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Classical Hypothesis Testing Theory Adapted from Alexander Senf 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. Determine a test statistic • 7/31/2008 a quantity calculated from the data 2 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 • • 7/31/2008 If it is, reject H0 Otherwise, accept H0 3 Overview of Ch. 9 – Simple Fixed-Sample-Size Tests – Composite Fixed-Sample-Size Tests – The -2 log λ Approximation – The Analysis of Variance (ANOVA) – Multivariate Methods – ANOVA: the Repeated Measures Case – Bootstrap Methods: the Two-sample t-test – Sequential Analysis 7/31/2008 4 Simple Fixed-Sample-Size Tests 7/31/2008 5 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 • It is desirable to have a procedure that minimizes β given a fixed α – This would maximize the power of the test • 1-β, the probability of rejecting H0 when H1 is true 7/31/2008 6 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: f1 ( X 1 ) f1 ( X 2 ) f1 ( X n ) LR f0 ( X 1 ) f0 ( X 2 ) f0 ( X n ) – where • f0, f1 are completely specified density functions for H0,H1 • X1, X2, … Xn are iid random variables 7/31/2008 7 Neyman-Pearson Lemma – H0 is rejected when LR ≥ K – With a constant K chosen such that: P(LR ≥ K when H0 is true) = α – Let’s look at an example using the NeymanPearson Lemma! – Then we will prove it. 7/31/2008 8 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.” • “Average height of the team and the population in general is the same” 7/31/2008 9 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 ( x 76.5 ) 2 ( x 69.5 ) 2 • Sample size: 3 8 8 e e f1 ( x) f 0 ( x) 2 2 2 2 – Choose α: 5% 7/31/2008 10 Example • The two populations: f0 f1 p height (inches) 7/31/2008 11 Example – Our test statistic is the Likelihood Ratio, LR e ( x1 76.5 ) 2 8 f1 ( x1 ) f1 ( x2 ) f1 ( x3 ) 2 2 ( x) ( x 69.5 ) f 0 ( x1 ) f 0 ( x2 ) f 0 ( x3 ) 8 e 2 2 ( x2 76.5 ) 2 8 e 2 2 2 1 e e ( x2 69.5 ) 2 8 2 2 ( x3 76.5 ) 2 8 2 2 e ( x3 69.5 ) 2 8 2 2 3 e 1 ( xi 69.5) 2 ( xi 76.5) 2 8 i1 – Now we need to determine a significance point K at which we can reject H0, given α = 5% 7/31/2008 • P(Λ(x) ≥ K | H0 is true) = 0.05, determine K 12 Example – So we just need to solve for K’ and calculate K: f 0 ( x1 ) f 0 ( x2 ) f 0 ( x3 )dx1dx2 dx3 0.05 K1' K 2' K 3' • 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: f 0 ( x1 ) f 0 ( x2 ) f 0 ( x3 )dx1dx2 dx3 0.05 6871 K 3' • We get one result: K3’=71.0803 7/31/2008 13 Example – Then we can just plug it in to Λ and calculate K: 3 K e e 1 ( K i' 69.5) 2 ( K i' 76.5) 2 8 i1 1 ( 6869.5) 2 ( 6876.5) 2 ( 7169.5) 2 ( 7176.5) 2 ( 71.080369.5) 2 ( 71.080376.5) 2 8 1.663 *10 7 7/31/2008 14 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 3 ( X {83,81,71}) e 1 ( X i 69.5 ) 2 ( X i 76.5 ) 2 8 i 1 (83,81,71) 1.446 *1012 • 1.446*1012 > 1.663*10-7 • Therefore, our hypothesis that tallness is a factor in the recruitment of KU basketball players is true. 7/31/2008 15 Neyman-Pearson Proof • Let A define region in the joint range of X1, X2, … Xn such that LR ≥ K. A is the critical region. – If A is the only critical region of size α we are done L(H ) f (u ) f (u ) f (u )du du du A 0 0 1 0 2 0 n 1 2 n A – Let’s assume another critical region of size α, defined by B L(H ) f (u ) f (u ) f (u )du du du B 7/31/2008 0 0 1 0 2 0 n 1 2 n B 16 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: L(H ) f (u ) f (u ) f (u )du du du A 7/31/2008 1 1 1 1 2 1 n 1 2 n A 17 Proof – Define: Δ = ∫AL(H1) - ∫BL(H1) • The power of the test using A minus using B f1 (u1 ) f1 (un )du1 dun f1 (u1 ) f1 (un )du1 dun A B f1 (u1 ) f1 (un )du1 dun f1 (u1 ) f1 (un )du1 dun A\C B\C • 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 7/31/2008 18 Proof – So, in A\C we have: f1 (u1 ) f1 (u n ) K f 0 (u1 ) f 0 (u n ) f1 (u1 ) f1 (un ) Kf0 (u1 ) f 0 (un ) – While in B\C we have: f1 (u1 ) f1 (un ) Kf0 (u1 ) f 0 (un ) Why? 7/31/2008 19 Proof – Thus Kf 0 (u1 ) f 0 (un )du1 dun Kf 0 (u1 ) f 0 (un )du1 dun A\C B\C Kf 0 (u1 ) f 0 (un )du1 dun Kf 0 (u1 ) f 0 (un )du1 dun A B K K 0 – Which implies that the power of the test using A is greater than or equal to the power using B. 7/31/2008 20 Composite Fixed-Sample-Size Tests 7/31/2008 21 Not Identically Distributed • In most cases, random variables are not identically distributed, at least not in H1 – This affects the likelihood function, L – For example, H1 in the two-sample t-test is: m L i 1 1 e 2 ( x1i 1 ) 2 2 2 n i 1 1 e 2 ( x2 i 2 ) 2 2 2 – Where μ1 and μ2 are different 7/31/2008 22 Composite – Further, the hypotheses being tested do not specify all parameters – They are composite – This chapter only outlines aspects of composite test theory relevant to the material in this book. 7/31/2008 23 Parameter Spaces – The set of values the parameters of interest can take – Null hypothesis: parameters in some region ω – Alternate hypothesis: parameters in Ω – ω is usually a subspace of Ω • Nested hypothesis case – Null hypothesis nested within alternate hypothesis – This book focuses on this case • “if the alternate hypothesis can explain the data significantly better we can reject the null hypothesis” 7/31/2008 24 λ Ratio • Optimality theory for composite tests suggests this as desirable test statistic: Lmax ( ) Lmax () • 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) 7/31/2008 25 Example: t-tests • The next slides calculate the λ-ratio for the two sample t-test (with the likelihood) m L i 1 1 e 2 ( x1i 1 ) 2 2 2 n i 1 1 e 2 ( x2 i 2 ) 2 2 2 – t-tests later generalize to ANOVA and T2 tests 7/31/2008 26 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) – Alternate hypothesis H1: both unspecified 7/31/2008 27 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: {1 2 ,0 2 } – Region Ω is: { 1 , 2 ,0 2 } 7/31/2008 28 Equal Variance Two-Sided t-test • Derivation – H0: writing μ for the mean, when μ1= μ2, the maximum over likelihood ω is at ˆ X X 11 X 12 X 1m X 21 X 22 X 2 n mn – And the (common) variance σ2 is 2 ( X X ) ( X X ) i1 1i i1 2i m ̂ 02 7/31/2008 2 n mn 29 Equal Variance Two-Sided t-test – Inserting both into the likelihood function, L Lmax ( ) 7/31/2008 1 (2ˆ ) 2 0 m n 2 e m n 2 30 Equal Variance Two-Sided t-test – Do the same thing for region Ω ˆ1 X 1 X 11 X 12 X 1m m ˆ 2 X 2 2 2 ( X X ) ( X X ) 1 2 i 1 1i i 1 2i m ˆ12 X 21 X 22 X 2 n n n mn – Which produces this likelihood Function, L Lmax () 7/31/2008 1 (2ˆ ) 2 1 m n 2 e m n 2 31 Equal Variance Two-Sided t-test – The test statistic λ is then e m2 n m n 2 ˆ12 Lmax ( ) (2ˆ 02 ) 2 m2 n Lmax () e ˆ 0 2 m2 n (2ˆ1 ) mn 2 It’s the same function, just With different variances 7/31/2008 32 Equal Variance Two-Sided t-test – We can then use the algebraic identity m n m n ( X 1i X ) ( X 2i X ) ( X 1i X 1 ) ( X 2i X 2 ) 2 2 i 1 2 i 1 – To show that 2 i 1 1 1t 2 m n2 – Where t is (from Ch. 3) 7/31/2008 i 1 mn ( X 1 X 2 )2 mn mn 2 T ( X 1 X 2 ) mn S mn 33 Equal Variance Two-Sided t-test – t is the observed value of T – S is defined in Ch. 3 as m n 2 ( X X ) ( X X ) 1 2 1i 2i 2 S2 i 1 i 1 mn2 λ We can plot λ as a function of t: (e.g. m+n=10) 7/31/2008 t 34 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| sufficiently large to reject H0 can be determined 7/31/2008 35 http://www.socr.ucla.edu/Applets.dir/T-table.html Equal Variance Two-Sided t-test 7/31/2008 36 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 {1 2 ,0 2 } • If x1 x 2 then maximum likelihood estimates are the same as for the two-sided case 7/31/2008 37 Equal Variance One-Sided t-test • If x1 x 2 then the unconstrained maximum of the likelihood is outside of ω • The unique maximum is at ( x1 , x 2 ) , implying that the maximum in ω occurs at a boundary point in Ω • At this point estimates of μ1 and μ2 are equal ( x) • At this point the likelihood ratio is 1 and H0 is not rejected • Result: H0 is rejected in favor of H1 (μ1 ≥ μ2) only for sufficiently large positive values of t 7/31/2008 38 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 the original example) 7/31/2008 39 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 (78.3 69) 9 27.9 t 2.185 12.7673 5.2122 6 – t > tα, so we can reject H0! 7/31/2008 40 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 not hold 7/31/2008 41 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 • • • • • 7/31/2008 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 < +∞} Ω makes no constraints on values μ1, μ2, σ12, and σ22 42 Unequal Variance Two-Sided t-test – The likelihood function of (X11, X12, …, X1m, X21, X22, …, X2n) then becomes m i 1 1 e 2 1 ( x1i 1 ) 2 2 12 n i 1 1 2 2 e ( x21i 2 ) 2 2 22 – Under H0 (μ1 = μ2 = μ), this becomes: m i 1 7/31/2008 1 e 2 1 ( x1i ) 2 2 12 n i 1 1 2 2 e ( x21i ) 2 2 22 43 Unequal Variance Two-Sided t-test – Maximum likelihood estimates ̂ , ̂ 12 and ̂ 22 satisfy the simultaneous equations: (x 1i 2 1 ˆ ) ˆ 2i 2 2 ˆ ) ˆ 2 1 (x ˆ 22 (x ˆ 7/31/2008 (x 1i 0 ˆ ) 2 m 2i ˆ ) 2 n 44 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) 7/31/2008 45 The -2 log λ Approximation 7/31/2008 46 The -2 log λ Approximation • Used when the λ-ratio procedure does not lead to a test statistic whose H0 distribution is known – Example: Unequal Variance Two-Sided t-test • Various approximations can be used – But only if certain regularity assumptions and restrictions hold true 7/31/2008 47 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 → ∞” – Provides an asymptotically valid testing procedure 7/31/2008 48 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 • I skip the proof… 7/31/2008 49 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 • Observation in group 1 (KU): X1 = {83, 81, 71} • Observation in group 2: X2 = {65, 72, 70} 7/31/2008 50 Example – Revised Again – Using the Unequal Variance One-Sided t-Test – We get: 7/31/2008 51 The Analysis of Variance (ANOVA) 7/31/2008 52 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 procedure – Introduces one-way ANOVA as a generalization of the two-sample t-test – Two-way and multi-way ANOVA – Further generalizations of ANOVA 7/31/2008 53 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: X ij i Eij i 1,2 – Where the Eij are assumed to be NID(0,σ2) – H0 is μ1 = μ2 7/31/2008 54 Sum of Squares – This can also be written as: X ij i Eij i 1,2 • μ could be seen as overall mean • αj as deviation from μ in group j – This model is overparameterized • Uses more parameters than necessary • Necessitates the requirement m1 n 2 0 • (always assumed imposed) 7/31/2008 55 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) ( X 1 X 2 ) mn T S mn 7/31/2008 56 Sum of Squares – …can, after algebraic manipulations, be written as F B F ( m n 2) W – where X X m m n X2 1j 1 j 1 j 1 X2j X n mX 1 nX 2 mn mn B ( X 1 X 2 ) 2 m( X 1 X ) 2 n ( X 2 X ) 2 mn m n W ( X1 j X 1 ) ( X 2 j X 2 )2 2 j 1 7/31/2008 j 1 57 Sum of Squares – B: between (among) group sum of squares – W: within group sum of squares – B + W: total sum of squares • Can be shown to be: m (X i 1 n 2 2 X ) ( X X ) 1i 2i i 1 – Total number of degrees of freedom: m + n – 1 • Between groups: 1 • Within groups: m + n - 2 7/31/2008 58 Sum of Squares – This gives us the F statistic F B (m n 2) W – 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 7/31/2008 59 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 test our hypothesis 7/31/2008 60 F-Statistic • Significance points depend on degrees of freedom in B and W – In our case, 1 and (m + n – 2) 7/31/2008 http://www.ento.vt.edu/~sharov/PopEcol/tables/f005.html 61 Comments • The two-group case readily generalizes to any number of groups. • ANOVAs can be classified in various ways, e.g. – fixed effects models – mixed effects models – random effects model – Difference is discussed later – For now we consider fixed effect models • Parameter αi is fixed, but unknown, in group i 7/31/2008 X ij i Eij 62 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 models from Ch. 8 7/31/2008 63 One-Way ANOVA • One-Way fixed-effect ANOVA • Setup and derivation – Like two-sample t-test for g number of groups – Observations (ni observations, i=1,2,…,g) X i1 , X i 2 ,, X in – Using overparameterized model for X X ij i Eij j 1,2,, ni i 1,2, , g – Eij assumed NID(0,σ2), Σniαi = 0, αi fixed in group i 7/31/2008 64 One-Way ANOVA – Null Hypothesis H0 is: α1 = α2 = … = αg = 0 – Total sum of squares is g ni ( X i 1 j 1 ij X )2 – This is subdivided into B and W g g ni W ( X ij X i ) 2 B ni ( X i X ) 2 i 1 j 1 i 1 – with ni X ij j 1 ni Xi 7/31/2008 g ni X i 1 j 1 X ij N g N ni i 1 65 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 F B Ng * W g 1 – 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 – And accept or reject H0 7/31/2008 66 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: (73.66 83) 2 (73.66 81) 2 (73.66 71) 2 (73.66 65) 2 (73.66 72) 2 (73.66 70) 2 239.3336 – B (between groups sum of squares) g B ni ( X i X ) 2 3(78.33 76.33) 2 3(69 76.33) 2 130.57 i 1 7/31/2008 67 Example – W (within groups sum of squares) g ni W ( X ij X i ) 2 i 1 j 1 ((83 78.33) 2 (81 78.33) 2 (71 78.33) 2 ) ((65 69) 2 (72 69) 2 (70 69) 2 ) 108.667 – Degrees of freedom • Total: N-1 = 5 • dfB = g – 1 = 2 - 1 = 1 • dfW = N – g = 6 – 2 = 4 7/31/2008 68 Example – Table lookup for df 1 and 4 and α = 0.05: – Critical value: F = 7.71 – Calculate F from our data: F B N g 130.57 6 2 * * 4.806 W g 1 108.667 2 1 – So… 4.806 < 7.71 – With ANOVA we actually accept H0! • Seems to be the large variance in group 1 7/31/2008 69 Same Example – with Excel • Screenshots: 7/31/2008 70 Excel • Offers most of these tests, built-in 7/31/2008 71 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 – b different age groups – n individuals for each risk/age combination 7/31/2008 72 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 • Theory for unbalanced designs is more complicated and not covered in this book 7/31/2008 73 Two-Way ANOVA – The Xijk can be arranged in a table: Risk category 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 Age group j i Number of individuals in this risk/age group (aka “cell”) 7/31/2008 This is a two-way table 74 Two-Way ANOVA – The model adopted for each Xijk is X ijk i j ij Eijk i 1,2,, a • • • • • j 1,2,, b k 1,2,, n 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 7/31/2008 75 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: 7/31/2008 • • • • Risk class sum of squares Age group sum of squares Interaction sum of squares Within cells (“residual” or “error”) sum of squares 76 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 interaction • F-ratio used is ratio of interaction mean square and within-cells mean square 7/31/2008 77 Two-Way ANOVA – Example of interaction 7/31/2008 Age – No evidence of interaction Age • 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: Risk 1 2 1 4 12 2 7 15 1 2 1 4 15 2 11 6 78 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: X ijkm i j k ij ik jk ijk Eijkm 7/31/2008 79 Further generalizations of ANOVA • The 2m 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 – Can test for main effects and interactions – Need replicated experiments • n replications for each of the 2m experiments 7/31/2008 80 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 – – – – 7/31/2008 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 (Tabc Tab Tac Ta Tbc Tb T cT1 ) 2 8n (Tabc Tab Tac Ta Tbc Tb T cT1 ) 2 8n 81 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 – So, prefer a model without ABC, reduce data – There are ANOVA designs for that 7/31/2008 82 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 of microarrays 7/31/2008 83 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? – If we fail to test for it, the significance of our procedure is diminished. – Days are a random category, unlike risk and age! 7/31/2008 84 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 effects • Example: two-way mixed effect model with – Risk class a and days d and n values collected each day, the appropriate model is written: X ikl i Dl Gil Eikl 7/31/2008 85 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) – There tend to be several fixed and several random effects, which complicates the analysis – Many interactions simply assumed zero 7/31/2008 86 Multivariate Methods ANOVA: the Repeated Measures Case Bootstrap Methods: the Twosample t-test All skipped … 7/31/2008 87 Sequential Analysis 7/31/2008 88 Sequential Analysis • Sequential Probability Ratio – Sample size not known in advance – Depends on outcomes of successive observations – Some of this theory is in BLAST • Basic Local Alignment Search Tool – The book focuses on discreet random variables 7/31/2008 89 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 α And a value for the Type II error β Sampling then continues while P( y1 ; 1 ) P( y2 ; 1 ) P( yn ; 1 ) A B P( y1 ; 0 ) P( y2 ; 0 ) P( yn ; 0 ) 7/31/2008 90 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 A 7/31/2008 1 B 1 91 Sequential Analysis – It is also convenient to take logarithms, which gives us: P( yi ; 1 ) 1 log log log 1 P ( yi ; 0 ) i – Using S1, 0 ( y ) log – We can write 7/31/2008 P ( y; 1 ) P ( y; 0 ) 1 log S1,0 ( yi ) log 1 i 92 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 7/31/2008 1 log S1, 0 (Yi ) log 99 99 i (0.35)Yi (0.65) (1Yi ) S1, 0 (Yi ) log (0.25)Yi (0.75) (1Yi ) 93 Sequential Analysis – S can be seen as the support offered by Yi for H1 – The inequality can be re-written as 9.581 (Yi 0.2984) 9.581 i – This is actually a random walk with step sizes 0.7016 for a match and -0.2984 for a mismatch 7/31/2008 94 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 test ( ) 7/31/2008 * 1 1 * * 1 1 ( ( ) ( ) ) 95 Sequential Analysis – Where θ* is the unique non-zero solution to θ in P ( y; 1 ) 1 P ( y; ) yR P ( y; 0 ) – R is the range of values of Y – Equivalently, θ* is the unique non-zero solution to θ in P( y; )e S1, 0 ( y ) 1 yR – Where S is defined as before 7/31/2008 96 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 – < skipping the random walk part > 7/31/2008 97 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 the mean number of steps until the random walk stops 7/31/2008 98 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 1 S ( y ) ( 1 ( )) log ( ) log i 1,0 i 1 P(Yi ; 1 ) P(Yi ; 1 ) E ( S1,0 (Yi )) E log P(Yi ; ) log P(Yi ; 0 ) yR P(Yi ; 0 ) 7/31/2008 99 Sequential Analysis – So, the mean sample size is: (1 ( )) log( 1 ) ( ) log( 1 ) P ( y ;1 ) P ( y ; ) log yR P ( y ; 0 ) – 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 (1 ( )) log( 1 ) ( ) log( 1 ) P ( y ;1 ) Q ( y ) log yR P ( y ; 0 ) 7/31/2008 100 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 0.01 – Mean sample size equation is: 9.190( p) 4.595 13 p log 75 (1 p ) log 15 – Mean sample size is when H0 is true: 194 – Mean sample size is when H1 is true: 182 7/31/2008 101 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) – BLAST deals with this in a novel way -> see Ch. 10 7/31/2008 102