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(1)

Classical Hypothesis

Testing Theory

(2)

Review

5 steps of classical hypothesis testing

(Ch. 3)

1. Declare null hypothesis H0 and alternate

hypothesis H1

2. Fix a threshold α for Type I error (1% or 5%)

• Type I error (α): reject H0 when it is true

Type II error (β): accept H0 when it is false

(3)

Review

4. Determine what observed values of the test statistic should lead to rejection of H0

Significance point K (determined by α)

5. Test to see if observed data is more extreme than significance point K

• If it is, reject H0

(4)

Overview of Ch. 9

Simple Fixed-Sample-Size Tests

Composite Fixed-Sample-Size TestsThe -2 log λ Approximation

The Analysis of Variance (ANOVA)

Multivariate Methods

ANOVA: the Repeated Measures Case

Bootstrap Methods: the Two-sample t

(5)
(6)

The Issue

In the simplest case, everything is specified

– Probability distribution of H0 and H1 • Including all parameters

α (and K)

But: β is left unspecified

(7)

Most Powerful Procedure

Neyman-Pearson Lemma

States that the likelihood-ratio (LR) test is the

most powerful test for a given α

The LR is defined as:

where

• f0, f1 are completely specified density functions for

H0,H1

• X1, X2, … Xn are iid random variables

(8)

Neyman-Pearson Lemma

H0 is rejected when LRK

With a constant K chosen such that:

P(LR ≥ K when H0 is true) = α

Let’s look at an example using the

Neyman-Pearson Lemma!

(9)

Example

Basketball players seem to

be taller than average

– Use this observation to

formulate our hypothesis H1:

• “Tallness is a factor in the recruitment of KU basketball players”

– The null hypothesis, H0, could

be:

• “No, the players on KU’s team are a just average height compared to the population in the U.S.”

(10)

Example

Setup:

Average height of males in the US: 5’9 ½“Average height of KU players in 2008:

6’04 ½”

Assumption: both populations are

normal-distributed centered on their respective

averages (μ0 = 69.5 in, μ1 = 76.5 in) and σ = 2

Sample size: 3

(11)

Example

The two populations:

height (inches)

p

f0 f

(12)

Example

Our test statistic is the Likelihood Ratio, LR

Now we need to determine a significance

point K at which we can reject H , given α =

) ( ) ( ) ( ) ( ) ( ) ( ) ( 3 0 2 0 1 0 3 1 2 1 1 1 x f x f x f x f x f x f x         2 2 2 2 2 2 2 2 2 2 2 2 8 ) 5 . 69 ( 8 ) 5 . 69 ( 8 ) 5 . 69 ( 8 ) 5 . 76 ( 8 ) 5 . 76 ( 8 ) 5 . 76 ( 2 3 2 2 2 1 2 3 2 2 2 1              x x x x x x e e e e e e       3 1 2 2 ( 76.5)

(13)

Example

So we just need to solve for K’ and calculate K:

How to solve this? Well, we only need one set of

values to calculate K, so let’s pick two and solve for

the third:

• We get one result: K3’=71.0803



  

'

1 2' 3'

05 . 0 ) ( ) ( )

( 1 0 2 0 3 1 2 3

0

K K K

(14)

Example

Then we can just plug it in to Λ and

calculate K:

      3 1 2 ' 2

' 69.5) ( 76.5)

( 8 1

i i i

K K

e K

(68 69.5)2 (68 76.5)2 (71 69.5)2 (71 76.5)2 (71.0803 69.5)2 (71.0803 76.5)2

(15)

Example

With the significance point K = 1.663*10-7 we

can now test our hypothesis based on observations:

E.g.: Sasha = 83 in, Darrell = 81 in, Sherron = 71 in

1.446*1012 > 1.663*10-7

Therefore, our hypothesis that tallness is a factor in

the recruitment of KU basketball players is true.

(16)

Neyman-Pearson Proof

Let A

define region in the joint range

of

X

1

,

X

2

, …

X

n

such that LR ≥

K

.

A

is

the

critical region

.

If A is the only critical region of size α

we are done

Let’s assume another critical region of

(17)

Proof

H0 is rejected if the observed vector (x1,

x2, …, xn) is in A or in B.

Let A and B overlap in region C

Power of the test: rejecting H0 when H1 is

true

The Power of this test using A is:

n n

A L(H1)

f1(u1) f1(u2) f1(u )du1du2du

(18)

Proof

Define: Δ = AL(H1) - BL(H1)

The power of the test using A minus using B

Where A\C is the set of points in A but not in

C

And B\C contains points in B but not in C

 

 

  f1(u1) f1(un)du1dunf1(u1) f1(un)du1dun

A B

 

 

  f1(u1) f1(un)du1dunf1(u1) f1(un)du1dun

C

(19)

Proof

So, in A\C we have:

While in B\C we have:

) ( ) ( ) ( )

( 1 1 0 1 0

1 u f un Kf u f un

f   

) ( ) ( ) ( )

( 1 1 0 1 0

1 u f un Kf u f un

f   

(20)

Proof

Thus

Which implies that the power of the test

 

 

  Kf0(u1) f0(un)du1dunKf0(u1) f0(un)du1dun

 

 

  Kf0(u1) f0(un)du1dunKf0(u1) f0(un)du1dun

C

A\ B\C

A B

K

K  

0

(21)
(22)

Not Identically Distributed

In most cases, random variables are not

identically distributed, at least not in

H

1

This affects the likelihood function, L

– For example, H1 in the two-sample t-test is:

– Where μ1 and μ2 are different

       n i x m i

x i i

e e L 1 2 ) ( 1 2 ) ( 2 2 2 2 2 2 1 1 2 1 2

1

  

(23)

Composite

Further, the hypotheses being tested do

not specify all parameters

They are composite

This chapter only outlines aspects of

(24)

Parameter Spaces

The set of values the parameters of interest

can take

Null hypothesis: parameters in some region ωAlternate hypothesis: parameters in Ω

ω is usually a subspace of ΩNested hypothesis case

– Null hypothesis nested within alternate hypothesis – This book focuses on this case

“if the alternate hypothesis can explain the data

(25)

λ Ratio

Optimality theory for composite tests

suggests this as desirable test statistic:

• Lmax(ω): maximum likelihood when parameters

are confined to the region ω

• Lmax(Ω): maximum likelihood when parameters

are confined to the region Ω, defined by H1

• H0 is rejected when λ is sufficiently small (→

Type I error)

)

(

)

(

max max

L

L

(26)

Example: t-tests

The next slides calculate the

λ

-ratio

for the two sample

t

-test (with the

likelihood)

t-tests later generalize to ANOVA and T2

tests

       n i x m i

x i i

e e L 1 2 ) ( 1 2 ) ( 2 2 2 2 2 2 1 1 2 1 2

1

  

(27)

Equal Variance Two-Sided

t-test

Setup

– Random variables X11,…,X1m in group 1 are

Normally and Independently Distributed (μ1,σ2)

– Random variables X21,…,X2n in group 2 are

NID (μ2,σ2)

– X1i and X2j are independent for all i and j

– Null hypothesis H0: μ1= μ2 (= μ, unspecified)

(28)

Equal Variance Two-Sided

t-test

Setup (continued)

σ2 is unknown and unspecified in H0 and

H1

Is assumed to be the same in both

distributions

Region ω is:

Region {Ω is: , ,0 }

2 2

1         

  

   

} 0

,

{ 12  2  

   

(29)

Equal Variance Two-Sided

t-test

Derivation

H0: writing μ for the mean, when μ1= μ2,

the maximum over likelihood ω is at

And the (common) variance σ2 is n m X X X X X X

X m n

        

 11 12 1 21 22 2

ˆ    n m X X X

X in i

m i i    

1
(30)

Equal Variance Two-Sided

t-test

Inserting both into the likelihood

function, L

2 2

0 max

2

) ˆ 2 (

1 )

(

n m

e

L m n

 

(31)

Equal Variance Two-Sided

t-test

Do the same thing for region Ω

Which produces this likelihood Function,

L

m

X X

X

X 11 12 1m

1 1 ˆ      n X X X

X 21 22 2n

2 2 ˆ      n m X X X

X in i

m i i    

1

2 2 2 1 2 1 1 2 1 ) ( ) ( ˆ  2 2 1 max 2 ) ˆ 2 ( 1 ) ( n m e

L m n

(32)

Equal Variance Two-Sided

t-test

The test statistic λ is then

(33)

Equal Variance Two-Sided

t-test

We can then use the algebraic identity

To show that

Where t is (from Ch. 3)

2 2 1 2 1 2 2 2 1 1 1 2 1 2 2 1

1 ) ( ) ( ) ( ) ( )

( X X

n m mn X X X X X X X X n i i m i i n i i m i i                   2 2 1 2 1 n m n m t              n m S mn X X T  

(34)

Equal Variance Two-Sided

t-test

t is the observed value of TS is defined in Ch. 3 as

2

) (

) (

1

2 2 2

1

2 1 1

2

 

 

n m

X X

X X

S

n

i

i m

i

i

We can plot λ as a

(35)

Equal Variance Two-Sided

t-test

So, by the monotonicity argument, we can

use t2 or |t| instead of λ as test statistic

Small values of λ correspond to large

values of |t|

– Sufficiently large |t| lead to rejection of H0

– The H0 distribution of t is known

t-distribution with m+n-2 degrees of freedom

Significance points are widely available

• Once α has been chosen, values of |t|

(36)

Equal Variance Two-Sided

t-test

.s

oc

r.u

cl

a.

ed

u/

A

pp

le

ts

.d

ir/

T-ta

bl

e.

ht

m

(37)

Equal Variance One-Sided

t-test

Similar to Two-Sided t-test case

– Different region Ω for H1:

• Means μ1 and μ2 are not simply different, but

one is larger than the other μ1 ≥ μ2

If then maximum likelihood

estimates are the same as for the two-sided case

} 0

,

{ 12  2   

   

2

1 x

(38)

Equal Variance One-Sided

t-test

If then the unconstrained maximum

of the likelihood is outside of ω

The unique maximum is at , implying

that the maximum in ω occurs at a boundary point in Ω

• At this point estimates of μ1 and μ2 are equal

• At this point the likelihood ratio is 1 and H0 is

not rejected

• Result: H is rejected in favor of H (μ ≥ μ )

2

1 x

x

) , (x1 x2

(39)

Example - Revised

This scenario fits with our original

example:

H1 is that the average height of KU

basketball players is bigger than for the general population

One-sided test

We could assume that we don’t know the

averages for H0 and H1

We actually don’t know σ (I just guessed 2 in

(40)

Example - Revised

Updated example:

– Observation in group 1 (KU): X1 = {83, 81, 71}

– Observation in group 2: X2 = {65, 72, 70}

– Pick significance point for t from a table: tα =

2.132

t-distribution, m+n-2 = 4 degrees of freedom, α =

0.05

– Calculate t with our observations

185 . 2 7673 .

12 9 . 27 6

2122 .

5

9 ) 69 3

. 78 (

 

 

(41)

Comments

Problems that might arise in other cases

The λ-ratio might not reduce to a function of a

well-known test statistic, such as t

– There might not be a unique H0 distribution of λ

Fortunately, the t statistic is a pivotal quantity

• Independent of the parameters not prescribed by H0

– e.g. μ, σ

For many testing procedures this property does

(42)

Unequal Variance Two-Sided

t-test

Identical to Equal Variance Two-Sided t

-test

Except: variances in group 1 and group 2 are

no longer assumed to be identical

Group 1: NID(μ1, σ12)

Group 2: NID(μ2, σ22)

With σ12 and σ22 unknown and not assumed

identical

• Region ω = {μ1 = μ2, 0 < σ12, σ22 < +∞}

(43)

Unequal Variance Two-Sided

t-test

The likelihood function of (X11, X12, …,

X1m, X21, X22, …, X2n) then becomes

– Under H0 (μ1 = μ2 = μ), this becomes:

      n i x m i

x i i

e e 1 2 ) ( 2 1 2 ) ( 1 2 2 2 2 21 2 1 2 1 1 2 1 2 1        

      n i x m i

x i i

(44)

Unequal Variance Two-Sided

t-test

Maximum likelihood estimates ,

and satisfy the simultaneous equations: 0 ˆ ) ˆ ( ˆ ) ˆ ( 2 2 2 2 1 1           i i x x m x i    2 1 2 1 ) ˆ ( ˆ   n x i    2 2 2 2 ) ˆ ( ˆ  

(45)

Unequal Variance Two-Sided

t-test

–  cubic equation in

Neither the λ ratio, nor any monotonic

function has a known probability distribution when H0 is true!

This does not lead to any useful testing

statistic

The t-statistic may be used as reasonably close • However H0 distribution is still unknown, as it

depends on the unknown ratio σ12/σ22

In practice, a heuristic is often used (see Ch. 3.5)

(46)
(47)

The -2 log λ Approximation

Used when the

λ

-ratio procedure does

not lead to a test statistic whose

H

0

distribution is known

Example: Unequal Variance Two-Sided t

-test

Various approximations can be used

(48)

The -2 log λ Approximation

Best known approximation:

– If H0 is true, -2 log λ has an asymptotic

chi-square distribution,

with degrees of freedom equal to the

difference in parameters unspecified by H0

and H1, respectively.

λ is the likelihood ratio

“asymptotic” = “as the sample size → ∞”

(49)

The -2 log λ Approximation

Restrictions:

• Parameters must be real numbers that can take on values in some interval

• The maximum likelihood estimator is found at a turning point of the function

i.e. a “real” maximum, not at a boundary point

• H0 is nested in H1 (as in all previous slides)

These restrictions are important in the

proof

(50)

The -2 log λ Approximation

Instead:

Our original basketball example, revised

again:

Let’s drop our last assumption, that the variance

in the population at large is the same as in the group of KU basketball players.

All we have left now are our observations and

the hypothesis that μ1 > μ2

– Where μ1 is the average height of Basketball players

(51)

Example – Revised Again

Using the Unequal Variance One-Sided t

-Test

(52)
(53)

The Analysis of Variance

(ANOVA)

Probably the most frequently used

hypothesis testing procedure in

statistics

This section

Derives of the Sum of Squares

Gives an outline of the ANOVA procedureIntroduces one-way ANOVA as a

generalization of the two-sample t-test

(54)

Sum of Squares

New variables (from Ch. 3)

The two-sample t-test tests for equality of the means of two groups.

We could express the observations as:

– Where the Eij are assumed to be

NID(0,σ2)

ij i

ij

E

(55)

Sum of Squares

This can also be written as:

μ could be seen as overall mean • αj as deviation from μ in group j

This model is overparameterized

Uses more parameters than necessaryNecessitates the requirement

(always assumed imposed)

ij i

ij E

X

i1,2

0

2

1   

n

(56)

Sum of Squares

We are deriving a test procedure similar

to the two-sample two-sided t-test

Using |t| as test statistic

Absolute value of the T statistic

This is equivalent to using t2

Because it’s a monotonic function of |t|

The square of the t statistic (from Ch. 3)

mn X

X  )

(57)

Sum of Squares

…can, after algebraic manipulations, be

written as F

where

) 2 (  

m n

W B F    m j j m X X 1 1 1    n j j n X X 1 2 2 n m X n X m X    1 2

2 2 2 1 2 2

1 ) ( ) ( )

(X X m X X n X X n

m mn

B     

 

      n j j m j

j X X X

X W 1 2 2 2 1 2 1

1 ) ( )

(58)

Sum of Squares

B: between (among) group sum of squaresW: within group sum of squares

B + W: total sum of squaresCan be shown to be:

Total number of degrees of freedom: m + n

– 1

Between groups: 1

 

 

n

i

i m

i

i X X X

X

1

2 2

1

2

1 ) ( )

(59)

Sum of Squares

This gives us the F statistic

Our goal is to test the significance of the

difference between the means of two groups

B measures the difference

The difference must be measured relative to

the variance within the groups

W measures that

The larger F is, the more significant the

difference

) 2 (  

m n

(60)

The ANOVA Procedure

Subdivide observed total sum of

squares into several components

In our case, B and W

Pick appropriate significance point for

a chosen Type I error

α

from an

F

table

Compare the observed components to

(61)

F-Statistic

Significance points depend on

degrees of freedom in

B

and

W

(62)

Comments

The two-group case readily generalizes

to any number of groups.

ANOVAs can be classified in various

ways, e.g.

fixed effects modelsmixed effects modelsrandom effects model

(63)

Comments

Terminology

Although ANOVA contains the word

‘variance’

What we actually test for is a equality in

means between the groups

• The different mean assumptions affect the variance, though

ANOVAs are special cases of regression

(64)

One-Way ANOVA

One-Way fixed-effect ANOVASetup and derivation

Like two-sample t-test for g number of groups – Observations (ni observations, i=1,2,…,g)

Using overparameterized model for X

E assumed NID(0,σ2), Σn α = 0, α fixed in

in i

i X X

X 1, 2,,

ij i

ij E

(65)

One-Way ANOVA

– Null Hypothesis H0 is: α1 = α2 = … = αg =

0

Total sum of squares is

This is subdivided into B and W

with



   g i n j ij i X X 1 1 2 ) (

   g i i

i X X

n B 1 2 ) (



    g i n j i ij i X X W 1 1 2 ) (    i n j i ij i n X X

1   

(66)

One-Way ANOVA

Total degrees of freedom: N – 1

• Subdivided into dfB = g – 1 and dfW = N - g

This gives us our test statistic F

We can now look in the F-table for these

degrees of freedom to pick significance points for B and W

And calculate B and W from the observed data

1 *

  

g g N W

(67)

Example

Revisiting the Basketball example

Looking at it as a One-Way ANOVA

analysis

Observation in group 1 (KU): X1 = {83, 81,

71}

• Observation in group 2: X2 = {65, 72, 70}

Total Sum of Squares:

B (between groups sum of squares)

3336 . 239 ) 70 66 . 73 ( ) 72 66 . 73 ( ) 65 66 . 73 ( ) 71 66 . 73 ( ) 81 66 . 73 ( ) 83 66 . 73

( 2 2 2 2 2 2

57 . 130 ) 33 . 76 69 ( 3 ) 33 . 76 33 . 78 ( 3 )

( 2 2

1

2

   g i i i X X

(68)

Example

W (within groups sum of squares)

Degrees of freedom

Total: N-1 = 5

• dfB = g – 1 = 2 - 1 = 1

• dfW = N – g = 6 – 2 = 4

(69)

Example

Table lookup for df 1 and 4 and α = 0.05:Critical value: F = 7.71

Calculate F from our data:

So… 4.806 < 7.71

– With ANOVA we actually accept H0!

Seems to be the large variance in group 1

806 . 4 1

2 2 6 * 667 . 108

57 . 130 1

* 

  

  

g g N W

(70)

Same Example – with Excel

(71)

Excel

(72)

Two-Way ANOVA

Two-Way Fixed Effects ANOVA

Overview only (in the scope of this book)

More complicated setup; example:

– Expression levels of one gene in lung cancer patients

a different risk classes

E.g.: ultrahigh, very high, intermediate, low

(73)

Two-Way ANOVA

– Expression levels (our observations): Xijk

i is the risk class (i = 1, 2, …, a) • j indicates the age group

k corresponds to the individual in each group (k = 1, …, n)

– Each group is a possible risk/age combination

• The number of individuals in each group is the same, n

• This is a “balanced” design

(74)

Two-Way ANOVA

The Xijk can be arranged in a table:

1 2 3 4

1 n n n n

2 n n n n

3 n n n n

4 n n n n

5 n n n n

i j

Risk category

A

ge

g

ro

(75)

Two-Way ANOVA

– The model adopted for each Xijk is

• Where Eijk are NID(μ, α2)

• The mean of Xijk is μ + αi + βi + δij

• αi is a fixed parameter, additive for risk class i

• βi is a fixed parameter, additive for age group i

• δij is a fixed risk/age interaction parameter

Should be added is a possible group/group interaction exists

ijk ij

j i

ijk E

X      

a

(76)

Two-Way ANOVA

These constraints are imposed • Σiαi = Σiβi = 0

• Σiδij = 0 for all j

• Σjδij = 0 for all i

The total sum of squares is then subdivided

into four groups:

(77)

Two-Way ANOVA

Associated with each sum of squares

Corresponding degrees of freedom

Hence also a corresponding mean square

– Sum of squares divided by degrees of freedom

The mean squares are then compared using

F ratios to test for significance of various effects

First – test for a significant risk/age interactionF-ratio used is ratio of interaction mean square

(78)

Two-Way ANOVA

If such an interaction is used, it may not be

reasonable to test for significant risk or age differences

Example, μ in two risk classes, two age

groups:

– No evidence of interaction

1 2

1 4 12 2 7 15

1 2

1 4 15 Risk

A

ge

A

(79)

Multi-Way ANOVA

One-way and two-way fixed effects

ANOVAs can be extended to

multi-way ANOVAs

Gets complicated

Example: three-way ANOVA model:

ijkm ijk

jk ik

ij k

j i

ijkm E

(80)

Further generalizations of

ANOVA

The 2

m

factorial design

A particular form of the one-way ANOVA

Interactions between main effects

m “factors” taken at two “levels”

E.g. (1) Gender, (2) Tissue (lung, kidney), and

(3) status (affected, not affected)

2m possible combinations of levels/groups

(81)

Further generalizations of

ANOVA

Example, m = 3, denoted by A, B, C

8 groups, {abc, ab, ac, bc, a, b, c, 1}

• Write totals of n observations Tabc, Tab, …, T1 • The total between sum of squares can be

subdivided into seven individual sums of squares

Three main effects (A, B, C)

Three pair wise interactions (AB, AC, BC)One triple-wise interaction (ABC)

Example: Sum of squares for A, and for BC,

respectively n T T T T T T T

Tabc ab ac a bc b c

8 ) ( 2 1        T T T T T T T

Tabc ab ac a bc b c )

(82)

Further generalizations of

ANOVA

– If m ≥ 5 the number of groups becomes large

– Then the total number of observations, n2m is

large

It is possible to reduce the number of

observations by a process …

Confounding

– Interaction ABC probably very small and not interesting

(83)

Further generalizations of

ANOVA

Fractional Replication

Related to confounding

Sometimes two groups cannot be

distinguished from each other, then they are aliases

• E.g. A and BC

This reduces the need to experiments and

data

Ch. 13 talks more about this in the context

(84)

Random/Mixed Effect

Models

So far: fixed effect models

– E.g. Risk class, age group fixed in previous example

• Multiple experiments would use same categories • But: what if we took experimental data on several

random days?

• The days in itself have no meaning, but a

“between days” sum of squares must be extracted

– What if the days turn out to be important?

(85)

Random/Mixed Effect

Models

Mixed Effect Models

If some categories are fixed and some

are random

Symbols used:

Greek letters for fixed effects

Uppercase Roman letters for random effectsExample: two-way mixed effect model with

Risk class a and days d and n values collected

each day, the appropriate model is written:

ikl il

l i

ikl D G E

(86)

Random/Mixed Effect

Models

Random effect model have no fixed

categories

The details on the ANOVA analysis depend

on which effects are random and which are fixed

In a microarray context (more in Ch. 13)

(87)

Multivariate Methods

ANOVA: the Repeated

Measures Case

Bootstrap Methods: the

Two-sample t-test

(88)
(89)

Sequential Analysis

Sequential Probability Ratio

Sample size not known in advanceDepends on outcomes of successive

observations

Some of this theory is in BLAST

Basic Local Alignment Search Tool

The book focuses on discreet random

(90)

Sequential Analysis

Consider:

Random variable Y with distribution P(y;ξ)Tests usually relate to the value of

parameter ξ

• H0: ξ is ξ0 • H1: ξ is ξ1

We can choose a value for the Type I error αAnd a value for the Type II error β

(91)

Sequential Analysis

A and B are chosen to correspond to an α

and β

– Sampling continues until the ratio is less

than A (accept H0) or greater than B (reject

H0)

– Because these are discreet variables,

boundary overshoot usually occurs

We don’t expect to exactly get values α and β

Desired values for α and β approximately

achieved by using

 

 

1 A

 

1

(92)

Sequential Analysis

It is also convenient to take logarithms,

which gives us:

Using

We can write

(93)

Sequential Analysis

Example: sequence matching

– H0: p0 = 0.25 (probability of a match is 0.25)

– H1: p1 = 0.35 (probability of a match is 0.35)

– Type I error α and Type II error β chosen 0.01

– Yi: 1 if there is a match at position i,

otherwise 0

– Sampling continues while

– with

  i i Y

S ( ) log99 99

1

log 1,0

(94)

Sequential Analysis

S can be seen as the support offered by

Yi for H1

The inequality can be re-written as

This is actually a random walk with step

sizes 0.7016 for a match and -0.2984 for a mismatch

 

 

i i

Y 0.2984) 9.581 (

(95)

Sequential Analysis

Power Function for a Sequential Test

Suppose the true value of the parameter

of interest is ξ

We wish to know the probability that H1

is accepted, given ξ

This probability is the power Ρ(ξ) of the

(96)

Sequential Analysis

– Where θ* is the unique non-zero solution

to θ in

R is the range of values of Y

– Equivalently, θ* is the unique non-zero

solution to θ in

        R

y P y

y P y P 1 ) ; ( ) ; ( ) ; ( 0 1    

  R y y S e y
(97)

Sequential Analysis

This is very similar to Ch. 7 – Random

Walks

The parameter θ* is the same as in Ch. 7

And it will be the same in Ch 10 – BLAST

(98)

Sequential Analysis

Mean Sample Size

The (random) number of observations

until one or the other hypothesis is accepted

Find approximation by ignoring

boundary overshoot

Essentially identical method used to find

(99)

Sequential Analysis

Two expressions are calculated for

ΣiS1,0(Yi)

One involves the mean sample size

By equating both expressions, solve for

mean sample size

                 

  

 ( )log 1

(100)

Sequential Analysis

So, the mean sample size is:

Both numerator and denominator

depend on Ρ(ξ), and so also on θ*

A generalization applies if Q(y) of Y has

different distribution than H0 and H1 –

relevant to BLAST

      R

y P y

y P

y

P (( ;; ))

(101)

Sequential Analysis

Example

Same sequence matching example as

before

• H0: p0 = 0.25 (probability of a match is 0.25)

• H1: p1 = 0.35 (probability of a match is 0.35)

Type I error α and Type II error β chosen 0.01Mean sample size equation is:

– Mean sample size is when H0 is true: 194

– Mean sample size is when H is true: 182

15 13 7

5 (1 )log

log

595 . 4 ) ( 190 . 9

p p

p

 

(102)

Sequential Analysis

Boundary Overshoot

– So far we assumed no boundary overshoot

– In practice, there will almost always be, though

• Exact Type I and Type II errors different from α and β

Random walk theory can be used to assess how

significant the effects of boundary overshoot are

– It can be shown that the sum of Type I and Type II errors is always less than α + β (also individually)

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