STATISTICAL INFERENCEPART VI

HYPOTHESIS TESTING

1

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

21

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

15

Ho: versus H1: 122

21

122

21

21.0)9,5,05.0(F05081.018.194

867.9

s

sF

22

21

Reject Ho. Group discussion reduces the variability in appraisals.

TEST OF HYPOTHESIS

HOW TO DERIVE AN APPROPRIATE TEST

16

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

17

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 .

18

Solution

19

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

20

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.

22

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.

23

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)=.

24

EXAMPLE

• X~Exp()

H0:0

H1:>0

Find UMPT of size .

25

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.

26

H0:0

H1: 1

27

GENERALIZED LIKELIHOOD RATIO TEST (GLRT)

;,,;,;;,,, 21

..

21 n

sr

n xfxfxfxxxfL

Let

ˆmaxˆ LLL

and

00

0ˆmaxˆ

LLL

MLE of

MLE of under H0

28

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 .

29

ASYMPTOTIC DISTRIBUTION OF −2ln

30

• 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

31

.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.