370 13 Testing Hypotheses 13.8 Applications of LR Tests: Contingency Tables, Goodness-of-Fit Tests Now we turn to a slightly different testing hypotheses problem, where the LR is also appropriate. We consider an r. experiment which may result in k possibly different outcomes denoted by O j , j = 1, , k. In n independent repetitions of the experiment, let p j be the (constant) probability that each one of the trials will result in the outcome O j and denote by X j the number of trials which result in O j , j = 1, , k. Then the joint distribution of the X’s is the Multinomial distribution, that is, PX x X x n xx pp kk k x k x k 11 1 1 1 == () = ⋅⋅⋅ ⋅⋅⋅, , ! !! , where x j ≥ 0, j = 1, , k, Σ k j = 1 x j = n and ΩΩθθ== () ′ >= = ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ = ∑ pppj kp kj j j k 1 1 01 1, , ; , , , , . We may suspect that the p’s have certain specified values; for example, in the case of a die, the die may be balanced. We then formulate this as a hypothesis and proceed to test it on the basis of the data. More generally, we may want to test the hypothesis that θ lies in a subset ωω ωω ω of ΩΩ ΩΩ Ω. Consider the case that H : θθ θθ θ∈ ωω ωω ω= { θθ θθ θ 0 } = {(p 10 , , p k0 )′}. Then, under ωω ωω ω, L n xx pp k x k x k ˆ ! !! ,ωω () = ⋅⋅⋅ ⋅⋅⋅ 1 10 0 1 while, under ΩΩ ΩΩ Ω, L n xx pp k x k x k ˆ ! !! ˆˆ ,ΩΩ () = ⋅⋅⋅ ⋅⋅⋅ 1 1 1 where p ˆ j = x j /n are the MLE’s of p j , j = 1, , k (see Example 11, Chapter 12). Therefore λ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ∏ n p x n j j j k x j 0 1 and H is rejected if −2 log λ > C. The constant C is determined by the fact that −2 log λ is asymptotically χ 2 k−1 distributed under H, as it can be shown on the basis of Theorem 6, and the desired level of significance α . Now consider r events A i , i = 1, , r which form a partition of the sample space S and let {B j , j = 1, . . . , s} be another partition of S. Let p ij = P(A i ∩ B j ) and let pppp iijjij i r j s ,.== == ∑∑ 11 13.3 UMP Tests for Testing Certain Composite Hypotheses 371 Then, clearly, p i. = P(A i ), p .j = P(B j ) and pp p ij ij j s i r j s i r .== = ==== ∑∑∑∑ 1 1111 Furthermore, the events {A 1 , , A r } and {B 1 , , B s } are independent if and only if p ij = p i. p .j , i = 1, . . . , r, j = 1, , s. A situation where this set-up is appropriate is the following: Certain experimental units are classified according to two characteristics denoted by A and B and let A 1 , , A r be the r levels of A and B 1 , , B r be the J levels of B. For instance, A may stand for gender and A 1 , A 2 for male and female, and B may denote educational status comprising the levels B 1 (el- ementary school graduate), B 2 (high school graduate), B 3 (college graduate), B 4 (beyond). We may think of the rs events A i ∩ B j being arranged in an r × s rectangular array which is known as a contingency table; the event A i ∩ B j is called the (i, j)th cell. Again consider n experimental units classified according to the character- istics A and B and let X ij be the number of those falling into the (i, j)th cell. We set XX XX iij jij i r j s .== == ∑∑ and 11 It is then clear that XXn ij j s i r .== == ∑∑ 11 Let θθ θθ θ= (p ij , i = 1, . . . , r, j = 1, , s)′. Then the set ΩΩ ΩΩ Ω of all possible values of θθ θθ θ is an (rs −1)-dimensional hyperplane in ޒ rs . Namely, ΩΩ ΩΩ Ω= { θθ θθ θ= (p ij , i = 1, , r, j = 1, , s)′∈ ޒ rs ; p ij > 0, i = 1, , r, j = 1, , s, Σ r i = 1 Σ s j=1 p ij = 1}. Under the above set-up, the problem of interest is that of testing whether the characteristics A and B are independent. That is, we want to test the existence of probabilities p i , q j , i = 1, . . . , r, j = 1, . . . , s such that H : p ij = p i q j , i = 1, , r, j = 1, , s. Since for i = 1, . . . , r − 1 and j = 1, , s − 1 we have the r + s − 2 independent linear relationships pp pq ij i ij j i r j s == == ∑∑ ,, 11 it follows that the set ωω ωω ω, specified by H, is an (r + s − 2)-dimensional subset of ΩΩ ΩΩ Ω. Next, if x ij is the observed value of X ij and if we set xxxx iijjij i r j s ,,== == ∑∑ 11 13.8 Applications of LR Tests: Contingency Tables, Goodness-of-Fit Tests 371 372 13 Testing Hypotheses the likelihood function takes the following forms under ΩΩ ΩΩ Ω and ωω ωω ω, respectively. Writing Π i,j instead of Π r i=1 Π s j = 1 , we have L n x p L n x pq n x pq n x pq ij ij ij x ij ij ij ij x ij ij ij i x j x ij ij ij i x i j x j ij ij ij ij i j ΩΩ ωω () = () = () == ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ∏ ∏ ∏∏ ∏ ∏ ∏∏∏ ! ! , ! ! ! ! ! ! , , ,, , , , . . since pq pq pq q pqq pq i x j x i x j x i xx i s x jiij i x i x s x i i x i j x j ij ij ij ij ii is i i is i j = = ⋅⋅⋅ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅⋅⋅ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⋅ ⋅⋅ ⋅ ∏∏∏∏ ∏∏ ∏∏ ! , . 1 1 1 Now the MLE’s of p ij , p i and q i are, under ΩΩ ΩΩ Ω and ωω ωω ω, respectively, ˆ , ˆ , ˆ , ,, . , . p x n p x n q x n ij ij i i j j W === ωωωω as is easily seen (see also Exercise 13.8.1). Therefore L n x x n L n x x n x n ij ij ij ij x ij ij i x i j x j ij i j ˆ ! ! , ˆ ! ! , , , . . . . ΩΩωω () = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ () = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ∏ ∏ ∏ ∏∏ and hence λ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ∏∏ ∏ ∏∏ ∏ ⋅ ⋅ ⋅ x n x n x n xx nx i x i j x j ij x ij i x i j x j n ij x ij i j ij i j ij . . , . , . . . It can be shown that the (unspecified) assumptions of Theorem 6 are fulfilled in the present case and therefore −2log λ is asymptotically χ 2 f , under ωω ωω ω, where f = (rs − 1) − (r + s − 2) = (r − 1)(s − 1) according to Theorem 6. Hence the test for H can be carried out explicitly. Now in a multinomial situation, as described at the beginning of this section and in connection with the estimation problem, it was seen (see Section 12.9, Chapter 12) that certain chi-square statistics were appropriate, in a sense. Recall that χ 2 2 1 = − () = ∑ Xnp np jj j j k . 13.3 UMP Tests for Testing Certain Composite Hypotheses 373 This χ 2 r.v. can be used for testing the hypothesis Hpp k : , , ,θθωωθθ∈= {} = () ′ ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ 010 0 where θθ θθ θ= (p 1 , , p k )′. That is, we consider χ ωω 2 0 2 0 1 = − () = ∑ xnp np jj j j k and reject H if χ 2 ωω ωω ω is too large, in the sense of being greater than a certain constant C which is specified by the desired level of the test. It can further be shown that, under ωω ωω ω, χ 2 ωω ωω ω is asymptotically distributed as χ 2 k−1 . In fact, the present test is asymptotically equivalent to the test based on −2log λ . For the case of contingency tables and the problem of testing indepen- dence there, we have χ ωω 2 2 = − () ∑ xnpq np q ij i j ij ij, , where ωω ωω ω is as in the previous case in connection with the contingency tables. However, χ 2 ωω ωω ω is not a statistic since it involves the parameters p i , q j . By replac- ing them by their MLE’s, we obtain the statistic χ ˆ ,, ,, , ˆˆ ˆˆ . ωω ωωωω ωωωω 2 2 = − () ∑ xnpp np q ij i j ij ij By means of χ ˆ ωω 2 , one can test H by rejecting it whenever χ ˆ ωω 2 > C. The constant C is to be determined by the significance level and the fact that the asymptotic distribution of χ ˆ ωω 2 , under ωω ωω ω, is χ 2 f with f = (r − 1)(s − 1), as can be shown. Once more this test is asymptotically equivalent to the corresponding test based on −2log λ . Tests based on chi-square statistics are known as chi-square tests or goodness-of-fit tests for obvious reasons. Exercises 13.8.1 Show that ˆ , p ij x ij n ΩΩ = , ˆ , . p i x i n ωω = , ˆ , q j x n j ωω = ⋅ as claimed in the discussion in this section. In Exercises 13.8.2–13.8.9 below, the test to be used will be the appropriate χ 2 test. 13.8.2 Refer to Exercise 13.7.2 and test the hypothesis formulated there at the specified level of significance by using a χ 2 -goodness-of-fit test. Also, compare the cut-off point with that found in Exercise 13.7.2(i). Exercises 373 374 13 Testing Hypotheses 13.8.3 A die is cast 600 times and the numbers 1 through 6 appear with the frequencies recorded below. 12 3 4 5 6 100 94 103 89 110 104 At the level of significance α = 0.1, test the fairness of the die. 13.8.4 In a certain genetic experiment, two different varieties of a certain species are crossed and a specific characteristic of the offspring can only occur at three levels A, B and C, say. According to a proposed model, the probabili- ties for A, B and C are 1 12 , 3 12 and 8 12 , respectively. Out of 60 offsprings, 6, 18, and 36 fall into levels A, B and C, respectively. Test the validity of the proposed model at the level of significance α = 0.05. 13.8.5 Course work grades are often assumed to be normally distributed. In a certain class, suppose that letter grades are given in the following manner: A for grades in [90, 100], B for grades in [75, 89], C for grades in [60, 74], D for grades in [50, 59] and F for grades in [0, 49]. Use the data given below to check the assumption that the data is coming from an N(75, 9 2 ) distribution. For this purpose, employ the appropriate χ 2 test and take α = 0.05. ABCDF 3121041 13.8.6 It is often assumed that I.Q. scores of human beings are normally distributed. Test this claim for the data given below by choosing appropriately the Normal distribution and taking α = 0.05. x ≤ 90 90 < x ≤ 100 100 < x ≤ 110 110 < x ≤ 120 120 < x ≤ 130 x > 130 10 18 23 22 18 9 (Hint: Estimate μ and σ 2 from the grouped data; take the midpoints for the finite intervals and the points 65 and 160 for the leftmost and rightmost intervals, respectively.) 13.8.7 Consider a group of 100 people living and working under very similar conditions. Half of them are given a preventive shot against a certain disease and the other half serve as control. Of those who received the treatment, 40 did not contract the disease whereas the remaining 10 did so. Of those not treated, 30 did contract the disease and the remaining 20 did not. Test effec- tiveness of the vaccine at the level of significance α = 0.05. 13.3 UMP Tests for Testing Certain Composite Hypotheses 375 13.8.8 On the basis of the following scores, appropriately taken, test whether there are gender-associated differences in mathematical ability (as is often claimed!). Take α = 0.05. Boys: 80 96 98 87 75 83 70 92 97 82 Girls: 82 90 84 70 80 97 76 90 88 86 (Hint: Group the grades into the following six intervals: [70, 75), [75, 80), [80, 85), [85, 90), [90, 100).) 13.8.9 From each of four political wards of a city with approximately the same number of voters, 100 voters were chosen at random and their opinions were asked regarding a certain legislative proposal. On the basis of the data given below, test whether the fractions of voters favoring the legislative pro- posal under consideration differ in the four wards. Take α = 0.05. WARD Totals 1234 Favor Proposal 37 29 32 21 119 Do not favor proposal 63 71 68 79 281 Totals 100 100 100 100 400 13.8.10 Let X 1 , , X n be independent r.v.’s with p.d.f. f(·; θθ θθ θ), θθ θθ θ∈ ΩΩ ΩΩ Ω⊆ ޒ r . For testing a hypothesis H against an alternative A at level of significance α , a test φ is said to be consistent if its power β φ , evaluated at any fixed θθ θθ θ∈ ΩΩ ΩΩ Ω, converges to 1 as n →∞. Refer to the previous exercises and find at least one test which enjoys the property of consistency. Specifically, check whether the consistency property is satisfied with regards to Exercises 13.2.3 and 13.3.2. 13.9 Decision-Theoretic Viewpoint of Testing Hypotheses For the definition of a decision, loss and risk function, the reader is referred to Section 6, Chapter 12. Let X 1 , , X n be i.i.d. r.v.’s with p.d.f. f(·; θθ θθ θ), θθ θθ θ∈ ΩΩ ΩΩ Ω⊆ ޒ r , and let ωω ωω ω be a (measurable) subset of ΩΩ ΩΩ Ω. Then the hypothesis to be tested is H : θθ θθ θ∈ ωω ωω ω against the alternative A : θθ θθ θ∈ ωω ωω ω c . Let B be a critical region. Then by setting z = (x 1 , , x n )′, in the present context a non-randomized decision function δ = δ (z) is defined as follows: δ z z () = ∈ ⎧ ⎨ ⎩ 1 0 , , if otherwise. B 13.9 Decision-Theoretic Viewpoint of Testing Hypotheses 375 376 13 Testing Hypotheses We shall confine ourselves to non-randomized decision functions only. Also an appropriate loss function, corresponding to δ , is of the following form: LL L c c θθ θθωωθθωω θθωω θθωω ; ,,. , ,, if and or and if and if and δ δδ δ δ () = ∈=∈= ∈= ∈= ⎧ ⎨ ⎪ ⎩ ⎪ 001 1 0 1 2 where L 1 , L 2 > 0. Clearly, a decision function in the present framework is simply a test function. The notation φ instead of δ could be used if one wished. By setting Z = (X 1 , , X n )′, the corresponding risk function is R θθθθθθΖΖ θθθθ ;; ; , δ () = () ∈ () + () ∈ () LPBLPB c 10Z or R if if θθ θθωω θθωω θθ θθ ; , ,. δ () = ∈ () ∈ ∈ () ∈ ⎧ ⎨ ⎪ ⎩ ⎪ LP B LP B cc 1 2 Z Z (44) In particular, if ωω ωω ω= { θθ θθ θ 0 }, ωω ωω ω c = { θθ θθ θ 1 } and P θθ θθ θ 0 (Z ∈ B) = α , P θθ θθ θ 1 (Z ∈ B) = β , we have R if if θθ θθθθ θθθθ ; , ,. δ α β () = = − () = ⎧ ⎨ ⎪ ⎩ ⎪ L L 10 21 1 (45) As in the point estimation case, we would like to determine a decision function δ for which the corresponding risk would be uniformly (in θθ θθ θ) smaller than the risk corresponding to any other decision function δ *. Since this is not feasible, except for trivial cases, we are led to minimax decision and Bayes decision functions corresponding to a given prior p.d.f. on ΩΩ ΩΩ Ω. Thus in the case that ωω ωω ω= { θθ θθ θ 0 } and ωω ωω ω c = { θθ θθ θ 1 }, δ is minimax if max R R max R R θθθθθθθθ 01 0 1 ;, ; ;*, ;* δδ δ δ ()() [] ≤ ()() [] for any other decision function δ *. Regarding the existence of minimax decision functions, we have the result below. The r.v.’s X 1 , , X n is a sample whose p.d.f. is either f(·; θθ θθ θ 0 ) or else f(·; θθ θθ θ 1 ). By setting f 0 = f(·; θθ θθ θ 0 ) and f 1 = f(·; θθ θθ θ 1 ), we have Let X 1 , , X n be i.i.d. r.v.’s with p.d.f. f(·; θθ θθ θ), θθ θθ θ∈ ΩΩ ΩΩ Ω= { θθ θθ θ 0 , θθ θθ θ 1 }. We are interested in testing the hypothesis H : θθ θθ θ= θθ θθ θ 0 against the alternative A : θθ θθ θ= θθ θθ θ 1 at level α . Define the subset B of ޒ n as follows: B = {z ∈ ޒ n ; f(z; θθ θθ θ 1 ) > Cf(z; θθ θθ θ 0 )} and assume that there is a determination of the constant C such that LP B LP B c 12 01 01 θθθθ θθθθZZ∈ () =∈ () () = () () equivalently, R R ;;. δδ (46) Then the decision function δ defined by THEOREM 7 13.3 UMP Tests for Testing Certain Composite Hypotheses 37713.9 Decision-Theoretic Viewpoint of Testing Hypotheses 377 δ z z () = ∈ ⎧ ⎨ ⎩ 1 0 , , if otherwise, B (47) is minimax. PROOF For simplicity, set P 0 and P 1 for P θθ θθ θ 0 and P θθ θθ θ 1 , respectively, and similarly R(0; δ ), R(1; δ ) for R( θθ θθ θ 0 ; δ ) and R( θθ θθ θ 1 ; δ ). Also set P 0 (Z ∈ B) = α and P 1 (Z ∈ B c ) = 1 − β . The relation (45) implies that R and R 011 12 ;;. δα δ β () = () =− () LL Let A be any other (measurable) subset of ޒ n and let δ * be the corresponding decision function. Then R and R 01 10 21 ;* ;* . δδ () =∈ () () =∈ () LP A LP A c ZZ Consider R(0; δ ) and R(0; δ *) and suppose that R(0; δ *) ≤ R(0; δ ). This is equivalent to L 1 P 0 (Z ∈ A) ≤ L 1 P 0 (Z ∈B), or PA 0 Z ∈ () ≤ α . Then Theorem 1 implies that P 1 (Z ∈A) ≤ P 1 (Z ∈B) because the test defined by (47) is MP in the class of all tests of level ≤ α . Hence PAPB LPALPB cc c c 1 1 21 21 ZZ Z Z∈ () ≥∈ () ∈ () ≥∈ () ,,or or equivalently, R(1; δ *) ≥ R(1; δ ). That is, if R R then R R 00 11;* ; , ; ;*. δδ δδ () ≤ () () ≤ () (48) Since by assumption ޒ (0; δ ) = ޒ (1; δ ), we have max R R R R max R R 01 1 1 01;*, ;* ;* ; ; , ; , δδ δ δ δδ ()() [] = () ≥ () = ()() [] (49) whereas if R(0; δ ) < R(0; δ *), then maxR R R R maxR R 01 0 0 01;*, ;* ;* ; ; , ; . δδ δ δ δδ ()() [] ≥ () > () = ()() [] (50) Relations (49) and (50) show that δ is minimax, as was to be seen. ▲ REMARK 7 It follows that the minimax decision function defined by (46) is an LR test and, in fact, is the MP test of level P 0 (Z ∈ B) constructed in Theorem 1. We close this section with a consideration of the Bayesian approach. In connection with this it is shown that, corresponding to a given p.d.f. on ΩΩ ΩΩ Ω= { θθ θθ θ 0 , θθ θθ θ 1 }, there is always a Bayes decision function which is actually an LR test. More precisely, we have 378 13 Testing Hypotheses Let X 1 , , X n be i.i.d. r.v.’s with p.d.f. f(·; θθ θθ θ), θθ θθ θ∈ ΩΩ ΩΩ Ω= { θθ θθ θ 0 , θθ θθ θ 1 } and let λ 0 = {p 0 , p 1 } (0 < p 0 < 1) be a probability distribution on ΩΩ ΩΩ Ω. Then for testing the hypothesis H : θθ θθ θ= θθ θθ θ 0 against the alternative A : θθ θθ θ= θθ θθ θ 1 , there exists a Bayes decision function δ λ 0 corresponding to λ 0 = {p 0 , p 1 }, that is, a decision rule which minimizes the average risk R( θθ θθ θ 0 ; δ )p 0 + R( θθ θθ θ 1 ; δ )p 1 , and is given by δ λ 0 1 0 z z () = ∈ ⎧ ⎨ ⎩ , , if otherwise, B where B = {z ∈ R n ; f(z; θθ θθ θ 1 ) > Cf(z; θθ θθ θ 0 )} and C = p 0 L 1 /p 1 L 2 . PROOF Let R λ 0 ( δ ) be the average risk corresponding to λ 0 . Then by virtue of (44), and by employing the simplified notation used in the proof of Theorem 7, we have RLPBpLPBp pLP B pL P B pL pLP B pLP B c λ δ 0 10 0 21 1 010 12 1 12 010 121 1 () =∈ () +∈ () =∈ () +−∈ () [] =+ ∈ () −∈ () [] ZZ ZZ ZZ (51) and this is equal to pL pLf pL f d B 12 01 0 12 1 + () − () [] ∫ zzz; ; θθθθ for the continuous case and equal to pL pLf pL f B 12 01 0 12 1 + () − () [] ∈ ∑ zz z ; ; θθθθ for the discrete case. In either case, it follows that the δ which minimizes R λ 0 ( δ ) is given by δ λ 0 10 0 01 0 12 1 z zz () = () − () < ⎧ ⎨ ⎪ ⎩ ⎪ , , if ; ; otherwise; pLf pLfθθθθ equivalently, δ λ 0 1 0 z z () = ∈ ⎧ ⎨ ⎩ , , if otherwise, B where Bf pL pL f n =∈ () > () ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ zz z ޒ ;, ; ; θθθθ 1 01 12 0 as was to be seen. ▲ REMARK 8 It follows that the Bayesian decision function is an LR test and is, in fact, the MP test for testing H against A at the level P 0 (Z ∈ B), as follows by Theorem 1. THEOREM 8 13.3 UMP Tests for Testing Certain Composite Hypotheses 379 The following examples are meant as illustrations of Theorems 7 and 8. Let X 1 , , X n be i.i.d. r.v.’s from N( θ , 1). We are interested in determining the minimax decision function δ for testing the hypothesis H : θ = θ 0 against the alternative A : θ = θ 1 . We have f f nx n z z ; ; exp exp θ θ θθ θθ 1 0 10 1 2 0 2 1 2 () () = − () [] − () ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ , so that f(z; θ 1 ) > Cf(z; θ 0 ) is equivalent to exp exp ornxCn xC θθ θ θ 10 1 2 0 2 0 1 2 − () [] >− () ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ > , where C C n 010 10 1 2 =+ () + − () > () θθ θθ θθ log for 12 . Then condition (46) becomes LPXC LPXC 1020 01 θθ > () =≤ () . As a numerical example, take θ 0 = 0, θ 1 = 1, n = 25 and L 1 = 5, L 2 = 2.5. Then LPXC LPXC 1020 01 θθ > () =< () becomes PX C PX C θθ 10 00 2< () => () , or PnX C PnX C θθ θθ 10 10 00 512 5− () <− () [] =− () > [] , or ΦΦΦΦ55215 25 55 1 0000 CCCC− () =− () [] () −− () =,or Hence C 0 = 0.53, as is found by the Normal tables. Therefore the minimax decision function is given by δ z () = > ⎧ ⎨ ⎩ 1053 0 ,. , if otherwise. x The type-I error probability of this test is PX PN θ 0 0 53 0 1 0 53 5 1 2 65 1 0 996 0 004> () = () >× [] =− () =− =.,. Φ EXAMPLE 13 13.9 Decision-Theoretic Viewpoint of Testing Hypotheses 379 [...]... which may arise is that, because a and b′ are used instead of a and b, the resulting α ′ and 1 − β ′ are too small compared to α and 1 − β, respectively As a consequence, we would be led to taking a much larger number of observations than would actually be needed to obtain α and β It can be argued that this does not happen REMARK 3 Exercise 14.2.1 Derive inequality ( 28) by using arguments similar to... that EN = ∑∞ P(N ≥ n) n=1 14.1.2 In Theorem 2, assume that P(Zj < 0) > 0 and arrive at relation (13) 14.2 Sequential Probability Ratio Test Although in the point estimation and testing hypotheses problems discussed in Chapter 12 and 13, respectively (as well as in the interval estimation problems to be dealt with in Chapter 15), sampling according to a stopping time is, in general, profitable, the mathematical. .. consideration cannot be repeated at will, in many other cases this is, indeed, the case In the latter case, as a rule, it is advantageous not to fix the sample size in advance, but to keep sampling and terminate the experiment according to a (random) stopping time DEFINITION 1 Let {Zn} be a sequence of r.v.’s A stopping time (defined on this sequence) is a positive integer-valued r.v N such that, for each... First, we consider the case that θ is a real-valued parameter and proceed to define what is meant by a random interval and a confidence interval DEFINITION 1 A random interval is a finite or in nite interval, where at least one of the end points is an r.v DEFINITION 2 Let L(X1, , Xn) and U(X1, , Xn) be two statistics such that L(X1, , Xn) ≤ U(X1, , Xn) We say that the r interval [L(X1, , Xn),... Basic Theorems ofProbability Ratio Test 14.2 Sequential Sequential Sampling 389 We shall use the same notation λn for λn (x1, , xn; 0, 1), where x1, , xn are the observed values of X1, , Xn For testing H against A, consider the following sequential procedure: As long as a < λn < b, take another observation, and as soon as λn ≤ a, stop sampling and accept H and as soon as λn ≥ b, stop sampling... da gn b or Thus (8) becomes a2 gn (a) = b2gn(b) By means of this result and (6), it follows that a and b are determined by () () a 2 gn a = b2 gn b and a g n (t )dt = 1 − α b (9) For the numerical solution of (9), tables are required Such tables are available (see Table 6 78 in R F Tate and G W Klett, “Optimum confidence intervals for the variance of a normal distribution,” Journal of the American Statistical... Z1, , Zn alone In certain circumstances, a stopping time N is also allowed to take the value ∞ but with probability equal to zero In such a case and when forming EN, the term ∞ · 0 appears, but that is interpreted as 0 and no problem arises REMARK 1 Next, suppose we observe the r.v.’s Z1, Z2, one after another, a single one at a time (sequentially), and we stop observing them after a specified... regardless of the true underlying density In the formulation of the proposition above, the determination of a and b was postponed until later At this point, we shall see what is the exact determination of a and b, at least from theoretical point of view However, the actual identification presents difficulties, as will be seen, and the use of approximate values is often necessary To start with, let α and... specified event occurs In connection with such a sampling scheme, we have the following definition DEFINITION 2 382 A sampling procedure which terminates according to a stopping time is called a sequential procedure 14.1 Some Basic Theorems of Sequential Sampling 383 Thus a sequential procedure terminates with the r.v ZN, where ZN is defined as follows: () the value of Z N at s ∈ S is equal to Z N ( s ) s... 0 For a numerical application, take α = 0.01 and 1 − β = 0.05 Then the cut-off points a and b are approximately equal to a and b′, respectively, where a and b′ are given by (30) In the present case, a = 0.05 0.05 0.95 = ≈ 0.0505 and b′ = = 95 1 − 0.01 0.99 0.01 For the cut-off points a and b′, the corresponding error probabilities α ′ and 1 − β′ are bounded as follows according to (32): a ≤ 0.01 . claimed!). Take α = 0.05. Boys: 80 96 98 87 75 83 70 92 97 82 Girls: 82 90 84 70 80 97 76 90 88 86 (Hint: Group the grades into the following six intervals: [70, 75), [75, 80 ), [80 , 85 ), [85 ,. As long as a < λ n < b, take another observation, and as soon as λ n ≤ a, stop sampling and accept H and as soon as λ n ≥ b, stop sampling and reject H. By letting N stand for the smallest. (as well as in the interval estimation problems to be dealt with in Chapter 15), sampling according to a stopping time is, in general, profitable, the mathematical machinery involved is well beyond