STATISTICAL INFERENCE PART VI
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Transcript of STATISTICAL INFERENCE PART VI
STATISTICAL INFERENCEPART VI
HYPOTHESIS TESTING
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Inference About the Difference of Two Population Proportions
Population 1 Population 2
PARAMETERS:p1
PARAMETERS:p2
Statistics: Statistics:
Sample size: n1Sample size: n2
1p̂2p̂
Independent populations
SAMPLING DISTRIBUTION OF
• A point estimator of p1-p2 is
• The sampling distribution of is
if nipi 5 and niqi 5, i=1,2.
1 2ˆ ˆp p
1 2ˆ ˆp p
1 1 2 21 2 1 2
1 2
p q p qˆ ˆp p ~ N(p p , )
n n
2
2
1
121 n
x
n
xp̂p̂
STATISTICAL TESTS• Two-tailed testHo:p1=p2
HA:p1p2
Reject H0 if z < -z/2 or z > z/2. • One-tailed testsHo:p1=p2
HA:p1 > p2
Reject H0 if z > z
1 2 1 2
1 2
1 2
ˆ ˆ(p p ) 0 x xˆz where p=
n n1 1ˆ ˆpqn n
Ho:p1=p2
HA:p1 < p2
Reject H0 if z < -z
EXAMPLE
• A manufacturer claims that compared with his closest competitor, fewer of his employees are union members. Of 318 of his employees, 117 are unionists. From a sample of 255 of the competitor’s labor force, 109 are union members. Perform a test at = 0.05.
SOLUTIONH0: p1 = p2
HA: p1 < p2
and , so pooled
sample proportion is
Test Statistic:
11
1
x 117p̂
n 318 2
22
x 109p̂
n 255
1 2
1 2
x x 117 109p̂ 0.39
n n 318 255
(117 / 318 109 / 255) 0z 1.4518
1 1(0.39)(1 0.39)
318 255
• Decision Rule: Reject H0 if z < -z0.05=-1.96.
• Conclusion: Because z = -1.4518 > -z0.05=-1.96, not reject H0 at =0.05. Manufacturer is wrong. There is no significant difference between the proportions of union members in these two companies.
Example• In a study, doctors discovered that aspirin seems to help
prevent heart attacks. Half of 22,000 male participants took aspirin and the other half took a placebo. After 3 years, 104 of the aspirin and 189 of the placebo group had heart attacks. Test whether this result is significant.
• p1: proportion of all men who regularly take aspirin and suffer from heart attack.
• p2: proportion of all men who do not take aspirin and suffer from heart attack
1
2
104ˆ .009455
11000189
ˆ .01718 ;11000
104+189ˆ sample proportion: p= .01332
11000+11000
p
p
Pooled
Test of Hypothesis for p1-p2
H0: p1-p2 = 0
HA: p1-p2 < 0• Test Statistic:
Conclusion: Reject H0 since p-value=P(z<-5.02) 0
1 2
1 2
ˆ ˆp pz=
1 1ˆ ˆpqn n
.009455 .01718 5.02
1 1(.01332)(.98668)
11,000 11,000
Confidence Interval for p1-p2
A 100(1-C.I. for pp is given by:
1 1 2 21 2 / 2
1 2
ˆ ˆ ˆ ˆp q p qˆ ˆ(p p )
n n z
1 1 2 21 2 / 2
1 2
1 2
ˆ ˆ ˆ ˆp q p qˆ ˆ(p p ) .077 1.96*.00156
n n
.08 .074
z
p p
Inference About Comparing Two Population Variances
Population 1 Population 2
PARAMETERS: PARAMETERS:
Statistics: Statistics:
Sample size: n1Sample size: n2
Independent populations
21
22
21s 2
2s
SAMPLING DISTRIBUTION OF
• For independent r.s. from normal populations
• (1-α)100% CI for
22
21 /
)1n,1n(F~/s
/s212
222
21
21
])1n,1n,2/(F
1[
s
s]
)1n,1n,2/1(F
1[
s
s
2122
21
22
21
2122
21
22
21 /
STATISTICAL TESTS• Two-tailed testHo: (or ) HA: Reject H0 if F < F/2,n1-1,n2-1 or F > F 1- /2,n1-1,n2-1. • One-tailed testsHo:HA: Reject H0 if F > F 1- , n1-1,n2-1
Ho:
HA: Reject H0 if F < F,n1-1,n2-1
22
21
22
21
122
21
22
21
s
sF
22
21
22
21
22
21
22
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Example
• A scientist would like to know whether group discussion tends to affect the closeness of judgments by property appraisers. Out of 16 appraisers, 6 randomly selected appraisers made decisions after a group discussion, and rest of them made their decisions without a group discussion. As a measure of closeness of judgments, scientist used variance.
• Hypothesis: Groups discussion will reduce the variability of appraisals.
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Example, cont.Appraisal values (in thousand $) Statistics
With discussion 97, 101,102,95,98,103 n1=6 s1²=9.867
Without discussion 118, 109, 84, 85, 100, 121, 115, 93, 91, 112
n2=10 s2²=194.18
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Ho: versus H1: 122
21
122
21
21.0)9,5,05.0(F05081.018.194
867.9
s
sF
22
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Reject Ho. Group discussion reduces the variability in appraisals.
TEST OF HYPOTHESIS
HOW TO DERIVE AN APPROPRIATE TEST
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Definition: A test which minimizes the Type II error (β) for fixed Type I error (α) is called a most powerful test or best test of size α.
MOST POWERFUL TEST (MPT)H0:=0 Simple Hypothesis
H1:=1 Simple Hypothesis
Reject H0 if (x1,x2,…,xn)C
The Neyman-Pearson Lemma:
Reject H0 if
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kLL
xxxC n1
021 :,,,
kLL
L 1
0
1
0
1
kLP
kkLP
Proof: Available in text books (e.g. Bain & Engelhardt, 1992, p.g.408)
EXAMPLES
• X~N(, 2) where 2 is known.H0: = 0
H1: = 1
where 0 > 1.
Find the most powerful test of size .
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Solution
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c2
)(
)(n
klnX
cesin)(n
kln2)(X2
k)]}()(X2[2
1exp{
]})X()X[(2
1exp{
)X|(L
)X|(L
0for})X(2
1exp{)
2
1()X|(L
10
01
2
1010
2
01
n
1i
20
2110i2
n
1i
20i
21i2
0
1
n
1i
2i2
n
Solution, cont.• What is c?: It is a constant that satisfies
since X~N(, 2).For a pre-specified α, most powerful test says,Reject Ho if
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nZc
)n/
c
n/
X(P)|cX(P
0
000
Z
n
X
nZcX
0
0
Examples
• Example2: See Bain & Engelhardt, 1992, p.g.410 Find MPT of
Ho: p=p0 vs H1: p=p1 > p0 • Example 3: See Bain & Engelhardt, 1992,
p.g.411 Find MPT of Ho: X~Unif(0,1) vs H1: X~Exp(1)
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UNIFORMLY MOST POWERFUL (UMP) TEST• If a test is most powerful against every possible value in
a composite alternative, then it will be a UMP test.• One way of finding UMPT is find MPT by Neyman-
Pearson Lemma for a particular alternative value, and then show that test does not depend the specific alternative value.
• Example: X~N(, 2), we reject Ho if Note that this does not depend on particular value of μ1, but only on
the fact that 0 > 1. So this is a UMPT of H0: = 0 vs H1: < 0.
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Z
n
X 0
UNIFORMLY MOST POWERFUL (UMP) TEST
• To find UMPT, we can also use Monotone Likelihood Ratio (MLR).
• If L=L(0)/L(1) depends on (x1,x2,…,xn) only through the statistic y=u(x1,x2,…,xn) and L is an increasing function of y for every given 0>1, then we have a monotone likelihood ratio (MLR) in statistic y.
• If L is a decreasing function of y for every given 0>1, then we have a monotone likelihood ratio (MLR) in statistic −y.
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UNIFORMLY MOST POWERFUL (UMP) TEST
• Theorem: If a joint pdf f(x1,x2,…,xn;) has MLR in the statistic Y, then a UMP test of size for H0:0 vs H1:>0 is to reject H0 if Yc where P(Y c0)=.
• for H0:0 vs H1:<0 is to reject H0 if Yc where P(Y c0)=.
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EXAMPLE
• X~Exp()
H0:0
H1:>0
Find UMPT of size .
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GENERALIZED LIKELIHOOD RATIO TEST (GLRT)
• GLRT is the generalization of MPT and provides a desirable test in many applications but it is not necessarily a UMP test.
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H0:0
H1: 1
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GENERALIZED LIKELIHOOD RATIO TEST (GLRT)
;,,;,;;,,, 21
..
21 n
sr
n xfxfxfxxxfL
Let
ˆmaxˆ LLL
and
00
0ˆmaxˆ
LLL
MLE of
MLE of under H0
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GENERALIZED LIKELIHOOD RATIO TEST (GLRT)
Ratio Likelihood dGeneralize The
ˆL
ˆL 0
GLRT: Reject H0 if 0
EXAMPLE
• X~N(, 2)H0: = 0
H1: 0
Derive GLRT of size .
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ASYMPTOTIC DISTRIBUTION OF −2ln
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• GLRT: Reject H0 if 0
• GLRT: Reject H0 if -2ln>-2ln0=c
2
.
0H
~ln2 kasympt
under
where k is the number of parameters to be tested.
Reject H0 if -2ln> 2,k
TWO SAMPLE TESTS
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.s.r,p,nBin~Y
.s.r,p,nBin~X
22
11
211
0210
pp:H
ppp:H
Derive GLRT of size , where X and Y are independent; p0, p1 and p2 are unknown.