Inference for the mean vector. Univariate Inference Let x 1, x 2, …, x n denote a sample of n from...

Inference for the mean vector

Univariate InferenceLet x1, x2, … , xn denote a sample of n from the normal distribution with mean and variance 2.Suppose we want to test

H0: = 0 vs

HA: ≠ 0

The appropriate test is the t test:

The test statistic:

Reject H0 if |t| > t/2

0xt n

s

The multivariate TestLet denote a sample of n from the p-variate normal distribution with mean vector and covariance matrix .

Suppose we want to test

1 2, , , nx x x

0 0

0

: vs

:A

H

H

Example

For n = 10 students we measure scores on – Math proficiency test (x1),

– Science proficiency test (x2),

– English proficiency test (x3) and

– French proficiency test (x4)

The average score for each of the tests in previous years was 60. Has this changed?

The data

Student Math Science Eng French

1 81 89 73 742 73 79 73 743 61 86 81 814 55 70 76 735 61 71 61 666 52 70 56 587 56 74 56 568 65 87 73 699 54 76 69 72

10 48 71 62 63

Summary Statistics

60.677.368.068.6

x

S

102.044 56.689 41.222 39.48956.689 56.456 42.000 35.35641.222 42.000 75.778 65.11139.489 35.356 65.111 61.378

the mean vector

the sample covariance matrix

0

60606060

Roy’s Union- Intersection PrincipleThis is a general procedure for developing a multivariate test from the corresponding univariate test.

1

i.e. observation vector

p

X

X

X

1. Convert the multivariate problem to a univariate problem by considering an arbitrary linear combination of the observation vector.

1 1 p pU a X a X a X

arbitrary linear combination of the observations

2. Perform the test for the arbitrary linear combination of the observation vector.

3. Repeat this for all possible choices of

1

p

a

a

a

4. Reject the multivariate hypothesis if H0 is rejected for any one of the choices for

5. Accept the multivariate hypothesis if H0 is accepted for all of the choices for

6. Set the type I error rate for the individual tests so that the type I error rate for the multivariate test is .

.a

.a

Let denote a sample of n from the p-variate normal distribution with mean vector and covariance matrix .


1 2, , , nx x x

0 0

0

: vs

:A

H

H

Application of Roy’s principle to the following situation

1 1Let i i i p piu a x a x a x

Then u1, …. un is a sample of n from the normal distribution with mean and variance .a a aΣ

to test

0 0

0

: vs

:

a

aA

H a a

H a a

we would use the test statistic:

0a

u

u at n

s

1 1

1 1Now

n n

i ii i

u u a xn n

1 1

1 1n n

i ii i

a x a x a xn n

and

222

1 1

1 1

1 1

n n

u i ii i

s u u a x a xn n

2

1

1

1

n

ii

a x xn

1

1

1

n

i ii

a x x x x an

1

1

1

n

i ii

a x x x x a a an

S

Thus

00

a a x a nt n a x

a aa a

SS

We will reject 0 0:aH a a

if 0 / 2

a nt a x t

a a

S

2

2 0 2

/ 2or a

n a xt t

a a

S

We will reject

0 0 0: in favour of :AH H

Using Roy’s Union- Intersection principle:

2

2 0 2

/ 2if for at least one a

n a xt t a

a a

S

We accept0 0:H

2

2 0 2

/ 2if for all a

n a xt t a

a a

S

We reject

0 0:H

i.e.

2

0 2

/ 2if max

a

n a xt

a a

S

We accept0 0:H

2

0 2

/ 2if max

a

n a xt

a a

S

Consider the problem of finding:

2

0max max

a a

n a xh a

a a

S

where

2

0 0 0n a x a x x a

h a na a a a

S S

0 0 0 0

2

2 20

a a x x a a x x a ah a

na a a

S S

S

0 0or a a x a x a S S

thus 2

0max

opt

aopt opt

n a xh a

a a

S

1 10 0

0

or opt

a aa x k x a

a x

SS S

21

0 0

2 1 10 0

n k x x

k x x

S

S SS

10 0n x x S

We reject 0 0:H Thus Roy’s Union- Intersection principle states:

1 20 0 / 2

if n x x t

S

We accept 0 0:H

1 20 0 / 2

if n x x t

S

2 10 0The statistic T n x x S

is called Hotelling’s T2 statistic

We reject 0 0:H Choosing the critical value for Hotelling’s T2 statistic

2 1 20 0 / 2

if T n x x t

S

2

/ 2To determine t

, we need to find the sampling distribution of T2 when H0 is true.

It turns out that if H0 is true than

2 1

0 0 1 1

n p nn pF T x x

p n p n

S

has an F distribution with 1 = p and 2 = n - p

We reject 0 0:H

Thus

Hotelling’s T2 test

2 1 20 0

1, a

p nT n x x F p n p T

n p

S

2 ,1

n pF T F p n p

p n

or if

f x

Another derivation of Hotelling’s T2 statistic

Another method of developing statistical tests is the Likelihood ratio method.

Suppose that the data vector, , has joint densityx

Suppose that the parameter vector, , belongs to the set . Let denote a subset of .

Finally we want to test 0 : vs

:A

H

H

ˆmax max

ˆmaxmax

Lf x L

Lf x L

The Likelihood ratio test rejects H0 if

ˆwhere the MLE of

0

ˆand the MLE of when is true.H

The situationLet denote a sample of n from the p-variate normal distribution with mean vector and covariance matrix .


1 2, , , nx x x

0 0

0

: vs

:A

H

H

The Likelihood function is:

1

1

1

2

/ 2 / 2

1, e

2

n

i ii

x x

np nL

and the Log-likelihood function is:

, ln , l L

1

1

1ln 2 ln

2 2 2

n

i ii

np nx x

and

the Maximum Likelihood estimators of

are

1

1ˆ n

ii

x xn

and

1

1 1ˆ n

i ii

nx x x x S

n n

and the Maximum Likelihood estimators of

when H 0 is true are:

0

ˆ ˆ

and

0 01

1ˆ n

i ii

x xn

The Likelihood function is:

1

1

1

2

/ 2 / 2

1, e

2

n

i ii

x x

np nL

now

11 1

1 1

ˆ ˆˆ n n

ni i i in

i i

x x x x S x x

11

1

n

ni in

i

tr x x S x x

1

11

n

ni in

i

tr S x x x x

11

1

n

ni in

i

tr S x x x x

1 11 = 1 = n nn ntr n I n p np

Thus 2

/ 2/ 2 1

1ˆ ˆ, 2

np

nnp nn

L eS

similarly

2/ 2

/ 2

1ˆ ˆˆ ˆ, ˆ2

np

nnp

L e

and

/ 2 / 21 1

/ 2 / 2

0 01

ˆ ˆˆ ˆ,

ˆ ˆ ˆ, 1ˆ

n nn nn n

n nn

i ii

L S S

Lx x

n

/ 2

/ 2

0 01

1

n

nn

i ii

n S

x x

Note:11 12

21 22

A A u wA

A A w V

Let

111 22 21 11 12

122 11 12 22 21

A A A A AA

A A A A A

1

1u V ww

u

V u w V w

11Thus u V ww V u w V w

u

and1

1

1V ww

w V wuV u

/ 2

/ 2

0 01

1

n

nn

i ii

n S

x x

Now

and

2/

0 01

1 n

n

i ii

n S

x x

Also

0 0 0 01 1

= n n

i i i ii i

x x x x x x x x

01 1

=n n

i i ii i

x x x x x x x

0 0 01

n

ii

x x x n x x

0 01

=n

i ii

x x x x n x x

0 01

=n

i ii

x x x x n x x

0 0= 1n S n x x

Thus

2/

0 01

1 n

n

i ii

n S

x x

0 0

1

1

n S

n S n x x

0 0

1

S

nS x x

n

Thus 0 02/ 1

n

nS x x

n

S

using 1

1

1V ww

w V wuV u

0

1,

and

u n

V S

w n x

Then 1

0 02/ 1 1

nn x S x

n

Thus to reject H0 if < 2/i.e. n n

2/or n n

10 0

and 1 1

nn x S x

n

10 0or 1 -1 nn x S x n

This is the same as Hotelling’s T2 test if

2/ 11 -1 , n p n

n T F p n pn p

Example






The data


1 81 89 73 742 73 79 73 743 61 86 81 814 55 70 76 735 61 71 61 666 52 70 56 587 56 74 56 568 65 87 73 699 54 76 69 72

10 48 71 62 63

Summary Statistics

60.677.368.068.6

x

S

102.044 56.689 41.222 39.48956.689 56.456 42.000 35.35641.222 42.000 75.778 65.11139.489 35.356 65.111 61.378

0.0245 -0.0255 0.0195 -0.0218-0.0255 0.0567 -0.0405 0.02670.0195 -0.0405 0.1782 -0.1783-0.0218 0.0267 -0.1783 0.2040

1

: S

Note

2 10 0 151.135T n x S x

0.05 0.05 0.05

1 4 9 4 9, 4,6 = 4.53 27.18

6 6

p nT F p n p F

n p

0

60606060

Inference for the mean vector

Univariate InferenceLet x1, x2, … , xn denote a sample of n from the normal distribution with mean and variance 2.Suppose we want to test

H0: = 0 vs

HA: ≠ 0


The test statistic:

Reject H0 if |t| > t/2

0xt n

s

We reject 0 0:H Hotelling’s T2 statistic and test

2 1 20 0 / 2

if T n x x t

S

0: offavour in

AH

2T

pnpFpn

npT

,1

where 2

Example






The data


1 81 89 73 742 73 79 73 743 61 86 81 814 55 70 76 735 61 71 61 666 52 70 56 587 56 74 56 568 65 87 73 699 54 76 69 72

10 48 71 62 63

Summary Statistics

60.677.368.068.6

x

S

102.044 56.689 41.222 39.48956.689 56.456 42.000 35.35641.222 42.000 75.778 65.11139.489 35.356 65.111 61.378

0.0245 -0.0255 0.0195 -0.0218-0.0255 0.0567 -0.0405 0.02670.0195 -0.0405 0.1782 -0.1783-0.0218 0.0267 -0.1783 0.2040

1

: S

Note

2 10 0 151.135T n x S x

0.05 0.05 0.05

1 4 9 4 9, 4,6 = 4.53 27.18

6 6

p nT F p n p F

n p

0

60606060

The two sample problem

Univariate Inference

Let x1, x2, … , xn denote a sample of n from the normal distribution with mean x and variance 2.

Let y1, y2, … , ym denote a sample of n from the normal distribution with mean y and variance 2.


H0: x = y vs

HA: x ≠ y


The test statistic:

Reject H0 if |t| > t/2 d.f. = n + m -2

1 1pooled

x yt

sn m

2 21 1

2x y

pooled

n s m ss

n m

The multivariate TestLet denote a sample of n from the p-variate normal distribution with mean vector and covariance matrix .

1 2, , , nx x x

x

0 : vs

:

x y

A x y

H

H


Let denote a sample of m from the p-variate normal distribution with mean vector and covariance matrix .

1 2, , , my y y

y

Hotelling’s T2 statistic for the two sample problem

2 111 1 pooledT x y x y

n m

S

if H0 is true than

21

2

n m pF T

p n m

has an F distribution with 1 = p and

2 = n +m – p - 1

1 1

2 2pooled x y

n m

n m n m

S S S

We reject 0 : x yH

Thus


21if , 1

2

n m pF T F p n m p

p n m

2 11with

1 1 pooledT x y x y

n m

S

1 1

2 2pooled x y

n m

n m n m

S S S

Example 2Annual financial data are collected for firms approximately 2 years prior to bankruptcy and for financially sound firms at about the same point in time. The data on the four variables

• x1 = CF/TD = (cash flow)/(total debt), • x2 = NI/TA = (net income)/(Total assets), • x3 = CA/CL = (current assets)/(current liabilties, and • x4 = CA/NS = (current assets)/(net sales) are given in

the following table.

The data are given in the following table:

Bankrupt Firms Nonbankrupt Firms x1 x2 x3 x4

x1 x2 x3 x4

Firm CF/TD NI/TA CA/CL CA/NS Firm CF/TD NI/TA CA/CL CA/NS 1 -0.4485 -0.4106 1.0865 0.4526 1 0.5135 0.1001 2.4871 0.5368 2 -0.5633 -0.3114 1.5314 0.1642 2 0.0769 0.0195 2.0069 0.5304 3 0.0643 0.0156 1.0077 0.3978 3 0.3776 0.1075 3.2651 0.3548 4 -0.0721 -0.0930 1.4544 0.2589 4 0.1933 0.0473 2.2506 0.3309 5 -0.1002 -0.0917 1.5644 0.6683 5 0.3248 0.0718 4.2401 0.6279 6 -0.1421 -0.0651 0.7066 0.2794 6 0.3132 0.0511 4.4500 0.6852 7 0.0351 0.0147 1.5046 0.7080 7 0.1184 0.0499 2.5210 0.6925 8 -0.6530 -0.0566 1.3737 0.4032 8 -0.0173 0.0233 2.0538 0.3484 9 0.0724 -0.0076 1.3723 0.3361 9 0.2169 0.0779 2.3489 0.3970 10 -0.1353 -0.1433 1.4196 0.4347 10 0.1703 0.0695 1.7973 0.5174 11 -0.2298 -0.2961 0.3310 0.1824 11 0.1460 0.0518 2.1692 0.5500 12 0.0713 0.0205 1.3124 0.2497 12 -0.0985 -0.0123 2.5029 0.5778 13 0.0109 0.0011 2.1495 0.6969 13 0.1398 -0.0312 0.4611 0.2643 14 -0.2777 -0.2316 1.1918 0.6601 14 0.1379 0.0728 2.6123 0.5151 15 0.1454 0.0500 1.8762 0.2723 15 0.1486 0.0564 2.2347 0.5563 16 0.3703 0.1098 1.9914 0.3828 16 0.1633 0.0486 2.3080 0.1978 17 -0.0757 -0.0821 1.5077 0.4215 17 0.2907 0.0597 1.8381 0.3786 18 0.0451 0.0263 1.6756 0.9494 18 0.5383 0.1064 2.3293 0.4835 19 0.0115 -0.0032 1.2602 0.6038 19 -0.3330 -0.0854 3.0124 0.4730 20 0.1227 0.1055 1.1434 0.1655 20 0.4875 0.0910 1.2444 0.1847 21 -0.2843 -0.2703 1.2722 0.5128 21 0.5603 0.1112 4.2918 0.4443 22 0.2029 0.0792 1.9936 0.3018 23 0.4746 0.1380 2.9166 0.4487 24 0.1661 0.0351 2.4527 0.1370 25 0.5808 0.0371 5.0594 0.1268


A graphical explanation



n m

S

1 1where

2 2pooled x y

n m

n m n m

S S S

2

2 2max max1 1a a

pooled

a x yT t a

a an m

S

: 1 1

pooled

a x a yt a

a an m

Note

S

is the test statistic for testing:

0 : vs :x y A x yH a a a H a a a

Popn A

Popn B

X1

X2


Popn A

Popn B

X1

X2

Univariate test for X1

Popn A

Popn B

X1

X2

Univariate test for X2

Popn A

Popn B

X1

X2

Univariate test for a1X1 + a2X2

Mahalanobis distance

A graphical explanation

22

1

,p

i ii

d a b a b a b a b

Euclidean distance

a

points equidistant

from a

2 ,Md a b a b a b

Mahalanobis distance: , a covariance matrix

a

points equidistant

from a


2 1 21 1, ,pooled M pooledT x y x y d x y

n m

S S


n m

S

1pooled

nmx y x y

n m

S

2 , ,M pooled

n md x y

nm

S

Popn A

Popn B

X1

X2

Case I

Popn A

Popn B

X1

X2

Case II

Popn A

Popn B

X1

X2

Case I

Popn A

Popn B

X1

X2

Case II

In Case I the Mahalanobis distance between the mean vectors is larger than in Case II, even though the Euclidean distance is smaller. In Case I there is more separation between the two bivariate normal distributions

Discrimination and Classification

Discrimination

Situation:

We have two or more populations 1, 2, etc

(possibly p-variate normal).

The populations are known (or we have data from each population)

We have data for a new case (population unknown) and we want to identify the which population for which the new case is a member.

Examples Population 1 and 2 Measured variables X1, X2, X3, ... , Xn

1. Solvent and distressed Total assets, cost of stocks and bonds, property-liability market value of stocks and bonds, loss insurance companies expenses, surplus, amount of premiums written. 2. Nonulcer dyspeptics (those Measures of anxiety, dependence, guilt, with stomach problems) and perfectionism. controls ("normal") 3. Federalist papers written by Frequencies of different words and length James Madison and those of sentences. written by Alexander Hamilton 4. Good and poor credit risks. Income, age, number of credit cards, family size education 5. Succesful and unsuccessful Entrance examination scores, high-school grade- (fail to graduate) college point average, number of high-school activities students 6. Purchasers and Non purchasers. Income, Education, family size, previous of a home computer purchase of other home computers, Occupation 7. Two species of chickweed Sepal length, Petal length, petal cleft depth, bract length, sreious tip length, sacrious tip length, pollen diameter

The Basic Problem

Suppose that the data from a new case x1, … , xp has joint density function either :

1: f(x1, … , xn) or

2: g(x1, … , xn)

We want to make the decision to

D1: Classify the case in 1 (f is the correct distribution) or

D2: Classify the case in 2 (g is the correct distribution)

The Two Types of Errors

1. Misclassifying the case in 1 when it actually lies in 2.

Let P[1|2] = P[D1|2] = probability of this type of error

2. Misclassifying the case in 2 when it actually lies in 1.

Let P[2|1] = P[D2|1] = probability of this type of error

This is similar Type I and Type II errors in hypothesis testing.

Note:

1. C1 = the region were we make the decision D1.

(the decision to classify the case in 1)

A discrimination scheme is defined by splitting p –dimensional space into two regions.

2. C2 = the region were we make the decision D2.

(the decision to classify the case in 2)

1. Set up the regions C1 and C2 so that one of the probabilities of misclassification , P[2|1] say, is at some low acceptable value . Accept the level of the other probability of misclassification P[1|2] = .

There can be several approaches to determining the regions C1 and C2. All concerned with taking into account the probabilities of misclassification P[2|1] and P[1|2]

2. Set up the regions C1 and C2 so that the total probability of misclassification:

P[Misclassification] = P[1] P[2|1] + P[2]P[1|2]

is minimized

P[1] = P[the case belongs to 1]


3. Set up the regions C1 and C2 so that the total expected cost of misclassification:

E[Cost of Misclassification]

= c2|1P[1] P[2|1] + c1|2 P[2]P[1|2]

is minimized



c2|1= the cost of misclassifying the case in 2 when the case belongs to 1.

c1|2= the cost of misclassifying the case in 1 when the case belongs to 2.

4. Set up the regions C1 and C2 The two types of error are equal:

P[2|1] = P[1|2]

Computer security:

P[2|1] = P[identifying a valid user as an imposter]

P[2] = P[imposter]

1: Valid users

2: Imposters

c1|2= the cost of identifying the user as a valid user when the user is an imposter.

P[1|2] = P[identifying an imposter as a valid user ]

P[1] = P[valid user]

c2|1= the cost of identifying the user as an imposter when the user is a valid user.

This problem can be viewed as an Hypothesis testing problem

P[2|1] =

H0:1 is the correct population

HA:2 is the correct population

P[1|2] =

Power = 1 -

The Neymann-Pearson Lemma Suppose that the data x1, … , xn has joint density function

f(x1, … , xn ;)

where is either 1 or 2.Let

g(x1, … , xn) = f(x1, … , xn ;1) and

h(x1, … , xn) = f(x1, … , xn ;2)

We want to test

H0: = 1 (g is the correct distribution) against

HA: = 2 (h is the correct distribution)

The Neymann-Pearson Lemma states that the Uniformly Most Powerful (UMP) test of size is to reject H0 if:

2 1

1 1

, ,

, ,n

n

L h x xk

L g x x

and accept H0 if:

2 1

1 1

, ,

, ,n

n

L h x xk

L g x x

where k is chosen so that the test is of size .

Proof: Let C be the critical region of any test of size . Let

1*

11

, ,, ,

, ,n

nn

h x xC x x k

g x x

*

1 1, , n n

C

g x x dx dx

1 1, , n n

C

g x x dx dx

Note: * * *C C C C C

* *C C C C C

We want to show that

*

1 1, , n n

C

h x x dx dx

1 1, , n n

C

h x x dx dx

hence *

1 1, , n n

C

g x x dx dx

1 1, , n n

C

g x x dx dx and

*

1 1, , n n

C C

g x x dx dx

*

1 1, , n n

C C

g x x dx dx

*

1 1, , n n

C C

g x x dx dx

*

1 1, , n n

C C

g x x dx dx

Thus *

1 1, , n n

C C

g x x dx dx

*

1 1, , n n

C C

g x x dx dx

*C*C C*C C

C

*C C

*

1 1, , n n

C C

g x x dx dx

*

1 1, , n n

C C

g x x dx dx

and

*

1 1, , n n

C C

g x x dx dx

*

1 1, , n n

C C

g x x dx dx

*

1 1

1, , n n

C C

h x x dx dxk

*1 1

1since , , , , in .n ng x x h x x C

k

*

1 1

1, , n n

C C

h x x dx dxk

*1 1

1since , , , , in .n ng x x h x x C

k

Thus *

1 1, , n n

C C

h x x dx dx

*

1 1, , n n

C C

h x x dx dx

*

1 1, , n n

C

h x x dx dx

1 1, , n n

C

h x x dx dx

and

when we add the common quantity

*

1 1, , n n

C C

h x x dx dx

to both sides.Q.E.D.

Fishers Linear Discriminant Function.

Suppose that x1, … , xp is either data from a p-variate Normal distribution with mean vector:

111 12

/ 2 1/ 2

1

2

x x

pf x e

The covariance matrix is the same for both populations 1 and 2.

1 2 or

112 22

/ 2 1/ 2

1

2

x x

pg x e

111 12

112 22

/ 2 1/ 2

/ 2 1/ 2

1

21

2

x x

p

x x

p

ef x

g x e

The Neymann-Pearson Lemma states that we should classify into populations 1 and 2 using:

1 11 12 2 1 12 2x x x xe

That is make the decision

D1 : population is 1

if ≥ k

1 11 12 2 1 12 2or ln lnx x x x k

or 1 12 2 1 1 2lnx x x x k

1 1 12 2 22x x x

1 1 11 1 12 2lnx x x k

1 1 111 2 1 1 2 22lnx k

or

and

a x K

1 1 111 2 1 1 2 22 and lna K k

Finally we make the decision

D1 : population is 1

if

where

11 2a x x

The function

Is called Fisher’s linear discriminant function

11 2a x x K

1

21

2

11 2a x x x S x

In the case where the populations are unknown but estimated from data

Fisher’s linear discriminant function

1201008060402000

100

200

A Pictorial representation of Fisher's procedure for two populations

x

x

1

2Classify as

Classify as

1

2

1 2

Example 1

1 : Riding-mower owners 2 : Nonowners

x1 (Income x2 (Lot size x1 (Income x2 (Lot size in $1000s) in 1000 sq ft) in $1000s) in 1000 sq ft) 20.0 9.2 25.0 9.8 28.5 8.4 17.6 10.4 21.6 10.8 21.6 8.6 20.5 10.4 14.4 10.2 29.0 11.8 28.0 8.8 36.7 9.6 16.4 8.8 36.0 8.8 19.8 8.0 27.6 11.2 22.0 9.2 23.0 10.0 15.8 8.2 31.0 10.4 11.0 9.4 17.0 11.0 17.0 7.0 27.0 10.0 21.0 7.4

403020104

8

12

Riding Mower ownersNon ownwers

Income (in thousands of dollars)

Lot

Siz

e (i

n th

ousa

nds

of s

quar

e fe

et)

Example 2Annual financial data are collected for firms approximately 2 years prior to bankruptcy and for financially sound firms at about the same point in time. The data on the four variables

• x1 = CF/TD = (cash flow)/(total debt), • x2 = NI/TA = (net income)/(Total assets), • x3 = CA/CL = (current assets)/(current liabilties, and • x4 = CA/NS = (current assets)/(net sales) are given in

the following table.

The data are given in the following table:

Bankrupt Firms Nonbankrupt Firms x1 x2 x3 x4

x1 x2 x3 x4

Firm CF/TD NI/TA CA/CL CA/NS Firm CF/TD NI/TA CA/CL CA/NS 1 -0.4485 -0.4106 1.0865 0.4526 1 0.5135 0.1001 2.4871 0.5368 2 -0.5633 -0.3114 1.5314 0.1642 2 0.0769 0.0195 2.0069 0.5304 3 0.0643 0.0156 1.0077 0.3978 3 0.3776 0.1075 3.2651 0.3548 4 -0.0721 -0.0930 1.4544 0.2589 4 0.1933 0.0473 2.2506 0.3309 5 -0.1002 -0.0917 1.5644 0.6683 5 0.3248 0.0718 4.2401 0.6279 6 -0.1421 -0.0651 0.7066 0.2794 6 0.3132 0.0511 4.4500 0.6852 7 0.0351 0.0147 1.5046 0.7080 7 0.1184 0.0499 2.5210 0.6925 8 -0.6530 -0.0566 1.3737 0.4032 8 -0.0173 0.0233 2.0538 0.3484 9 0.0724 -0.0076 1.3723 0.3361 9 0.2169 0.0779 2.3489 0.3970 10 -0.1353 -0.1433 1.4196 0.4347 10 0.1703 0.0695 1.7973 0.5174 11 -0.2298 -0.2961 0.3310 0.1824 11 0.1460 0.0518 2.1692 0.5500 12 0.0713 0.0205 1.3124 0.2497 12 -0.0985 -0.0123 2.5029 0.5778 13 0.0109 0.0011 2.1495 0.6969 13 0.1398 -0.0312 0.4611 0.2643 14 -0.2777 -0.2316 1.1918 0.6601 14 0.1379 0.0728 2.6123 0.5151 15 0.1454 0.0500 1.8762 0.2723 15 0.1486 0.0564 2.2347 0.5563 16 0.3703 0.1098 1.9914 0.3828 16 0.1633 0.0486 2.3080 0.1978 17 -0.0757 -0.0821 1.5077 0.4215 17 0.2907 0.0597 1.8381 0.3786 18 0.0451 0.0263 1.6756 0.9494 18 0.5383 0.1064 2.3293 0.4835 19 0.0115 -0.0032 1.2602 0.6038 19 -0.3330 -0.0854 3.0124 0.4730 20 0.1227 0.1055 1.1434 0.1655 20 0.4875 0.0910 1.2444 0.1847 21 -0.2843 -0.2703 1.2722 0.5128 21 0.5603 0.1112 4.2918 0.4443 22 0.2029 0.0792 1.9936 0.3018 23 0.4746 0.1380 2.9166 0.4487 24 0.1661 0.0351 2.4527 0.1370 25 0.5808 0.0371 5.0594 0.1268

Inference for the mean vector. Univariate Inference Let x 1, x 2, …, x n denote a sample of n from...

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