) ϕ(x)) 0. Joonpyo Kim Ch7 June 24, / 63

33 Example: MP test Example. Let X 1,, X n be random sample from N(µ, σ 2 ), where σ 2 > 0 is known. Consider testing hypotheses H 0 : µ = µ 0 vs H 1 : µ = µ 1 where µ 0 and µ 1 are known. Also assume that µ 0 < µ 1. Then, likelihood ratio is so test p µ1 (x) ( p µ0 (x) = exp n ) 2σ 2 (µ 1 µ 0 )(µ 1 + µ 0 2x) ϕ (x) = { 1 x > k 0 x < k with E µ0 ϕ (X) = α is MP at level α (0 < α < 1). It can be easily shown that k = µ 0 + z α/2 σ/ n. Joonpyo Kim Ch7 June 24, / 63

34 Example: UMP test Example. Continued from previous example. (i) First, consider H 0S : µ = µ 0 vs H 1 : µ > µ 0. For any alternative H 1 (µ 1 ) : µ = µ 1 for µ 1 > µ 0, ϕ is MP at level α for testing H 0S vs H 1 (µ 1 ). In addition, ϕ does not depend on µ 1, so for any level α test ϕ, E µ ϕ (X) E µ ϕ(x) µ > µ 0 holds. Therefore, ϕ is UMP at level α for testing H 0S vs H 1. Joonpyo Kim Ch7 June 24, / 63

35 Example: UMP test Example. (Cont d) (ii) Next, consider Note that H 0C : µ µ 0 vs H 1 : µ > µ 0. sup E µ ϕ (X) = E µ0 ϕ (X) = α µ µ 0 so ϕ is also level α for testing H 0C vs H 1. Since ϕ was the best among level α tests for H 0S, so it is also the best among level α tests for H 0C. Therefore, ϕ is UMP at level α for testing H 0C vs H 1. Joonpyo Kim Ch7 June 24, / 63

36 Example: UMP test Remark. It is well known that, UMP test for H 0 : µ = µ 0 vs H 1 : µ µ 0 does not exist. For another optimality, we find the best test among special class of tests, which is called unbiased test, and such optimal one is Uniformly Most Powerful Unbiased test. Joonpyo Kim Ch7 June 24, / 63

37 Asymptotics of LRT In this section we see an asymptotic test. First recall that n(ˆθ θ) = [I(θ)] 1 n l(θ) + o p (1). Then by Taylor s theorem, there exists θ such that and by LLN holds under H 0. l(θ 0 ) = l(ˆθ) + l(ˆθ) t (θ 0 ˆθ) (θ 0 ˆθ) t l(θ )(θ 0 ˆθ) 1 n l(θ ) p n I(θ 0) Joonpyo Kim Ch7 June 24, / 63

38 Asymptotics of LRT Further under H 0 l(ˆθ) l(θ 0 ). Thus likelihood ratio test statistics of is H 0 : θ = θ 0 vs H 1 : θ θ 0 2(l(ˆθ) l(θ 0 )) n(ˆθ θ 0 ) t I(θ 0 ) n(ˆθ θ 0 ) d n χ2 (k). Joonpyo Kim Ch7 June 24, / 63

39 Asymptotics of LRT Theorem. Let θ Ω R k and consider hypotheses Under regularity conditions, Under H 0, H 0 : θ = θ 0 vs H 1 : θ θ 0. 2(l(ˆθ) l(θ 0 )) d n χ2 (k). Joonpyo Kim Ch7 June 24, / 63

40 Asymptotics of LRT Theorem. (Cont d) Under H 0, 2(l(ˆθ) l(θ 0 )) = n(ˆθ θ 0 ) t I(θ 0 )(ˆθ θ 0 ) + o p (1) =: W n (θ 0 ) + o p (1). W n (θ 0 ) is called Wald s statistics. Under H 0, 2(l(ˆθ) l(θ 0 )) = n l(θ 0 ) t [I(θ 0 )] 1 l(θ0 ) + o p (1) =: R n (θ 0 ) + o p (1). R n (θ 0 ) is called Rao s statistics. Joonpyo Kim Ch7 June 24, / 63

41 Asymptotics of LRT Theorem. Let Ω R k be an open set. Let θ = (ξ t, η t ) t. Consider Ω 0 = {(ξ t 0, ηt ) t η R k 0 } and hypotheses H 0 : ξ = ξ 0 vs H 1 : ξ ξ 0. Then under H 0, 2(l(ˆθ Ω ) l(ˆθ Ω 0 )) d n χ2 (k k 0 ). 2(l(ˆθ Ω ) l(ˆθ Ω 0 )) = n(ˆθ Ω ˆθ Ω 0 ) t I(θ 0 )(ˆθ Ω ˆθ Ω 0 ) + o p (1). 2(l(ˆθ Ω ) l(ˆθ Ω 0 )) = n l(ˆθ Ω 0 ) t [I(θ 0 )] 1 l(ˆθω 0 ) + o p (1). Joonpyo Kim Ch7 June 24, / 63

42 Asymptotics of LRT Example. Consider X 1,, X n Exp(θ) and H 0 : θ = θ 0 vs H 1 : θ θ 0. Then W n (θ 0 ) = R n (θ 0 ) = n(x θ 0) 2. θ 2 0 Joonpyo Kim Ch7 June 24, / 63

43 Asymptotics of LRT Example. Consider X 1,, X n Multi(1, p) and H 0 : p = p 0 vs H 1 : p p 0 for p 0 = (p 01,, p 0k ) t, p p 0k = 1. Let X i = (X i1,, X ik ) t and X j = n i=1 X ij. Define θ = (p 1,, p k 1 ) t. Then log-likelihood is l(θ) = k k 1 x j log p j = x j log p j + x k log(1 p 1 p k 1 ) j=1 and we can easily show that j=1 ˆp j Ω = x j n. Joonpyo Kim Ch7 June 24, / 63

44 Asymptotics of LRT Example. (Cont d) Also we can show that p p I(θ) = t p k 0 0 p k 1 so W n (θ 0 ) = k j=1 (X j np 0j ) 2 np 0j. Joonpyo Kim Ch7 June 24, / 63

45 Asymptotics of LRT Example. (Cont d) Similarly we can show that R n (θ 0 ) = k j=1 In here the fact that for vectors v, w may be used. (I + vw t ) 1 = I (X j np 0j ) 2 np 0j w t v vwt Joonpyo Kim Ch7 June 24, / 63

46 Asymptotics of LRT Example. Consider a random sample X 1,, X n Bin(n, p) and H 0 : p = p 0, vs H 1 : p p 0. We already know that asymptotic test with size α has a rejection region X np 0 np0 q 0 z α/2. Joonpyo Kim Ch7 June 24, / 63

47 Asymptotics of LRT Example. (Cont d) It is equivalent to (S np 0 ) 2 np 0 + (F nq 0) 2 nq 0 = (X np 0) 2 np 0 + (X np 0) 2 = (X np 0) 2 np 0 q 0 nq 0 χ 2 α(1) where S and F are the number of success and fail respectively. So test statistics can be denoted as 2 (O c Ê0 c ) 2. c=1 Ê 0 c Joonpyo Kim Ch7 June 24, / 63

48 Asymptotics of LRT Example. Similarly, we can test hypotheses H 0 : p ij = p i p j vs H 1 : else on the r c contingency table with model (X ij ) Multi(n, (p ij )), r c p ij = 1 i=1 j=1 and its test statistics is r i=1 j=1 For this show the textbook. c (O ij Ê0 ij )2 Ê 0 ij χ 2 ((r 1)(c 1)). H 0 Joonpyo Kim Ch7 June 24, / 63

49 Asymptotics of LRT Example. Now we see some other case. Suppose that for i = 1, 2,, k, X i1,, X ini Ber(p i ). Our goal is to test H 0 : p 1 = = p k vs H 1 : not H 0. First it s easy to show that 2[l(ˆp) l(ˆp 0 )] = 2 k i=1 ( n i ˆp i log ˆp i ˆp 0 i + n iˆq i log ˆq ) i ˆq i 0 where MLE ˆp i and MLE under H 0 ˆp 0 i ˆp i = 1 n i n i j=1 X ij, ˆp 0 i = 1 n are defined as k n i X ij, n = i=1 j=1 k n i. i=1 Joonpyo Kim Ch7 June 24, / 63

50 Asymptotics of LRT Example. (Cont d) Now let X i = n i j=1 X ij and consider the situation that n i, n i n n k γ i, 0 < γ i < 1, i = 1, 2,, k. (1) Then under H 0 if we define p 1 = p 2 = = p k =: p 0. then ˆp i p 0 and ˆp 0 i p0 with assumption (1) so ˆp i /ˆp 0 i 1 and hence the approximation Joonpyo Kim Ch7 June 24, / 63

51 Asymptotics of LRT 2[l(ˆp) l(ˆp 0 )] = 2 k ( n i ˆp i log ˆp i ˆp 0 i i=1 ( k 2 n i ˆp 0 i i=1 k = 2 i=1 k = + n i ˆq i log ˆq ) i ˆq i 0 ( ˆp i ˆp ( ) ) 2 ˆpi i 2 ˆp n i ˆq i 0 i n i ( ˆp i ˆp 0 i + ˆp2 i 2ˆp 0 i + ˆq2 i ˆq 0 i ) 1 = ˆp i ˆp0 i + ˆq i ˆq 0 i + ˆq2 i 2ˆq 0 i k ( ˆq i ˆq i ( ) )) 2 ˆqi 2 ˆq i 0 1 ) ˆq i ˆq0 i i=1 ( ˆp 2 n i i ˆp 0 i i=1 ˆq i 0 n ˆp2 i ˆp0 i ˆq0 i + ˆp0 i ˆq2 i i ˆp 0 i ˆq0 i k ˆq i 0 = i n2 i ˆp0 i ˆq0 i + ˆp0 i (n i X i ) 2 n i=1 i ˆp 0 i ˆq0 i k (1 ˆp 0 i = i n2 i ˆp0 i (1 ˆp0 i ) + ˆp0 i (n2 i 2n ix i + Xi ) 2 n i=1 i ˆp 0 i ˆq0 i 2 k = X i nˆp 0 i i=1 n i ˆp 0 i ˆq0 i Joonpyo Kim Ch7 June 24, / 63

52 Asymptotics of LRT Example. (Cont d) might be used. In here the fact that (1 + x) log(1 + x) x x2 if x 0 used. Now define Z = (Z 1,, Z k ) t and w = (w 1,, w k ) t where Z i := n i X i p 0 p 0 q 0, w i := n i. Then Joonpyo Kim Ch7 June 24, / 63

53 Asymptotics of LRT Example. (Cont d) k X i nˆp 0 i n i ˆp 0 i ˆq0 i i=1 2 = = k i=1 k i=1 ( ) 2 X i ˆp 0 i ni p 0 q 0 ( ) 2 X i p 0 ni p 0 q X p 0 n i 0 p 0 q 0 k (Z i Y i ) 2 i=1 where Y = w(w t w) 1 w t Z. Joonpyo Kim Ch7 June 24, / 63

54 Asymptotics of LRT Example. (Cont d) Therefore k X i nˆp 0 i n i ˆp 0 i ˆq0 i i=1 2 Z t (I w(w t w) 1 w t )Z χ 2 (k 1) holds and using this fact we can find an asymptotic test of given hypotheses. Joonpyo Kim Ch7 June 24, / 63

55 Confidence Region Definition. Suppose that statistics S(X) based on X satisfies inf P θ(q(θ) S(X)) 1 α. θ Θ Then S(X) is called confidence region for q(θ) of confidence level 1 α. Joonpyo Kim Ch7 June 24, / 63

56 Duality Theorem. Let A(q(θ 0 )) be an acceptance region of level α for testing H 0 (θ 0 ) : q(θ) = q(θ 0 ) for fixed but arbitrary θ 0 Θ. Then S(X) defined by q(θ) S(X) x A(q(θ)) is a level (1 α) confidence region of q(θ). Proof. For any θ 0 Θ, P θ0 (q(θ 0 ) S(X)) = P θ0 (X A(q(θ 0 ))) 1 α holds. Joonpyo Kim Ch7 June 24, / 63

57 Confidence Region Example. Consider a random sample X 1,, X n N(µ, σ 2 ) and hypotheses Then acceptance region of LRT is A(µ 0 ) = H 0 : µ = µ 0 vs H 1 : µ µ 0. { x : x µ 0 s/ n } t α/2(n 1) so [ x t α/2 (n 1) s, x + t n α/2 (n 1) s ] n is a level 1 α confidence region for µ. Joonpyo Kim Ch7 June 24, / 63

58 Confidence Region Example. (Cont d) Now consider H 0 : µ = µ 0 vs H 1 : µ > µ 0. Then acceptance region of LRT is A(µ 0 ) = { x : x µ } 0 s/ n t α(n 1) and from x A(µ 0 ) µ 0 [ x t α (n 1) s ), n we get the (one-sided) confidence interval for µ [ x t α (n 1) s ),. n Joonpyo Kim Ch7 June 24, / 63

59 Simultaneous Confidence Region Example. Bonferroni approach. Consider a linear regression model y i = β 0 + x i1 β x ik β k + ɛ i, i = 1, 2,, n. Suppose that we carry out the k tests of H 0j : β j = 0, j = 1, 2,, k. Let E j be the event that the jth test rejects H 0j even if it is true, where P (E j ) = α j. The overall size α f can be defined as k α f = P (reject at least one H 0j when all H 0j are true) = P Note that if α j = 0.05 for all j, actual size of this simultaneous testing is 1 (1 0.05) k, which is not small enough. For example, if k = 10, α f 0.40, which makes our testing procedure unreliable. j=1 E j. Joonpyo Kim Ch7 June 24, / 63

60 Simultaneous Confidence Region Example. (Cont d) To overcome this problem, if α is desired overall error rate, using k P j=1 E j k P (E j ) j=1 k α j, making k j=1 α j = α, we obtain α f α. Our choice is α j = α/k. Such method is called Bonferroni correction. Using Bonferroni correction, we get the simultaneous confidence interval, j=1 β j ( ˆβ j ± t α/2k (n k 1)s g jj ) j, where g jj = V ar( ˆβ j )/σ 2. Joonpyo Kim Ch7 June 24, / 63

61 Simultaneous Confidence Region Example. Scheffe s method. Now consider the null H 0 : a t β = 0 a R p. In the same way, denoting E a = {reject H 0 (a) : a t β = 0 when H 0 (a) is true}, we should make ( ) P a R p E a = α. Joonpyo Kim Ch7 June 24, / 63

62 Simultaneous Confidence Region Example. (Cont d) Note that, if we use F a = (at ˆβ) t (a t (X t X) 1 a) 1 a t ˆβ ˆσ 2 = (at ˆβ) t (a t ˆβ) s 2 a t (X t X) 1 a as a test statistic for H 0 (a), we get P ( a R p E a ) = P ( max a (a t ˆβ a t β) 2 a t Sa c ), (2) where S 2 = ˆσ 2 (X t X) 1. It s known that max a (a t ˆβ a t β) 2 a t Sa = (ˆβ β) t S 1 (ˆβ β), and hence we can find suitable c that makes (2) equals α. Joonpyo Kim Ch7 June 24, / 63

63 Simultaneous Confidence Region Example. (Cont d) Therefore, we get the simultaneous confidence intervals for all possible linear functions a t ˆβ as a t β ( ) a t ˆβ ± s (k + 1)Fα (k + 1, n k 1)a t (X t X) 1 a a R k. Such method is called Scheffe s method. Joonpyo Kim Ch7 June 24, / 63

Summary of Chapters 7-9

Summary of Chapters 7-9 Chapter 7. Interval Estimation 7.2. Confidence Intervals for Difference of Two Means Let X 1,, X n and Y 1, Y 2,, Y m be two independent random samples of sizes n and m from two