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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES BY Neyman Pearson 2013-01-28
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESESTable of contents 1 Introductory
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESESTable of contents 1 Introductory 2 Outline of General Theory Two types errors of hypothesis testing The likelihood principle
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESESTable of contents 1 Introductory 2 Outline of General Theory Two types errors of hypothesis testing The likelihood principle 3 Simple Hypotheses General problems and solutions Illustrate Examples
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESESTable of contents 1 Introductory 2 Outline of General Theory Two types errors of hypothesis testing The likelihood principle 3 Simple Hypotheses General problems and solutions Illustrate Examples 4 Composite Hypotheses-H0 has One Degree of Freedom General problems and solutions Illustrative Examples
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESESTable of contents 1 Introductory 2 Outline of General Theory Two types errors of hypothesis testing The likelihood principle 3 Simple Hypotheses General problems and solutions Illustrate Examples 4 Composite Hypotheses-H0 has One Degree of Freedom General problems and solutions Illustrative Examples 5 Composite Hypotheses with C Degrees of Freedom General problems and solutions Illustrative Examples
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESESTable of contents 1 Introductory 2 Outline of General Theory Two types errors of hypothesis testing The likelihood principle 3 Simple Hypotheses General problems and solutions Illustrate Examples 4 Composite Hypotheses-H0 has One Degree of Freedom General problems and solutions Illustrative Examples 5 Composite Hypotheses with C Degrees of Freedom General problems and solutions Illustrative Examples 6 Conclusion
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES IntroductoryIntroductory The problem of testing statistical is an old one.We appear to ﬁnd disagreement between statisticians of ”Whether there is eﬃcient tests or not?” In this article,we will talk about this problem and accept the words ” an eﬃcient test of the hypothesis H0 ”. Which I have to refer is that, here, ”eﬃcient” means ”without hoping to know whether each separate hypothesis is true or false, we may search for rules to givern our behaviours with regard to them, in the long run of experience, we shall not be too often wrong.”
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Outline of General Theory Outline of General Theory
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Outline of General TheoryOutline of General Theory Two types errors of hypothesis testing The likelihood principle
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Outline of General Theory Two types errors of hypothesis testing Deﬁnition (Type I error) Occurs when the null hypothesis (H0 ) is true, but is rejected. Deﬁnition (Type II error) Occurs when the null hypothesis is false, but erroneously fails to be rejected. Table: summary of possible results of any hypothesis test Reject H0 Don t reject H0 H0 T ype I Error Right decision H1 Right decision T ype II Error
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Outline of General Theory The likelihood principle Suppose that the nature of an Event,E, is exactly described by the values of n variable, x1 , x2 , ...xn (2.1) Suppose now that there exists a certain hypothesis, H0 , con- cerning the origin of the event which is such as to determine the probability of occurrence of every possible event E. Let p0 = p0 (x1 , x2 , ...xn ) (2.2) To obtain the probability that the event will give a sample point falling into a particular region, say ω, we shall have either to take the sum of (2.1) over all sample points included in ω, or to calculate the integral, P0 (ω) = ... p0 (x1 , x2 , ...xn )dx1 dx2 ...dxn (2.3) ω
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Outline of General Theory The likelihood principle we shall assume that the sample points may fall anywhere within a continuous sample space (which may be limited or not),which we shall denote by W. It will follow that P0 (W ) = 1 (2.4)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Outline of General Theory The likelihood principlethe criterion of likelihood If H0 is a simple hypothesis,denoted Ω the set of all simple hy- potheses. Take any sample point,Σ, with co-ordinates (x1 , x2 , ...xn ) and con- sider the set AΣ , of probabilities pH (x1 , x2 , ...xn ) corresponding to this sample point and determined by diﬀerent simple hypotheses belonging to Σ. We shall suppose that whatever be the sample point the set AΣ is bounded. Denote pΩ (x1 , x2 , ...xn ) the upper bound of the set AΣ , then if H0 determine the elementary probabil- ity p0 (x1 , x2 , ...xn ), we have deﬁned its likelihood to be: p0 (x1 , x2 , ...xn ) λ= (2.5) pΩ (x1 , x2 , ...xn )
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Outline of General Theory The likelihood principlethe criterion of likelihood If H0 is a composite hypothesis, then it will be possible to specify a part of the set Ω, say ω, such that every simple hypothesis belonging to the sub-set ω Analogously, we denote AΣ (ω)the sub-set of AΣ , corresponding to the set ω of simple hypotheses belonging to H0 , then the likelihood is pω (x1 , x2 , ...xn ) λ= (2.6) pΩ (x1 , x2 , ...xn )
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Outline of General Theory The likelihood principle The use of the principle of likelihood in testing hypotheses, consists in accepting for critical regions those determined by the inequality: λ ≤ C = const (2.7)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Outline of General Theory The likelihood principle Example of sample space of two dimensions Fig:2.1
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Outline of General Theory The likelihood principle Key Point As we know,if there are may alternative tests for the same hypothesis, the diﬀerence between them consists in the diﬀerence in critical regions.That is to say, our problem would become picking out the best critical regions for the same P0 (ω) = ε, the solution will be the maximum of P1 (ω) For example, assume ω1 and ω2 are two critical regions described in the Fig:2.1 , if, P1 (ω2 ) > P1 (ω1 ) Then, ω2 is better than ω1 .
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses Simple Hypotheses
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses General problems and solutions We shall now consider how to ﬁnd the best critical region for H0 , with regard to a single alternative H1 , this will be the region ω0 for which P1 (ω0 ) is a maximum subject to the condition that, P0 (ω0 ) = (3.1) Suppose the probability laws for H0 and H1 , namely, p0 (x1 , x2 , ...xn ) and p1 (x1 , x2 , ...xn ),exist, are continuous and not negative through- out the whole sample space W ; further that, P0 (W ) = P1 (W ) = 1 (3.2) The problem will consist in ﬁnding an unconditioned minimum of the expression P0 (ω0 ) − k ∗ P1 (ω0 ) = ... {p0 (x1 , x2 , ...xn ) − k ∗ p1 (x1 , x2 , ...xn )}dx1 dx2 ...dxn ω0 (3.3) where k being a constant afterwards to be determined by the con- dition(3.1)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses General problems and solutions We shall now show that the necessary and suﬃcient condition for a region ω0 , being the best critical region for H0 , which is deﬁned by the inequality, p0 (x1 , x2 , ...xn ) ≤ k ∗ p1 (x1 , x2 , ...xn ) (3.4)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses General problems and solutions Denote by ω0 the region deﬁned by (3.4) and let ω1 be any other region satisfying the condition P0 (ω1 ) = P0 (ω0 ) = ε. These regions may have a common part ω01 , which represented in Fig:3.1 Fig:3.1 It follow that, P0 (ω0 − ω01 ) = ε − P0 (ω01 ) = P0 (ω1 − ω01 ) (3.5) and consequently, k ∗ P1 (ω0 − ω01 ) ≥ P0 (ω0 − ω01 ) = P0 (ω1 − ω01 ) ≥ k ∗ P1 (ω1 − ω01 ) (3.6) If we add k ∗ P1 (ω01 ) to both side,we obtain, P1 (ω0 ) ≥ P1 (ω1 ) (3.7)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses Illustrate Examples Case of the Normal Distribution )Example 1 Suppose that it is known that a sample of n individuals,x1 , x2 , ...xn , has been drawn randomly from some normally distribution with standard deviation σ = σ0 . Let x and s be the mean and standard deviation of the sample. It is desired to test the hypothesis H0 that whether the mean in the sampled population is a = a0 or not. Then the probabilities of its occurrence determined by H0 and by H1 will be, (x − a0 )2 + s2 −n 1 2σ0 2 p0 (x1 , x2 , ...xn ) = ( √ )n e (3.8) σ0 2π (x − a1 )2 + s2 −n 1 2σ0 2 p1 (x1 , x2 , ...xn ) = ( √ )n e (3.9) σ0 2π
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses Illustrate Examples The criterion in this case becomes, (x − a0 )2 + (x − a1 )2 −n p0 2 2σ0 =e =k (3.10) p1 From this it follows that the best critical region for H0 with regard to H1 , deﬁned by the inequality (2.7), 1 σ2 (a0 − a1 )x ≤ (a2 − a2 ) + 0 log k = (a0 − a1 )x0 (say) (3.11) 0 1 2 n The critical value for the best critical value will be, x = x0 (3.12)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses Illustrate Examples Now two cases will ariseµ (1)a1 < a0 , then the region is deﬁned by, x ≤ x0 (2)a1 > a0 , then the region is deﬁned by, x ≥ x0 where x0 the statistic which is easy to decide in the practice. Conclusion The test obtained by ﬁnding the best critical region is in fact the ordinary test for the signiﬁcance of a variation in the mean of a sample; but the method of approach helps to bring out clearly the relation of the two critical regions x ≤ x0 and x ≥ x0 . Further, the test of this hypothesis could not be improved by using any other form of criterion or critical region.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses Illustrate Examples Case of the Normal Distribution )Example 2 Suppose that it is known that a sample of n individuals,x1 , x2 , ...xn , has been drawn randomly from some normally distribution with mean a = a0 = 0. Let x and s be the mean and standard deviation of the sample. It is desired to test the hypothesis H0 that whether the standard deviation in the sampled population is σ = σ0 or not. Analogously,it is easy to show that the best critical region with regard to H1 is deﬁned by the inequality, n 1 (x2 )(σ0 − σ1 ) = (x2 + s2 )(σ0 − σ1 ) ≤ v 2 (σ0 − σ1 ) (3.13) i 2 2 2 2 2 2 n i=1 where v is a constant depending on ε, σ0 , σ1
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses Illustrate Examples Again two cases will arise, (1)σ1 < σ0 , then the region is deﬁned by, x2 + s2 ≤ v 2 (2)σ1 > σ0 , then the region is deﬁned by, x2 + s2 ≥ v 2 In this case, there are two ways to deﬁne the criterion,suppose that m2 = x2 + s2 and ψ = χ2 .Then, m2 = σ0 ψ/n, d = n 2 (3.14) s2 = σ0 ψ/n − 1, d = n − 1 2 (3.15)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses Illustrate Examples It is of interest to compare the relative eﬃciency of the two criterions m2 and s2 in reducing errors of the second type. If H0 is false, suppose the true hypothesis to be H1 relating to a population in which σ1 = hσ0 > σ0 (3.16) In testing H0 with regard to the class of alternatives for which σ > σ0 , we should determine the critical value of ψ0 , so that +∞ P0 (ψ ≥ ψ0 ) = p(ψ)dψ = ε (3.17) ψ0
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses Illustrate Examples When H1 is true, the Type II Error will be P1 (ψ ≤ ψ0 ), and this value will be diﬀerent according to the two diﬀerent criterions. Now suppose that ε = 0.01 and n=5, the position is shown in Fig:3.2 Fig:3.1
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Simple Hypotheses Illustrate Examples (1)Using m2 , with degrees of freedom 5, ﬁnd that ψ0 = 15.086, thus the Type II Error will be, if h = 2, σ1 = 2σ0 , P1 (ψ ≤ ψ0 ) = 0.42 (3.18) if h = 3, σ1 = 3σ0 , P1 (ψ ≤ ψ0 ) = 0.11 (2)Usings2 , with degrees of freedom 4, ﬁnd that ψ0 = 13.277, thus the Type II Error will be, if h = 2, σ1 = 2σ0 , P1 (ψ ≤ ψ0 ) = 0.49 (3.19) if h = 3, σ1 = 3σ0 , P1 (ψ ≤ ψ0 ) = 0.17 Conclusion The second test is better, and the choice of criterion can distinguish test.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom General problems and solutions Composite Hypotheses-H0 has One Degree of Freedom
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom General problems and solutions Deﬁnition Suppose that W a common sample space for any admissible hypothesis Ht , p0 (x1 , x2 , ...xn ) is the probability law for this sample who is dependent upon the value of c+d parameters α(1) , α(2) , ...α(c) ; α(c+1) ...α(c+d) (d parameters are speciﬁed and c unspeciﬁed), ω is a sub-set of simple hypothesis for testing a composite hypothesis H0 , such that P0 (ω) = ... p0 (x1 , x2 , ...xn )dx1 dx2 ...dxn = constant = ε ω (4.1) and P0 (ω) is independent of all the values of α(1) , α(2) , ...α(c) . We shall speak of ω as a region of ”size” ε, similar to W with regard to the c parameters.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom General problems and solutionsSimilar regions We shall commence the problem with simple case for which the degree of freedom of parameters is one. we suppose that the set Ω of admissible hypotheses deﬁnes the functional form of the probability law for a given sample, namely p(x1 , x2 , ...xn ) (4.2) but that this law is dependent upon the values of 1+d parameters (2) (d+1) α(1) ; α0 , ...α0 ; (4.3) A composite hypothesis, H0 , of one degrees of freedom, its proba- bility law is denoted by p0 = p0 (x1 , x2 , ...xn ) (4.4)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom General problems and solutionsConditions of the problem of similar regions We have been able to solve the problem of similar regions only under very limiting conditions concerning p0 . These are as follows : (a) p0 is indeﬁnitely diﬀerentiable with regard to α(1) for all values of α(1) and in every point of W, except perhaps in points forming a ∂ k p0 set of measure zero. That is to say, we suppose that exists ∂(α(1) )k for any k = 1, 2, ...and is integrable over the region W.Denote by, ∂ log p0 1 ∂p0 ∂φ φ= = ;φ = (4.5) ∂α(1) p0 ∂α(1) ∂α(1) (b)The function p0 satisﬁes the equation φ = A + Bφ, (4.6) where the coeﬃcients A and B are functions of α(1) but are inde- pendent of x1 , x2 , ...xn .
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom General problems and solutionsSimilar regions for this case If the probability law p0 satisﬁes the two conditions (a) and (b), then it follows that a necessary and suﬃcient condition for ω to be similar to W with regard to α(1) is that ∂ k P0 (ω) ∂ k p0 = ... dx1 dx2 ...dxn , k = 1, 2, ... (4.7) ∂(α(1) )k ω ∂(α(1) )k When k=1, ∂p0 = p0 φ (4.8) ∂α(1) then, ∂P0 (ω) = ... p0 φdx1 dx2 ...dxn = 0 (4.9) ∂α(1) ω
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom General problems and solutions By the same way, when k=2 ∂ 2 p0 ∂ 2 (1) )2 = (1) (p0 φ) = p0 (φ2 + φ ), (4.10) ∂(α ∂α ∂ 2 P0 (ω) = ... p0 (φ2 + φ )dx1 dx2 ...dxn = 0 (4.11) ∂(α(1) )2 ω Using (4.6),we may have ∂ 2 P0 (ω) = ... p0 (φ2 + A + Bφ)dx1 dx2 ...dxn = 0 (4.12) ∂(α(1) )2 ω Having regard to (4.6) and (4.10), it follows that ... p0 φ2 dx1 dx2 ...dxn = −Aε = εψ2 (α(1) )(say) (4.13) ω
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom General problems and solutions When k=3, by diﬀerentiating(4.12) ∂ 3 P0 (ω) = ... p0 (φ3 + 3Bφ2 + (3A + B 2 + B )φ + A + AB)dx1 dx2 ...dxn = 0 ∂(α(1) )3 ω (4.14) Again,having regard to (4.6),(4.10) and (4.13), it follows that ... p0 φ3 dx1 dx2 ...dxn = (3AB−A −AB)ε = εψ3 (α(1) )(say) ω (4.15) Using the method of induction, en general, we shall ﬁnd, ... p0 φk dx1 dx2 ...dxn = εψk (α(1) )(say) (4.16) ω where ψk is a function of α(1) but independent of the sample. Since ω is similar to W, it follows that, 1 ... p0 φk dx1 dx2 ...dxn = ... p0 φk dx1 dx2 ...dxn , k = 1, 2, ε ω W (4.17)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom General problems and solutions Now we consider the signiﬁcance above in the following way. Suppose φ = constant = φ1 Then if P0 (ω(φ)) = ... p0 dω(φ1 ) (4.18) ω(φ1 ) P0 (W (φ)) = ... p0 dW (φ1 ) (4.19) W (φ1 ) represent the integral of p0 taken over the common parts, ω(φ) and W (φ) of φ = φ1 and ω and W respectively, it follows that if ω be similar to W and of size ε, P0 (ω(φ)) = εP0 (W (φ)) (4.20)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom General problems and solutionsChoice of the best critical region Let Ht be an alternative simple hypothesis to H0 . We shall assume that regions similar to W with regard to α(1) do exist. Then ω0 , the best critical region for H0 with regard to Ht , must be determined to maximise Pt (ω0 ) = ... p0 (x1 , x2 , ...xn )dx1 dx2 ...dxn (4.21) ω subject to the condition (4.20) holding for all values of φ.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom General problems and solutions Then the problem of ﬁnding the best critical region ω0 is reduce to ﬁnd parts ω0 (φ) of W (φ), which will maximise P (ω(φ)) subject to the condition (4.20). This is the same problem that we have treated already when dealing with the case of a simple hypothesis except that instead of the regions ω0 and W, what we have are the regions ω0 (φ) and W (φ). The inequality pt ≥ k(φ)p0 (4.22) will therefore determine the region ω0 (φ), where k(φ) is a constant chosen subject to the condition (4.20)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom Illustrative Examples Example 5 A sample of n has been drawn at random from normal population, and H0 is the composite hypothesis that the mean in this population is a = a0 , σ being unspeciﬁed. (x − a)2 + s2 1 −n p(x1 , x2 , ...xn ) = ( √ )n e 2σ 2 (4.23) σ 2π For H0 , α(1) = σ, α(2) = a0 (4.24) ∂p0 n (x − a)2 + s2 φ= =− +n (4.25) ∂σ σ 4σ 3 n (x − a)2 + s2 2n 3 φ = 2 + 3n 4 =− 2 − φ (4.26) σ σ σ σ (1) (2) H0 is an alternative hypothesis that αt = σ, αt = a0 .
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom Illustrative Examples It is seen that the condition pt ≥ kp0 , becomes (n[x − a)2 + s2 ]) (n[x − a0 )2 + s2 ]) − 1 2σt2 1 − ne ≥ k ne 2σ 2 (4.27) σt σ This can be reduces to, 1 2 x(at −a0 ) ≥ σ log k+0.5(a2 −a2 ) = (at −a0 )k1 (φ)(say) (4.28) n t t 0 Two cases must be distinguished in determining ω0 (φ) (a)at > a0 , then x ≥ k1 (φ) (b)at < a0 , then x ≤ k1 (φ) where k1 (φ) has to be chosen so that (4.20) is satisﬁed.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom Illustrative Examples In the case n=3, the position of ω0 (φ) is indicated in Figµ4.1 Fig:4.1 Then there will be two families of these cones containing the best critical regions: (a)For the class of hypotheses at > a0 ; the cones will be in the quadrant of positive x’s. (b)For the class of hypotheses at < a0 ; the cones will be in the quadrant of negative x’s.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses-H0 has One Degree of Freedom Illustrative Examples conclusion In the case n > 3,we may either appeal to the geometry of multiple space,the calculation of this situation may be a little complicated, but we can get similar result.Further, it has now been shown that starting with information in the form supposed, there can be no better test for the hypothesis under consideration.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom General problems and solutions Composite Hypotheses with C Degrees of Freedom
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom General problems and solutionsSimilar regions We shall now consider a probability function depending upon c pa- rameters α,this will correspond to a composite hypothesis H0 , with c degrees of freedom. Deﬁnition Let ω be any region in the sample space W,and denote by P ({α(1) , α(2) , ...α(c) }ω) the integral of p(x1 , x2 , ...xn ) over the (1) (2) (c) region ω. Fix any system of values of the α’s,say αA , αA , ...αA , If the region ω has the property, that (1) (2) (c) P ({αA , αA , ...αA }ω) = ε = constant (5.1) whatever be the system A, we shall say that it is similar to the sample space W with regard to the set of parameters α(1) , α(2) , ...α(c) and of size ε.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom General problems and solutionsConditions of the problem of similar regions Deﬁnition Let fi (α, x1 , x2 , ...xn ) = Ci (i = 1, 2, ...k < n), (5.2) be the equations of certain hypersurfaces, α and Ci being parameters. Denote by S(α, C1 , C2 , ...Ck ) the intersection of these hypersurfaces, and F (α) denote the family of S(α, C1 , C2 , ...Ck ) corresponding to a ﬁxed values α’ and to diﬀerent systems of values of the C’s.Take any S(α(1) , C1 , C2 , ...Ck ) from any family F (α1 ). If whatever be α2 it is possible to ﬁnd suitable values of the C’s, for example C1 , C2 , ...Ck , such that S(α(1) , C1 , C2 , ...Ck ) is identical with S(α(1) , C1 , C2 , ...Ck ), then we shall say that the family F (α) is independent of α.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom General problems and solutionsConditions of the problem of similar regions It is now possible to solve the problem of ﬁnding regions similar to W with regard to a set of parameters α(1) , α(2) , ...α(c) , we are at still only able to do so under rather limiting conditions. ∂ k p0 (a) We shall assume the existence of in every point of the ∂(α(i) )k sample space. (b)Writing ∂ log p0 1 ∂p0 ∂φi φi = = ;φ = ; (5.3) ∂α(i) p0 ∂α(i) i ∂α(i) it will be assumed that for every i = 1, 2, ... c. φ i = Ai + B i φ i , (5.4) Ai and Bi being independent of the x’s.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom General problems and solutions (c)Further there will need to be conditions concerning the φi = const.Denote by S(α(i) , C1 , C2 , ...Ci−1 ) the intersection that φj = Cj for any j=1,2,...i-1, corresponding to ﬁxed values of the α’s and C’s. F (α(i) ) will denote the family of S(α(i) , C1 , C2 , ...Ci−1 ) corre- sponding to ﬁxed values of the α’s and to diﬀerent systems of values of the C’s. We shall assume that any family F (α(i) ) is independent of α(i) for i =2, 3, ...c.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom General problems and solutionsSimilar regions Property If the above conditions are satisﬁed, then the necessary and suﬃcient condition for ω being similar to W with regard to the set of parameters α(1) , α(2) , ...α(c) , and of any given size ε is that ... pdω(φ1 , φ2 , ...φc ) = ε ... pdW (φ1 , φ2 , ...φc ) ω(φ1 ,φ2 ,...φc ) W (φ1 ,φ2 ,...φc ) (5.5) where W (φ1 , φ2 , ...φc ) means the intersection in W of φi = Ci for any j=1,2,...i-1, corresponding to ﬁxed values of the α’s and C’s. The symbol ω(φ1 , φ2 , ...φc ) means the part of W (φ1 , φ2 , ...φc ) included in ω.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom General problems and solutionsThe Determination of the best critical region The choice of the best critical regions in this case is similar with we already discuss before, it is simple to identify that best critical region ω0 of size ε with regard to a simple alternative Ht must satisfy the following conditions: (1) ω0 must be similar to W with regard to the set of parameters α(1) , α(2) , ...α(c) ,that is to say, P0 (ω0 ) = ε (5.6) must be independent of the values of the α’s. (2)If ν be any other region of size ε similar to W with regard to the same parameters, Pt (ω0 ) ≥ Pt (ν) (5.7)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom General problems and solutions In this way the problem of ﬁnding the best critical region for testing H0 is reduced to that of maximising Pt (ω(φ1 , φ2 , ...φc )) (5.8) under the condition (5.5) for every set of values φ1 = C1 , φ2 = C2 , ...φc = Cc , (5.9)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom Illustrative Examples Example 6 - Test for the signiﬁcance of the diﬀerence between two variances We suppose that two samples, (1) Σ1 of size n1 ,mean = x1 ,standard deviation = s1 . (2) Σ2 of size n2 ,mean = x2 ,standard deviation = s2 . have been drawn at random from normal distribution.If this is so, the most general probability law for the observed event may be written (x1 − a1 )2 + s2 2 2 1 −n (x2 − a2 ) + s2 −n1 2 1 N 1 2σ 2 2σ 2 p(x1 , x2 , ...xn ; xn +1 , xn +2 , ...xN ) = ( √ ) n n e 1 2 1 1 1 2π σ 1σ 2 1 2 (5.10) where n1 + n1 = N , and a1 , σ1 are the mean and standard deviation of the ﬁrst, a2 , σ2 are the mean and standard deviation of the second sampled distribution. H0 is the composite hypothesis that σ1 = σ2 . For an alternative Ht ,the parameters may be deﬁned as σ2 α1 = a1 ; α2 = a2 − a1 = bt ; α3 = σ1 ; α4 = = θt . (5.11) σ1
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom Illustrative Examples For the hypothesis to be tested, H0 : α1 = a; α2 = b; α3 = σ; α4 = 1. (5.12) H0 it will be seen, is a composite hypothesis with 3 degrees of freedom, a, b and σ being unspeciﬁed.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom Illustrative Examples We shall now consider whether the conditions (A), (B) and (C) of the above theory are satisﬁed. (A) The ﬁrst condition is obviously satisﬁed. (B) Making use of (5.10),we ﬁnd that √ 1 log p0 = −N log 2π − N log σ − {n1 (x1 − a)2 + n2 (x2 − a − b)2 + n1 s2 + n2 s2 } 1 2 2σ 2 (5.13) ∂ log p0 1 φ1 = = 2 {n1 (x1 − a) + n2 (x2 − a − b)} (5.14) ∂a σ ∂ log p0 1 φ2 = = 2 n2 (x2 − a − b) (5.15) ∂b σ ∂ log p0 N 1 φ3 = = − + N 3 {n1 (x1 − a)2 + n2 (x2 − a − b)2 + n1 s2 + n2 s2 } (5.16) 1 2 ∂σ σ σ
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom Illustrative Examples We see that φ1 = A1 , φ2 = A2 , φ3 = A3 + B3 φ3 , (5.17) where the A’s and B’s are independent of the sample variates, so that the condition (B) is satisﬁed. (C) The equation S(α(2) , C1 ) namely,φ1 = C1 ,where C1 is an ar- bitrary constant, is obviously equivalent to n1 x1 + n2 x2 = C1 de- pending only upon one arbitrary parameter C1 . Hence F (α(2) ) is in- dependent of α(2) . Similarly, F (α(3) ) is independent of α(3) . Hence also the condition (C) is fulﬁlled.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom Illustrative Examples We may now attempt to construct the best critical region ω0 .Its elements ω0 (φ1 , φ2 , φ3 ) are parts of the W(φ1 , φ2 , φ3 ) satisfying the system of three equations φi = Ci (i = 1, 2, 3), within which pt ≥ k(x1 , x2 , sa )p0 (5.18) 1 where sa = n1 (s2 + n2 s2 ) and the value of k being determined 1 2 N for each system of values of x1 , x2 , sa , so that P0 (ω0 (φ1 , φ2 , φ3 )) = εP0 (W0 (φ1 , φ2 , φ3 )) (5.19) The condition (5.19) becomes −n1 [(x1 − a)2 + s2 ] + n2 [(x2 − a − b)2 + s2 ] 1 2 1 e 2σ 2 ≥ σN −n1 [(x1 − a1 )2 + s2 ] + n2 θ2 [(x2 − a1 − b1 )2 + s2 ] 1 2 k 2σ12 N e σ1 θn2 (5.20)
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom Illustrative Examples Since the region determined by (5.19) will be similar to W with regard to a, b and σ, we may put a = a1 , b = b1 , σ = σ1 , and the condition (5.20) will be found on taking logarithms to reduce to n2 {(x2 − a1 − b1 )2 + s2 }(1 − θ2 ) ≤ 2σ1 θ2 (log k − n2 log θ) (5.21) 2 2 This inequality contains only one variable s2 . Solving with regard 2 to s2 we ﬁnd that the solution will depend upon the sign of the 2 diﬀerence 1 − θ2 . Accordingly we shall have to consider separately the two classes of alternatives, σ2 θ= > 1; the B.C.R will be deﬁned by s2 ≥ k1 (x1 , x2 , sa ) 2 σ1 σ2 θ= < 1; the B.C.R will be deﬁned by s2 ≥ k2 (x1 , x2 , sa ) 2 σ1
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Composite Hypotheses with C Degrees of Freedom Illustrative Examples By some technical calculation, we can arrive at the criterion µ that, σ2 n2 s 2 2 θ= > 1; µ = ≥ µ0 (5.22) σ1 n1 s2 + n2 s2 1 2 σ2 n2 s 2 2 θ= < 1; µ = ≤ µ0 (5.23) σ1 n1 s2 + n2 s2 1 2 where The constant µ0 depends only upon n1 , n2 value of ε chosen. Conclusion (1) We see that the best critical regions are common for all the alternatives. (2) the criterion µ we have reached,is equivalent to that suggested on intuitive grounds by FISHER.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Conclusion Conclusion 1.The conception of the best critical region we talk about in this article,is the region with regard to the alternative hypothesis Ht , for a ﬁxed value of ε, assures the minimum chance of accepting H0 , when the true hypothesis is Ht . 2.To solve the same problem in the case where the hypothesis tested is composite, the solution of a further problem is required;this consists in determining what has been called a region similar to the sample space with regard to a parameter. We have been able to solve this problem only under certain limiting conditions; at present, therefore, these conditions also restrict the generality of the solution given to the problem of the best critical region for testing composite hypotheses.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES ConclusionReference [1] NEYMAN.”Contribution to the Theory of Certain Test Criteri- a”,Bull.Inst.int.Statist 1928. [2] FISHER.”Statistical methods for Research Workers”,London 1932. [3] PEARSON and NEYMAN.”On the Problem of Two Samples”,Bull. Acad. Polon. Sci. Lettres 1930. [4] Lehmann EL.”The Fisher, Neyman and Pearson theories of test- ing hypotheses: one theory or two? ” J Am Stat Assoc 1993. [5] Berkson J.”Tests of signiﬁcance considered as evidence”, J Am Stat Assoc 1942.
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THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES Conclusion THANK YOU!
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