Last Lecture. Wald Test

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1 Last Lecture Biostatistics Statistical Iferece Lecture 22 Hyu Mi Kag April 9th, 2013 Is the exact distributio of LRT statistic typically easy to obtai? How about its asymptotic distributio? For testig which ull/alterative hypotheses is the asymptotic distributio valid? What is a Wald Test? Describe a typical way to costruct a Wald Test Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Asymptotics of LRT Wald Test Theorem 1031 Cosider testig H 0 : θ = θ 0 vs H 1 : θ θ 0 Suppose X 1,, X are iid samples from f(x θ, ad ˆθ is the MLE of θ, ad f(x θ satisfies certai regularity coditios (eg see misc 1062, the uder H 0 : d 2 log λ(x χ 2 1 as Wald test relates poit estimator of θ to hypothesis testig about θ Defiitio Suppose W is a estimator of θ ad W AN (θ, σw 2 The Wald test statistic is defied as Z = W θ 0 S where θ 0 is the value of θ uder H 0 ad S is a cosistet estimator of σ W Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31

2 Advatage to reportig a test result via a p-value Coclusios from Hypothesis Testig Reject H 0 or accept H 0 If size of the test is (α small, the decisio to reject H 0 is covicig If α is large, the decisio may ot be very covicig Defiitio: p-value A p-value p(x is a test statistic satisfyig 0 p(x 1 for every sample poit x Small values of p(x give evidece that H 1 is true A p-value is valid if, for every θ Ω 0 ad every 0 α 1, Pr(p(X α θ α The size α does ot eed to be predefied Each reader ca choose the α he or she cosiders appropriate Ad the ca compare the reported p(x to α So that each reader ca idividually determie whether these data lead to acceptace or rejectio to H 0 The p-value quatifies the evidece agaist H 0 The smaller the p-value, the stroger, the evidece for rejectig H 0 A p-value reports the results of a test o a more cotiuous scale Rather tha just the dichotomous decisio Accept H 0 or Reject H 0 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Costructig a valid p-value Example : Two-sided ormal p-value Theorem 8327 Let W(X be a test statistic such that large values of W give evidece that H 1 is true For each sample poit x, defie p(x = sup Pr(W(X W(x θ θ Ω 0 The p(x is a valid p-value Let X = (X 1,, X be a radom sample from a N (θ, σ 2 populatio Cosider testig H 0 : θ = θ 0 versus H 1 : θ θ 0 1 Costruct a size α LRT test 2 Fid a valid p-value, as a fuctio of x Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31

3 Solutio - Costructig LRT Solutio - Costructig ad Simplifyig the Test Combiig the results together Ω = {(θ, σ 2 : θ R, σ 2 > 0} Ω 0 = {(θ, σ 2 : θ = θ 0, σ 2 > 0} λ(x = sup {(θ,σ 2 :θ=θ 0,σ 2 >0} L(θ, σ 2 x sup {(θ,σ 2 :θ R,σ 2 >0} L(θ, σ 2 x For the deomiator, the MLE of θ ad σ 2 are { ˆθ = X ˆσ 2 = (Xi X 2 = 1 s2 X For the umerator, the MLE of θ ad σ 2 are { ˆθ0 = θ 0 ˆσ 2 0 = (Xi θ 0 2 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 ( ˆσ 2 λ(x = LRT test rejects H 0 if ad oly if ˆσ 2 0 /2 ( ˆσ 2 /2 c ˆσ 2 0 ( (xi x 2 / (xi θ 0 2 / /2 c (xi x 2 (xi θ 0 2 c Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Solutio - Simplifyig the LRT Solutio - Obtaiig size α test LRT test rejects H 0 if size α test (xi x 2 (xi s 2 + (x θ 0 2 c x θ (x θ c 0 2 (xi x 2 (x θ 0 2 (xi x 2 c x θ 0 c c The ext step is specify c to get Uder H 0 X θ 0 T 1 ( X θ 0 Pr c = α Pr ( T 1 c = α c = t 1,α/2 Therefore, size α LRT test rejects H 0 if ad oly if x θ 0 t 1,α/2 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31

4 Solutio - p-value from two-sided test Example : Oe-sided ormal p-value For a test statistic W(X = X θ 0, uder H 0, regardless of the value of σ 2, W(X T 1 The, a valid p-value ca be defied by p(x = sup θ Ω 0 Pr(W(X W(x θ, σ 2 = Pr(W(X W(x θ 0, σ 2 = 2 Pr(T 1 W(x [ ] = 2 1 F 1 T 1 {W(x} Let X = (X 1,, X be a radom sample from a N (θ, σ 2 populatio Cosider testig H 0 : θ θ 0 versus H 1 : θ > θ 0 1 Costruct a size α LRT test 2 Fid a valid p-value, as a fuctio of x where F 1 T 1 ( is the iverse CDF of t-distributio with 1 degrees of freedom Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Costructig LRT test Obtaiig oe-sided p-value As show i previous lectures, the LRT size α test rejects H 0 if W(x = x θ 0 t 1,α Because the ull hypothesis cotais multiple possible θ θ 0, we first wat to show that the supreme i the defiitio of p-value p(x = sup θ Ω 0 Pr(W(X W(x θ, σ 2 always occurs at whe θ = θ 0, ad the value of σ does ot matter Cosider ay θ θ 0 ad ay σ ( X Pr(W(X W(x θ, σ 2 θ0 = Pr ( X θ = Pr = Pr W(x θ, σ2 W(x + θ 0 θ ( T 1 W(x + θ 0 θ Pr (T 1 W(x = Pr ( W(X W(x θ 0, σ 2 θ, σ2 θ, σ2 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31

5 Obtaiig oe-sided p-value (cot d by coditioig o o sufficiet statistic Thus, the p-value for this oe-side test is p(x = sup θ Ω 0 Pr(W(X W(x θ, σ 2 = Pr(W(X W(x θ 0, σ 2 = Pr(T 1 W(x = 1 F 1 T 1 [W(x] Suppose S(X is a sufficiet statistic for the model {f(x θ : θ Ω 0 } (ot ecessarily icludig alterative hypothesis If the ull hypothesis is true, the coditioal distributio of X give S = s does ot deped o θ Agai, let W(X deote a test statistic where large value give evidece that H 1 is true Defie p(x = Pr(W(X W(x S = S(x If we cosider oly the coditioal distributio, by Theorem 8327, this is a valid p-value, meaig that Pr(p(X α S = s α Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 by coditioig o sufficiet statistic (cot d Example - Fisher s Exact Test The for ay θ Ω 0, ucoditioally we have Pr(p(X α θ = Pr(p(X α S = s Pr(S = s θ s Thus, p(x is a valid p-value s α Pr(S = s θ = α Let X 1 ad X 2 be idepedet observatios with X 1 Biomial( 1, p 1, ad X 2 Biomial( 2, p 2 Cosider testig H 0 : p 1 = p 2 versus H 1 : p 1 > p 2 Fid a valid p-value fuctio Solutio Uder H 0, if we let p deote the commo value of p 1 = p 2 The the joi pmf of (X 1, X 2 is ( ( 1 f(x 1, x 2 p = p x 1 (1 p 1 x 1 2 p x 2 (1 p 2 x 2 x 1 x 2 ( ( 1 2 = p x 1+x 2 (1 p 1+ 2 x 1 x 2 x 1 x 2 Therefore S = X 1 + X 2 is a sufficiet statistic uder H 0 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31

6 Solutio - Fisher s Exact Test (cot d Exercise 81 Give the value of S = s, it is reasoable to use X 1 as a test statistic ad reject H 0 i favor of H 1 for large values of X 1,because large values of X 1 correspod to small values of X 2 = s X 1 The coditioal distributio of X 1 give S = s is a hypergeometric distributio f(x 1 = x 1 s = ( 1 ( 2 x 1 ( s s x 1 Thus, the p-value coditioal o the sufficiet statistic s = x 1 + x 2 is p(x 1, x 2 = mi( 1,s j=x 1 f(j s I 1,000 tosses of a coi, 560 heads ad 440 tails appear Is it reasoable to assume that the coi is fair? Justify your aswer Hypothesis Let θ (0, 1 be the probability of head 1 H 0 : θ = 1/2 2 H 1 : θ 1/2 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Two possible strategies Asymptotic size α test Performig size α Hypothesis Testig 1 Defie a level α test for a reasoably small α 2 Test whether the observatio rejects H 0 or ot 3 Coclude that H 0 is true or false at level α Obtaiig p-value 1 Obtai a p-value fuctio p(x 2 Compute p-value as a quatitative support for the ull hypothesis 1,000 tosses are large eough to approximate usig CLT ( θ(1 θ X AN θ, A two-sided Wald test statistic ca be defied by Z(X = At level α, the H 0 is rejected if ad oly if X θ 0 X(1 X Z(x > z α/2 = 3822 > z α/2 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31

7 Hypothesis Testig Usig p-value fuctio If the ormal approximatio is used, the p-value ca be obtaied as Sice z α/2 is 196, 257, ad 442 for α = 005, 001, ad 10 5, respectively, we ca coclude that the coi is biased at level 005 ad 001 However, at the level of 10 5, the coi ca be assumed to be fair Pr( Z(X Z(x = Pr( Z(X 3795 = So, uder the ull hypothesis, the size of test is less tha , suggestig a strog evidece for rejectig H 0 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Exercise 82 Costructig a test based o sufficiet statistic I a give city, it is assume that the umber of automobile accidets i a give year follows a Poisso distributio I past years, the average umber of accidets per year was 15, ad this year it was 10 Is it justified to claim that the accidet rate has dropped? Solutio - Hypothesis X 1 Poisso(λ 1, X 2 Poisso(λ 2 1 H 0 : λ 1 = λ 2 2 H 1 : λ 1 λ 2 Uder H 0, let λ 1 = λ 2 = λ f X (x 1, x 2 λ = Pr(X = x 1 λ Pr(X = x 2 λ = e 2λ λ x 1+x 2 x 1!x 2! Let S = X 1 + X 2 S is sufficiet statistic for λ uder H 0 S Poisso(2λ f S (s λ = Pr(S = s 2λ = e 2λ λ s s! Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31

8 Costructig a test based o sufficiet statistic (cot d Costructig a test based o sufficiet statistic (cot d The coditioal distributio of x give s is f(x 1, x 2 s = f X(x 1, x 2 λ f S (s λ = = e 2λ λ x 1 +x 2 x 1!x 2! e 2λ (2λ s s! s! 2 s x 1!x 2! = ( s x 1 2 s Let W(X = X 1, the the p-value coditioed o sufficiet statistic is p(x = Pr(W(X W(x S = S(x = Pr(X 1 x 1 S = s ( s s x 1 +x 2 ( x1 +x 2 x = 1 x 2 s = 1 2 x 1+x j=x 1 j=x 1 where x 1 = 15, x 2 = 10 Therefore, H 0 is ot rejected whe α < 05, ad it is ot reasoable to claim that the accidet rate has dropped Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31 Today p-value Fisher s Exact Test Examples of Hypothesis Testig Next Lectures Iterval Estimatio Cofidece Iterval Hyu Mi Kag Biostatistics Lecture 22 April 9th, / 31