557: MATHEMATICAL STATISTICS II HYPOTHESIS TESTING: EXAMPLES

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1 557: MATHEMATICAL STATISTICS II HYPOTHESIS TESTING: EXAMPLES Example Suppose that X,..., X n N, ). To test H 0 : 0 H : the most powerful test at level α is based on the statistic λx) f π) X x ) n/ exp } x i ) } f X x 0) π) n/ exp } exp x i n/ x i with critical region R given by x R if x i n > log k where k is defined by Pr[X R 0] α. We can convert this to a rejection region of the form X > c n. Now, given 0, X N0, /n), so Pr[X R 0] Pr[X > c n 0] Φ n c n ) α c n Φ α) n For α 0.05, Φ α).645. Hence we reject H 0 in favour of H if X >.645 n For example, for n 5, c n The power function β) is given by β) Pr[X R ] Pr[X > c n ] Φ nc n )) which we evaluate specifically at. Note that β) is an increasing function of so that as increases, the power to reject H 0 in favour of H increases. Example Suppose that X,..., X n Exp/). To test H 0 : H : > Let 0, Θ, ). The most powerful test of the hypotheses H 0 : 0 H : is given by the Neyman-Pearson Lemma to be λx) f X x ) ) n exp n } x i / f X x 0 ) 0 exp n } x i / 0 ) n exp [ x i ] } > k.

2 so that, in terms of the sufficient statistic, T X ) X i > log k n log/ ) say. Hence the critical region is of the form T X ) > c n, and as under H 0, T X ) Gamman, /), we require that Pr[T X ) > c n ] α c n q n,/ α) where q a,b α) is the inverse cdf for the Gammaa, b) distribution evaluated at α. Consider tests where R T t : t > c}; this test has power function β) Pr[T X ) > c ] c n Γn) tn e t/ dt ) which can be computed numerically. Now, note from equation ) that β) is a decreasing function of c, so therefore the most powerful test across all possible values of Θ that attain size/level α is the one with c c n. Below is a table of β) for different values of n and, when α 0.05 and c c n : n Example 3 Suppose that X,..., X n Bernoulli). A test of H 0 : 0 H : > 0 is required. The likelihood ratio for < for this model is λx) f X x ) f X x ) T x) ) T x) ) n T x) n T x) ) / ) T x) / ) where T ) X n X i. Thus λx) is a monotone increasing function of T x) as for < ) > ) and by the Karlin-Rubin theorem, the UMP test at level α is based on the critical region } R : T x) x i > t 0 x To find t 0, we need to solve Pr[T X ) > t 0 0 ] α. ) Now if 0, then T X ) Binomialn, 0 ), so t 0 need only take integer values on 0,..., n}. Note that the equation ) can not be solved for all α, as T X ) has a discrete distribution. ) n

3 Example 4 Consider the likelihood arising from a random sample X,..., X n following a one-parameter Exponential Family model: L x) hx)c)} n expw)t x)} where T ) X n tx i ) is a sufficient statistic. For < L x) L x) ) c ) n expw ) w ))T x c ) )}. This is a monotone function of T x) if w) is a monotone function; if w) is non-decreasing, then the test of the hypothesis H 0 : 0 H : > 0 that uses the rejection region R x : T x) t 0 }, where Pr[T X ) t 0 0 ] α, is the UMP α level test. Example 5 Suppose that X..., X n N, σ ) and Y,..., Y n N, σ ) are independent random samples. To test the likelihood ratio statistic is λx, y) H 0 :, σ unspecified H :, σ unspecified sup f X,Ỹ x, y, σ,σ ) Θ 0 ) sup f X,Ỹ x, y,,σ ) Θ Under H 0, the maximum likelihood estimators of and σ are σ 0 X i + n Y i,, σ ) L 0, σ 0 x, y) L,, σ x, y) n X + n Y n + n n + n [ n ] X i n + n ) n + Y i ) whereas under H, the maximum likelihood estimators of, and σ are Therefore Now λx, y) k is equivalent to X Y [ n ] σ X i n + n n ) + Y i ) σ σ 0 λx, y) σ σ 0 ) n +n )/ X i ) + n Y i ) n X i ) + n c Y i ) 3

4 say. In the denominator X i ) X i + ) X i X) + n X n X + n Y n + n X i X) + n n n + n ) X Y ) with an equivalent expression for n Y i ) n Y i Y ) + n n ) X Y n + n ) Therefore, after substitution into the inequality above, we have X i X) + n Y i Y ) X i X) + n Y i Y ) + n c n X Y ) n + n which is equivalent to the inequality n n X Y ) n + n X i X) + n c Y i Y ) or more familiarly say, where T X, Ỹ ) s P n + n X Y ) ) n + n )c c 3) [ n ] s n P X i X) + Y i Y ) n + n is the unbiased estimator of σ under H. The statistic on the left hand side of equation 3) is, under H 0, the square of a Student-t random variable with n + n degrees of freedom, and thus the likelihood ratio test is equivalent to the traditional two-sample t-test for the equality of means. The appropriate value of c can be computed using tables of that distribution; we have for a level α test c St n +n α/) where St n p) is the inverse cdf of the Student-t density with n degrees of freedom evaluated at probability p. Thus the rejection region R T is defined by R T t : t c) t c)}. Power Function: The power function β can be formed in terms of the difference δ, and a specific σ. We have βδ, σ) Pr[T X, Ỹ ) R T δ, σ] Pr[T X, Ỹ ) c δ, σ] + Pr[T X, Ỹ ) c δ, σ]. To compute these probabilities, we need to compute the distribution of T X, Ỹ ) when the difference between the means is δ. It turns out that this distribution is the non-central Student-t distribution: if Z Nµ, ) and V χ ν are independent random variables, then T Z V/ν Studentν, µ) ) 4

5 for which the pdf can be computed using standard methods from MATH 556. The statistic T X, Ỹ ) from equation 3) can be written in this fashion, with X i X) + n Y i Y ) n n X Y ) Z V n + n σ σ µ n n ) n n δ n + n σ n + n σ The term δ/σ is the standardized difference between and, and the form of µ indicates that we can look at power on this standardized scale for different sample sizes. In R, the functions pt and qt compute, respectively, the cdf and inverse cdf for both the Student-t and non-central Student-t distributions; for the probabilities required to compute β, σ) the R commands are n<-n+n alpha<-0.05 sigma<- delta<-seq-,,by0.0) cval<-qt-alpha/,n-) mu<-sqrtn*n/n+n)))*delta/sigma) beta.power<-pt-cval,dfn-,ncpmu)+-ptcval,dfn-,ncpmu) The plot below depicts βδ/σ) for α 0.05; note that the power is higher as n n + n increases, but that the power for n 0 is also higher if n n 0 than if n 5 and n 5. βδ) n 5, n 5 n 5, n 0 n 5, n 5 n 0, n 0 0 δ Example 6 Randomized Tests A test T with test function φ R T x)) taking values in 0, } with probability one) is termed a nonrandomized test; given the observed value of statistic T x), the null hypothesis is deterministically) rejected if φ R T x)), and is not rejected otherwise. For such a test E ft [φ R T x)) ] Pr[φ R T x)) ] Pr[T X ) R ] β). In the Neyman-Pearson Lemma, for testing parametric models f X and two possible values 0 and, at level α, the critical region R is defined by f X x ) > kf X x 0 ) x R f X x ) < kf X x 0 ) x R 5

6 where k is defined by noting the requirement Pr[X R 0 ] α. However, it may occur that f X x ) kf X x 0 ) in which case the result of the test is ambiguous. A potential resolution of the ambiguity is to construct a randomized test, T, where the decision to reject H 0 is potentially randomly chosen, but that matches the power of T. Consider the test function φ R x) defined by f X x ) > kf X x 0 ) φ Rx) γ f X x ) kf X x 0 ) 0 f X x ) < kf X x 0 ) for 0 γ, so that, with a non-zero probability, φ R x) takes a value not equal to zero or one. In this randomized test, the constant γ represents the probability with which H 0 is rejected in the case that f X x ) kf X x 0 ). Note that the requirement Pr[X R 0 ] α implies that we much choose γ so that E ft [φ RT X )) 0 ] Pr[φ RT X )) 0 ] + γ Pr[φ RT X )) γ 0 ] The final term needs some explanation; it is equal to the probability of the set under the model that assumes 0. A x : f X x ) kf X x 0 )} For example, suppose that X,..., X n Bernoulli) and consider a test of the simple hypotheses with values 0 <. Let T ) be defined by T ) X X n X i. If λ T x) f X x ) f X x 0 ) 0 ) T x) 0 ) n T x) then λ T x) is an increasing function of T x). Therefore, there exist constants c and γ such that a test T can be constructed with test function T x) > c φ Rx) γ 0 T x) c T x) c such that α E ft [φ RT X )) 0 ] Pr[φ RT X )) 0 ] + γ Pr[φ RT X )) γ 0 ] Pr[T X ) > c 0 ] + γ Pr[T X ) c 0 ] jc+ ) ) n j n 0 j 0) j + γ c c 0 0 ) c. The introduction of the random element allows this equation to be solved exactly, whatever the value of α; this was not possible under the non-randomized rule. For a specific numerical example, let n 0, and 0.5. For α 0.05, the probability distribution of T X ) is Binomialn, ), so that the probability Pr[T x) > c 0.3] can be computed: 6

7 c Pr[T x) c 0.3] Pr[T x) > c 0.3] Hence choosing c equal to 8 or 9 gives Pr[T x) > c 0.3] equal to 0.3 and respectively, so that α 0.05 cannot be matched exactly in a non-randomized test that is, if γ 0). However choosing c 9 and γ in the randomized test yields Pr[T X ) > c 0 ] + γ Pr[T X ) c 0 ] α so the randomized test that specifies x i > 9 Reject H 0 x i 9 Reject H 0 with probability γ x i < 9 Do Not Reject H 0 has size/level precisely α. The power function is β) jc+ ) ) n n j ) j + γ c ) c j c Example 7 Suppose that X,..., X n Uniform0, ). To test The likelihood ratio for < for this model is λx) f X x ) f X x ) H 0 : 0 H : > 0. ) n T X ) T X ) where T X ) X n) minx,..., X n }. Thus λx) is a non decreasing function of T x) as for <, and by the Karlin-Rubin theorem, the UMP test at level α is based on the critical region To find t 0, we need to solve Pr[X n) > t 0 0 ] R x : T x) x n) > t 0 }. t0 0 ) n α t 0 0 α) /n 7

8 with power function for > 0 ) β) ) n 0 α). Now consider the randomized test T with test function xn) > φ 0 Rx) α x n) 0 We have for > 0 that β ) E ft [φ RT X )) ] Pr[φ RT X )) ] + α Pr[φ RT X )) α ] Pr[X n) > 0 ] + α Pr[X n) 0 ] ) n ) n α ) n 0 α) thus matching the power of the UMP test described above. Therefore the UMP test is not unique. Note that for the hypotheses the likelihood ratio test statistic is λx) f X x 0 ) f X x ) H 0 : 0 H : 0 xn) 0 ) n x n) 0 0 x n) > 0 Therefore the likelihood ratio test λx) k is has rejection region To choose k, we require that the size/level is α; as X n) > 0 ) X n) / 0 k /n ) Pr[X n) > 0 ) X n) / 0 k /n ) 0 ] Pr[X n) k /n 0 0 ] kn 0 n 0 k we choose k α. The power function β) is Pr[X n) > 0 ) X n) / 0 < α /n ) ] α α) ) n 0 0 < < 0 ) n 0 > 0 8