This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MTH/STA 562 POWER OF TESTS AND NEYMANPEARSON 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
 Spring '08
 Cheng

Click to edit the document details