M 562 Section-10-10

M 562 Section-10-10 - MTH/STA 562 POWER OF TESTS AND...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MTH/STA 562 POWER OF TESTS AND NEYMAN-PEARSON LEMMA Definition 1. Suppose that W is the test statistic and RR is the rejection region for a test of a hypothesis concerning the value of an unknown parameter 6. Then the power of the test, denoted by power (6), is the probability that the test will lead to rejection of H0 when the actual parameter value is 6. That is, power (6) = P {W in RR when the parameter value is 6} . For testing H0 : 6 = 60 versus H, z 6 aé 60, a = P {rejecting H0 when 6 == 60} ,8 P {accepting H0 when 6 = 61} power (61) = P {rejecting H0 when 6 = 6,} H Hence, power (6a) = 1 — fl. A typical power curve for testing H0 : 6 = 60 versus Ha : 6 yé 60 is shown as follows. Ideal power curve for testing H0 : 6 = 60 versus H, : 6 # 60 should be as follows. Power (6) We adopt the procedure of selecting a (small) value for a and finding a rejection region RR to minimize fl for the fixed sample size, that is, we choose RR to max- imize power (9) for 9 in Ha. Definition 2. If a random sample is taken from a distribution with parameter 0, a hypothesis is said to be a simple hypothesis if the hypothesis uniquely specifies the distribution from which the sample is taken. Any hypothesis that is not a simple hypothesis is called a composite hypothesis. Example 1. Consider an exponential distribution with parameter A, that is, le_y/’\ for y > 0 . _ A f (y’ A) _ { 0 elsewhere. Then H0 : A = 2 is a simple hypothesis while Ha : A > 2 is a composite hypothesis. I Example 2. Consider a normal distribution with mean ,u and variance 02, that is, 1 0271' (y - M2 202 f (y; M, 02) = exp[— ] for —oo<y<oo. Then H0 : u = 2 and 02 = 1 is a simple hypothesis while H0 : ,u = 2 is a composite hypothesis. I Definition 3. For testing H0 : 0 = 00 versus H, : 6 = 6a, a test procedure with maximum power, power (0a), is called the most powerful test. Neyman—Pearson Lemma. Let Yl, Y2, . - -, Yn represent a random sample from a distribution with parameter 0. Let L (6) denote the likelihood of the sample when the value of the parameter is 6. Suppose that we wish to test H0 : 0 = 00 versus H, : t9 = 0a. For a given a, the test that maximizes the power at 9., has a rejection region determined by <: k (10.101) for an appropriate value k. Such a test will be a most powerful test for H0 : 0 = 00 versus H, : 0 = 0a. Example 3. Let Y represent a single observation from a population with the following probability density function _ _ Qyo“1 for0<y<1 f (y’g) _ { 0 elsewhere. ' Find the most powerful test with significance level or z 0.05 to test H0 : 9 = 2 versus H, : 0 = 1. Solution. The ration of likelihood in (1010.1) is 13(00), = f (y;00) = 2y“ L(9a) f (31%) 1111—1 By the Neyman—Pearson lemma, the form of rejection region for the most powerful test is given by = 2y. 2y<k. Equivalently, the rejection region R is y < k’ = k/ 2. At a = 0.05, the value of k’ is determined by 0.05 = P{reject H0 when H0 is true} = P{Y < 19’ when 0 = 2} k /2ydy = (k’)2. 0 Therefore, [6’ :2 v0.05, and the rejection region for the most powerful test is RR = {y : y > \/0.05}. so we have I Example 4. Let Y1, Y2, - - -, Yn denote a random sample from a normal distribution with mean ,u, and variance 02 given by 1 02% 1 exp [—335 (y - M2] for — 00 < y < oo. f (WW2) = Find the uniformly most powerful test for testing H0 : p = no versus Ha : ,u > no at significance level 04. Solution. The likelihood of the sample is given by l l 1 1 2 __ _ _ 2 . __ _ 2 L (“’0 ) — a 271' exp [ 202 (yl ll) l a 271' eXp i: 20‘2 (2/2 M) l 1 1 2 Zwexp[--2U2(yn—M)] || /"\ q p_\ N) >1 \_/ § £2 "0‘ I “1 '—\ N LEM: {S I 5 mL Applying the Neyman—Pearson lemma, we see that the most powerful test of H0 : ,u : no versus Ha : p, = ya, where ya > no, is given by L (#02 L (Ma) 2 “fl- H (D N 'U H‘H | MI N r——-""'1 M3 {E | 7; 8/ [\D | M: {S | t a i W /\ kn Taking natural logarithms and simplifying, we obtain 1 (yz' — #0)2 — - Ma)2 < Ink 2‘72 1:1 £21 or n n E (yz- — #0)2 — 2 (y,- — ,ua)2 > —2a21nk i=1 i=1 or n n — 27713110 ‘+ "#3 — Z + 27mm, — nu: > —202 In k i=1 i=1 0r 2n? (/4; — no) > —20’2 lnk — 7mg + nlui or __2 21 k_ 2 2 3(Ma—Mo) > —————" “ 2n”“°+”“a Since ,ua > Mo, .—2a2 In k — 1741(2) + nu: 2” (Na _ #0) Thus the rejection region of the most powerful test for H0 : ,u = no versus Ha : p = ,ua is given by y> =k’. RR = {y > k’} . The precise value of k’ is determined by fixing a and noting that a = P{reject H0 when H0 is true} = P {7 < 16' when n = #0} We now observe that the form of the rejection region does not depend upon the particular value assigned to ,ua. That is, any value of pa greater than #0 would lead to exactly the same rejection region. Thus, we have found the uniformly most powerful test for H0 : ,u = no versus H, : ,u > no. I ...
View Full Document

Page1 / 4

M 562 Section-10-10 - MTH/STA 562 POWER OF TESTS AND...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online