Reading Neyman's 1933

THE MOST EFFICIENT TESTS OF STATISTICAL HYPOTHESES

THE MOST EFFICIENT TESTS OFSTATISTICAL HYPOTHESES

BY Neyman Pearson

2013-01-28


Table of contents

1 Introductory

2 Outline of General TheoryTwo types errors of hypothesis testingThe likelihood principle

3 Simple HypothesesGeneral problems and solutionsIllustrate Examples

4 Composite Hypotheses-H ′0 has One Degree of FreedomGeneral problems and solutionsIllustrative Examples

5 Composite Hypotheses with C Degrees of FreedomGeneral problems and solutionsIllustrative Examples

6 Conclusion


Introductory

Introductory

The problem of testing statistical is an old one.We appear to finddisagreement between statisticians of ”Whether there is efficienttests or not?”In this article,we will talk about this problem and accept the words” an efficient test of the hypothesis H0 ”. Which I have to refer isthat, here, ”efficient” means ”without hoping to know whether eachseparate hypothesis is true or false, we may search for rules to givernour behaviours with regard to them, in the long run of experience,we shall not be too often wrong.”


Outline of General Theory





Two types errors of hypothesis testing

The likelihood principle



Two types errors of hypothesis testing

Definition (Type I error)

Occurs when the null hypothesis (H0) is true, but is rejected.

Definition (Type II error)

Occurs when the null hypothesis is false, but erroneously fails to berejected.

Table: summary of possible results of any hypothesis test

Reject H0 Don′t reject H0

H0 Type I Error Right decision

H1 Right decision Type II Error




Suppose that the nature of an Event,E, is exactly described bythe values of n variable,

x1, x2, ...xn (2.1)

Suppose now that there exists a certain hypothesis, H0, con-cerning the origin of the event which is such as to determinethe probability of occurrence of every possible event E. Let

p0 = p0(x1, x2, ...xn) (2.2)

To obtain the probability that the event will give a sample pointfalling into a particular region, say ω, we shall have either totake the sum of (2.1) over all sample points included in ω, orto calculate the integral,

P0(ω) =

∫...

∫ωp0(x1, x2, ...xn)dx1dx2...dxn (2.3)




we shall assume that the sample points may fall anywhere withina continuous sample space (which may be limited or not),whichwe shall denote by W. It will follow that

P0(W ) = 1 (2.4)




the criterion of likelihood

If H0 is a simple hypothesis,denoted Ω the set of all simple hy-potheses.

Take any sample point,Σ, with co-ordinates (x1, x2, ...xn) and con-sider the set AΣ, of probabilities pH(x1, x2, ...xn) corresponding tothis sample point and determined by different simple hypothesesbelonging to Σ. We shall suppose that whatever be the samplepoint the set AΣ is bounded. Denote pΩ(x1, x2, ...xn) the upperbound of the set AΣ, then if H0 determine the elementary probabil-ity p0(x1, x2, ...xn), we have defined its likelihood to be:

λ =p0(x1, x2, ...xn)

pΩ(x1, x2, ...xn)(2.5)




the criterion of likelihood

If H0 is a composite hypothesis, then it will be possible to specifya part of the set Ω, say ω, such that every simple hypothesisbelonging to the sub-set ω

Analogously, we denote AΣ(ω)the sub-set of AΣ, corresponding tothe set ω of simple hypotheses belonging to H0, then the likelihoodis

λ =pω(x1, x2, ...xn)

pΩ(x1, x2, ...xn)(2.6)




The use of the principle of likelihood in testing hypotheses, consistsin accepting for critical regions those determined by the inequality:

λ ≤ C = const (2.7)




Example of sample space of two dimensions

Fig:2.1




Key Point

As we know,if there are may alternative tests for the samehypothesis, the difference between them consists in the differencein critical regions.That is to say, our problem would becomepicking out the best critical regions for the same P0(ω) = ε, thesolution will be the maximum of P1(ω)

For example, assume ω1 and ω2 are two critical regions described inthe Fig:2.1 , if,

P1(ω2) > P1(ω1)

Then, ω2 is better than ω1.


Simple Hypotheses

Simple Hypotheses


Simple Hypotheses

General problems and solutions

We shall now consider how to find the best critical region for H0,with regard to a single alternative H1, this will be the region ω0 forwhich P1(ω0) is a maximum subject to the condition that,

P0(ω0) = ε (3.1)

Suppose the probability laws for H0 and H1, namely, p0(x1, x2, ...xn)and p1(x1, x2, ...xn),exist, are continuous and not negative through-out the whole sample space W ; further that,

P0(W ) = P1(W ) = 1 (3.2)

The problem will consist in finding an unconditioned minimum ofthe expression

P0(ω0)−k ∗P1(ω0) =

∫...

∫ω0

p0(x1, x2, ...xn)−k ∗p1(x1, x2, ...xn)dx1dx2...dxn

(3.3)

where k being a constant afterwards to be determined by the con-dition(3.1)


Simple Hypotheses


We shall now show that the necessary and sufficient condition for aregion ω0, being the best critical region for H0, which is defined bythe inequality,

p0(x1, x2, ...xn) ≤ k ∗ p1(x1, x2, ...xn) (3.4)


Simple Hypotheses


Denote by ω0 the region defined by (3.4) and let ω1 be any otherregion satisfying the condition P0(ω1) = P0(ω0) = ε. These regionsmay have a common part ω01, which represented in Fig:3.1

Fig:3.1

It follow that,

P0(ω0 − ω01) = ε− P0(ω01) = P0(ω1 − ω01) (3.5)

and consequently,

k ∗P1(ω0−ω01) ≥ P0(ω0−ω01) = P0(ω1−ω01) ≥ k ∗P1(ω1−ω01)(3.6)

If we add k ∗ P1(ω01) to both side,we obtain,

P1(ω0) ≥ P1(ω1) (3.7)


Simple Hypotheses

Illustrate Examples

Case of the Normal Distribution )Example 1

Suppose that it is known that a sample of nindividuals,x1, x2, ...xn, has been drawn randomly from somenormally distribution with standard deviation σ = σ0. Let x and sbe the mean and standard deviation of the sample.It is desired to test the hypothesis H0 that whether the mean inthe sampled population is a = a0 or not.

Then the probabilities of its occurrence determined by H0 and byH1 will be,

p0(x1, x2, ...xn) = (1

σ0

√2π

)ne−n

(x− a0)2 + s2

2σ20 (3.8)

p1(x1, x2, ...xn) = (1

σ0

√2π

)ne−n

(x− a1)2 + s2

2σ20 (3.9)


Simple Hypotheses

Illustrate Examples

The criterion in this case becomes,

p0

p1= e−n

(x− a0)2 + (x− a1)2

2σ20 = k (3.10)

From this it follows that the best critical region for H0 with regardto H1, defined by the inequality (2.7),

(a0 − a1)x ≤ 1

2(a2

0 − a21) +

σ20

nlog k = (a0 − a1)x0(say) (3.11)

The critical value for the best critical value will be,

x = x0 (3.12)


Simple Hypotheses

Illustrate Examples

Now two cases will ariseµ(1)a1 < a0, then the region is defined by, x ≤ x0

(2)a1 > a0, then the region is defined by, x ≥ x0

where x0 the statistic which is easy to decide in the practice.

Conclusion

The test obtained by finding the best critical region is in fact theordinary test for the significance of a variation in the mean of asample; but the method of approach helps to bring out clearly therelation of the two critical regions x ≤ x0 and x ≥ x0. Further, thetest of this hypothesis could not be improved by using any otherform of criterion or critical region.


Simple Hypotheses

Illustrate Examples

Case of the Normal Distribution )Example 2

Suppose that it is known that a sample of nindividuals,x1, x2, ...xn, has been drawn randomly from somenormally distribution with mean a = a0 = 0. Let x and s be themean and standard deviation of the sample.It is desired to test the hypothesis H0 that whether the standarddeviation in the sampled population is σ = σ0 or not.

Analogously,it is easy to show that the best critical region with regardto H1 is defined by the inequality,

1

n

n∑i=1

(x2i )(σ

20 − σ2

1) = (x2 + s2)(σ20 − σ2

1) ≤ v2(σ20 − σ2

1) (3.13)

where v is a constant depending on ε, σ0, σ1


Simple Hypotheses

Illustrate Examples

Again two cases will arise,(1)σ1 < σ0, then the region is defined by, x2 + s2 ≤ v2

(2)σ1 > σ0, then the region is defined by, x2 + s2 ≥ v2

In this case, there are two ways to define the criterion,suppose thatm2 = x2 + s2 and ψ = χ2.Then,

m2 = σ20ψ/n, d = n (3.14)

s2 = σ20ψ/n− 1, d = n− 1 (3.15)


Simple Hypotheses

Illustrate Examples

It is of interest to compare the relative efficiency of the two criterionsm2 and s2 in reducing errors of the second type. If H0 is false,suppose the true hypothesis to be H1 relating to a population inwhich

σ1 = hσ0 > σ0 (3.16)

In testing H0 with regard to the class of alternatives for which σ >σ0, we should determine the critical value of ψ0, so that

P0(ψ ≥ ψ0) =

∫ +∞

ψ0

p(ψ)dψ = ε (3.17)


Simple Hypotheses

Illustrate Examples

When H1 is true, the Type II Error will be P1(ψ ≤ ψ0), and thisvalue will be different according to the two different criterions. Nowsuppose that ε = 0.01 and n=5, the position is shown in Fig:3.2

Fig:3.1


Simple Hypotheses

Illustrate Examples

(1)Using m2, with degrees of freedom 5, find that ψ0 = 15.086,thus the Type II Error will be,

ifh = 2, σ1 = 2σ0, P1(ψ ≤ ψ0) = 0.42

ifh = 3, σ1 = 3σ0, P1(ψ ≤ ψ0) = 0.11(3.18)

(2)Usings2, with degrees of freedom 4, find that ψ0 = 13.277, thusthe Type II Error will be,

ifh = 2, σ1 = 2σ0, P1(ψ ≤ ψ0) = 0.49

ifh = 3, σ1 = 3σ0, P1(ψ ≤ ψ0) = 0.17(3.19)

Conclusion

The second test is better, and the choice of criterion candistinguish test.


Composite Hypotheses-H′0 has One Degree of Freedom


Composite Hypotheses-H ′0 has OneDegree of Freedom




Definition

Suppose that W a common sample space for any admissiblehypothesis Ht, p0(x1, x2, ...xn) is the probability law for thissample who is dependent upon the value of c+d parametersα(1), α(2), ...α(c);α(c+1)...α(c+d)(d parameters are specified and cunspecified), ω is a sub-set of simple hypothesis for testing acomposite hypothesis H ′0, such that

P0(ω) =

∫ ∫...

∫ωp0(x1, x2, ...xn)dx1dx2...dxn = constant = ε

(4.1)and P0(ω) is independent of all the values of α(1), α(2), ...α(c). Weshall speak of ω as a region of ”size” ε, similar to W with regardto the c parameters.




Similar regions

We shall commence the problem with simple case for which thedegree of freedom of parameters is one.we suppose that the set Ω of admissible hypotheses defines thefunctional form of the probability law for a given sample, namely

p(x1, x2, ...xn) (4.2)

but that this law is dependent upon the values of 1+d parameters

α(1);α(2)0 , ...α

(d+1)0 ; (4.3)

A composite hypothesis, H ′0, of one degrees of freedom, its proba-bility law is denoted by

p0 = p0(x1, x2, ...xn) (4.4)




Conditions of the problem of similar regions

We have been able to solve the problem of similar regions only undervery limiting conditions concerning p0. These are as follows :(a) p0 is indefinitely differentiable with regard to α(1) for all valuesof α(1) and in every point of W, except perhaps in points forming a

set of measure zero. That is to say, we suppose that∂kp0

∂(α(1))kexists

for any k = 1, 2, ...and is integrable over the region W.Denote by,

φ =∂ log p0

∂α(1)=

1

p0

∂p0

∂α(1);φ′ =

∂φ

∂α(1)(4.5)

(b)The function p0 satisfies the equation

φ′ = A+Bφ, (4.6)

where the coefficients A and B are functions of α(1) but are inde-pendent of x1, x2, ...xn.




Similar regions for this case

If the probability law p0 satisfies the two conditions (a) and (b),then it follows that a necessary and sufficient condition for ω to besimilar to W with regard to α(1) is that

∂kP0(ω)

∂(α(1))k=

∫ ∫...

∫ω

∂kp0

∂(α(1))kdx1dx2...dxn, k = 1, 2, ... (4.7)

When k=1,∂p0

∂α(1)= p0φ (4.8)

then,∂P0(ω)

∂α(1)=

∫ ∫...

∫ωp0φdx1dx2...dxn = 0 (4.9)




By the same way, when k=2

∂2p0

∂(α(1))2=

∂

∂α(1)

2

(p0φ) = p0(φ2 + φ′), (4.10)

∂2P0(ω)

∂(α(1))2=

∫ ∫...

∫ωp0(φ2 + φ′)dx1dx2...dxn = 0 (4.11)

Using (4.6),we may have

∂2P0(ω)

∂(α(1))2=

∫ ∫...

∫ωp0(φ2 +A+Bφ)dx1dx2...dxn = 0 (4.12)

Having regard to (4.6) and (4.10), it follows that∫ ∫...

∫ωp0φ

2dx1dx2...dxn = −Aε = εψ2(α(1))(say) (4.13)




When k=3, by differentiating(4.12)

∂3P0(ω)

∂(α(1))3=

∫ ∫...

∫ωp0(φ3 +3Bφ2 +(3A+B2 +B′)φ+A′ +AB)dx1dx2...dxn = 0

(4.14)

Again,having regard to (4.6),(4.10) and (4.13), it follows that∫ ∫...

∫ωp0φ

3dx1dx2...dxn = (3AB−A′−AB)ε = εψ3(α(1))(say)

(4.15)Using the method of induction, en general, we shall find,∫ ∫

...

∫ωp0φ

kdx1dx2...dxn = εψk(α(1))(say) (4.16)

where ψk is a function of α(1) but independent of the sample.Since ω is similar to W, it follows that,

1

ε

∫ ∫...

∫ωp0φ

kdx1dx2...dxn =

∫ ∫...

∫Wp0φ

kdx1dx2...dxn, k = 1, 2, 3...

(4.17)




Now we consider the significance above in the following way.Suppose φ = constant = φ1

Then if

P0(ω(φ)) =

∫ ∫...

∫ω(φ1)

p0dω(φ1) (4.18)

P0(W (φ)) =

∫ ∫...

∫W (φ1)

p0dW (φ1) (4.19)

represent the integral of p0 taken over the common parts, ω(φ) andW (φ) of φ = φ1 and ω and W respectively, it follows that if ω besimilar to W and of size ε,

P0(ω(φ)) = εP0(W (φ)) (4.20)




Choice of the best critical region

Let Ht be an alternative simple hypothesis to H0. We shall assumethat regions similar to W with regard to α(1) do exist. Then ω0, thebest critical region for H0 with regard to Ht, must be determinedto maximise

Pt(ω0) =

∫ ∫...

∫ωp0(x1, x2, ...xn)dx1dx2...dxn (4.21)

subject to the condition (4.20) holding for all values of φ.




Then the problem of finding the best critical region ω0 is reduce tofind parts ω0(φ) of W (φ), which will maximise P (ω(φ)) subject tothe condition (4.20).This is the same problem that we have treated already when dealingwith the case of a simple hypothesis except that instead of theregions ω0 and W, what we have are the regions ω0(φ) and W (φ).The inequality

pt ≥ k(φ)p0 (4.22)

will therefore determine the region ω0(φ), where k(φ) is a constantchosen subject to the condition (4.20)



Illustrative Examples

Example 5

A sample of n has been drawn at random from normal population,and H ′0 is the composite hypothesis that the mean in thispopulation is a = a0, σ being unspecified.

p(x1, x2, ...xn) = (1

σ√

2π)ne−n

(x− a)2 + s2

2σ2 (4.23)

For H ′0,α(1) = σ, α(2) = a0 (4.24)

φ =∂p0

∂σ= −n

σ+ n

(x− a)2 + s2

4σ3(4.25)

φ′ =n

σ2+ 3n

(x− a)2 + s2

σ4= −2n

σ2− 3

σφ (4.26)

H0 is an alternative hypothesis that α(1)t = σ, α

(2)t = a0.




It is seen that the condition pt ≥ kp0, becomes

1

σnte−

(n[x− a)2 + s2])

2σ2t ≥ k 1

σne−

(n[x− a0)2 + s2])

2σ2 (4.27)

This can be reduces to,

x(at−a0) ≥ 1

nσ2t log k+0.5(a2

t−a20) = (at−a0)k1(φ)(say) (4.28)

Two cases must be distinguished in determining ω0(φ)(a)at > a0, then x ≥ k1(φ)(b)at < a0, then x ≤ k1(φ)where k1(φ) has to be chosen so that (4.20) is satisfied.




In the case n=3, the position of ω0(φ) is indicated in Figµ4.1

Fig:4.1

Then there will be two families of these cones containing the bestcritical regions:(a)For the class of hypotheses at > a0; the cones will be in thequadrant of positive x’s.(b)For the class of hypotheses at < a0; the cones will be in thequadrant of negative x’s.




conclusion

In the case n > 3,we may either appeal to the geometry ofmultiple space,the calculation of this situation may be a littlecomplicated, but we can get similar result.Further, it has now beenshown that starting with information in the form supposed, therecan be no better test for the hypothesis under consideration.


Composite Hypotheses with C Degrees of Freedom


Composite Hypotheses with CDegrees of Freedom




Similar regions

We shall now consider a probability function depending upon c pa-rameters α,this will correspond to a composite hypothesis H ′0, withc degrees of freedom.

Definition

Let ω be any region in the sample space W,and denote byP (α(1), α(2), ...α(c)ω) the integral of p(x1, x2, ...xn) over the

region ω. Fix any system of values of the α’s,say α(1)A , α

(2)A , ...α

(c)A ,

If the region ω has the property, that

P (α(1)A , α

(2)A , ...α

(c)A ω) = ε = constant (5.1)

whatever be the system A, we shall say that it is similar to thesample space W with regard to the set of parametersα(1), α(2), ...α(c) and of size ε.





Definition

Letfi(α, x1, x2, ...xn) = Ci(i = 1, 2, ...k < n), (5.2)

be the equations of certain hypersurfaces, α and Ci beingparameters. Denote by S(α,C1, C2, ...Ck) the intersection of thesehypersurfaces, and F (α) denote the family of S(α,C1, C2, ...Ck)corresponding to a fixed values α’ and to different systems ofvalues of the C’s.Take any S(α(1), C ′1, C

′2, ...C

′k) from any family

F (α1). If whatever be α2 it is possible to find suitable values ofthe C’s, for example C ′′1 , C

′′2 , ...C

′′k , such that S(α(1), C ′′1 , C

′′2 , ...C

′′k )

is identical with S(α(1), C ′1, C′2, ...C

′k), then we shall say that the

family F (α) is independent of α.





It is now possible to solve the problem of finding regions similar toW with regard to a set of parameters α(1), α(2), ...α(c), we are atstill only able to do so under rather limiting conditions.

(a) We shall assume the existence of∂kp0

∂(α(i))kin every point of the

sample space.

(b)Writing

φi =∂ log p0

∂α(i)=

1

p0

∂p0

∂α(i);φ′i =

∂φi

∂α(i); (5.3)

it will be assumed that for every i = 1, 2, ... c.

φ′i = Ai +Biφi, (5.4)

Aiand Bi being independent of the x’s.




(c)Further there will need to be conditions concerning the φi =const.Denote by S(α(i), C1, C2, ...Ci−1) the intersection that φj =Cj for any j=1,2,...i-1, corresponding to fixed values of the α’s andC’s. F (α(i)) will denote the family of S(α(i), C1, C2, ...Ci−1) corre-sponding to fixed values of the α’s and to different systems of valuesof the C’s.We shall assume that any family F (α(i)) is independent of α(i) fori =2, 3, ...c.




Similar regions

Property

If the above conditions are satisfied, then the necessary andsufficient condition for ω being similar to W with regard to the setof parameters α(1), α(2), ...α(c), and of any given size ε is that∫

...

∫ω(φ1,φ2,...φc)

pdω(φ1, φ2, ...φc) = ε

∫...

∫W (φ1,φ2,...φc)

pdW (φ1, φ2, ...φc)

(5.5)

where W (φ1, φ2, ...φc) means the intersection in W of φi = Ci forany j=1,2,...i-1, corresponding to fixed values of the α’s and C’s.The symbol ω(φ1, φ2, ...φc) means the part of W (φ1, φ2, ...φc)included in ω.




The Determination of the best critical region

The choice of the best critical regions in this case is similar with wealready discuss before, it is simple to identify that best critical regionω0 of size ε with regard to a simple alternative Ht must satisfy thefollowing conditions:(1) ω0 must be similar to W with regard to the set of parametersα(1), α(2), ...α(c),that is to say,

P0(ω0) = ε (5.6)

must be independent of the values of the α’s.(2)If ν be any other region of size ε similar to W with regard to thesame parameters,

Pt(ω0) ≥ Pt(ν) (5.7)




In this way the problem of finding the best critical region for testingH ′0 is reduced to that of maximising

Pt(ω(φ1, φ2, ...φc)) (5.8)

under the condition (5.5) for every set of values

φ1 = C1, φ2 = C2, ...φc = Cc, (5.9)




Example 6 - Test for the significance of the difference between two variances

We suppose that two samples,

(1) Σ1 of size n1,mean = x1,standard deviation = s1.

(2) Σ2 of size n2,mean = x2,standard deviation = s2. have been drawn at random

from normal distribution.If this is so, the most general probability law for the observed

event may be written

p(x1, x2, ...xn1; xn1+1, xn1+2, ...xN ) = (

1√

2π)N 1

σn11σn22

e

−n1(x1 − a1)2 + s21

2σ21

−n2(x2 − a2)2 + s22

2σ22

(5.10)

where n1 + n1 = N , and a1, σ1 are the mean and standard deviation of the first, a2,

σ2 are the mean and standard deviation of the second sampled distribution. H′0 is the

composite hypothesis that σ1 = σ2. For an alternative Ht,the parameters may be

defined as

α1 = a1;α2 = a2 − a1 = bt;α3 = σ1;α4 =σ2

σ1= θt. (5.11)




For the hypothesis to be tested, H ′0:

α1 = a;α2 = b;α3 = σ;α4 = 1. (5.12)

H ′0 it will be seen, is a composite hypothesis with 3 degrees offreedom, a, b and σ being unspecified.




We shall now consider whether the conditions (A), (B) and (C) ofthe above theory are satisfied.(A) The first condition is obviously satisfied.(B) Making use of (5.10),we find that

log p0 = −N log√

2π−N log σ−1

2σ2n1(x1−a)2 +n2(x2−a− b)2 +n1s

21 +n2s

22

(5.13)

φ1 =∂ log p0

∂a=

1

σ2n1(x1 − a) + n2(x2 − a− b) (5.14)

φ2 =∂ log p0

∂b=

1

σ2n2(x2 − a− b) (5.15)

φ3 =∂ log p0

∂σ= −

N

σ+N

1

σ3n1(x1− a)2 +n2(x2− a− b)2 +n1s

21 +n2s

22 (5.16)




We see that

φ′1 = A1, φ′2 = A2, φ

′3 = A3 +B3φ3, (5.17)

where the A’s and B’s are independent of the sample variates, sothat the condition (B) is satisfied.(C) The equation S(α(2), C1) namely,φ1 = C1,where C1 is an ar-bitrary constant, is obviously equivalent to n1x1 + n2x2 = C ′1 de-pending only upon one arbitrary parameter C1. Hence F (α(2)) is in-dependent of α(2). Similarly, F (α(3)) is independent of α(3). Hencealso the condition (C) is fulfilled.




We may now attempt to construct the best critical region ω0.Itselements ω0(φ1, φ2, φ3) are parts of the W(φ1, φ2, φ3) satisfying thesystem of three equations φi = Ci(i = 1, 2, 3), within which

pt ≥ k(x1, x2, sa)p0 (5.18)

where sa =1

Nn1(s2

1 + n2s22) and the value of k being determined

for each system of values of x1, x2, sa, so that

P0(ω0(φ1, φ2, φ3)) = εP0(W0(φ1, φ2, φ3)) (5.19)

The condition (5.19) becomes

1

σNe

−n1[(x1 − a)2 + s21] + n2[(x2 − a− b)2 + s2

2]

2σ2 ≥

k

σN1 θn2e

−n1[(x1 − a1)2 + s21] + n2θ

2[(x2 − a1 − b1)2 + s22]

2σ21

(5.20)




Since the region determined by (5.19) will be similar to W withregard to a, b and σ, we may put a = a1, b = b1, σ = σ1, and thecondition (5.20) will be found on taking logarithms to reduce to

n2(x2− a1− b1)2 + s22(1− θ2) ≤ 2σ2

1θ2(log k−n2 log θ) (5.21)

This inequality contains only one variable s22. Solving with regard

to s22 we find that the solution will depend upon the sign of the

difference 1 − θ2. Accordingly we shall have to consider separatelythe two classes of alternatives,

θ =σ2

σ1> 1; the B.C.R will be defined by s2

2 ≥ k′1(x1, x2, sa)

θ =σ2

σ1< 1; the B.C.R will be defined by s2

2 ≥ k′2(x1, x2, sa)




By some technical calculation, we can arrive at the criterion µ that,

θ =σ2

σ1> 1;µ =

n2s22

n1s21 + n2s2

2

≥ µ0 (5.22)

θ =σ2

σ1< 1;µ =

n2s22

n1s21 + n2s2

2

≤ µ′0 (5.23)

where The constant µ0 depends only upon n1, n2 value of ε chosen.

Conclusion

(1) We see that the best critical regions are common for all thealternatives.(2) the criterion µ we have reached,is equivalent to that suggestedon intuitive grounds by FISHER.


Conclusion

Conclusion

1.The conception of the best critical region we talk about in thisarticle,is the region with regard to the alternative hypothesis Ht,for a fixed value of ε, assures the minimum chance of acceptingH0, when the true hypothesis is Ht.2.To solve the same problem in the case where the hypothesistested is composite, the solution of a further problem isrequired;this consists in determining what has been called a regionsimilar to the sample space with regard to a parameter.We have been able to solve this problem only under certainlimiting conditions; at present, therefore, these conditions alsorestrict the generality of the solution given to the problem of thebest critical region for testing composite hypotheses.


Conclusion

Reference

[1] NEYMAN.”Contribution to the Theory of Certain Test Criteri-a”,Bull.Inst.int.Statist 1928.[2] FISHER.”Statistical methods for Research Workers”,London 1932.[3] PEARSON and NEYMAN.”On the Problem of Two Samples”,Bull.Acad. Polon. Sci. Lettres 1930.[4] Lehmann EL.”The Fisher, Neyman and Pearson theories of test-ing hypotheses: one theory or two? ” J Am Stat Assoc 1993.[5] Berkson J.”Tests of significance considered as evidence”, J AmStat Assoc 1942.


Conclusion

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Reading Neyman's 1933

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