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. . . . .. . . .. . . .. . . .. . . .. . . Lecture 22 April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang April 9th, 2013 Hyun Min Kang p-Values Biostatistics 602 - Statistical Inference .. . . . .. . . ..

  1. . .. .. . . .. . . . . . .. . . .. . . .. .. . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang . . Definition: p-Value . . . Conclusions from Hypothesis Testing . p-Values . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . .. .. . . .. . . .. 5 / 1 . . .. . . .. . . • Reject H 0 or accept H 0 . • If size of the test is ( α ) small, the decision to reject H 0 is convincing. • If α is large, the decision may not be very convincing. A p-value p ( X ) is a test statistic satisfying 0 ≤ p ( x ) ≤ 1 for every sample point x . Small values of p ( X ) given evidence that H 1 is true. A p-value is valid if, for every θ ∈ Ω 0 and every 0 ≤ α ≤ 1 , Pr ( p ( X ) ≤ α | θ ) ≤ α

  2. • Each reader can choose the • And then can compare the reported p x to • So that each reader can individually determine whether these data lead • The p-value quantifies the evidence against H . • The smaller the p-value, the stronger, the evidence for rejecting H . • A p-value reports the results of a test on a more continuous scale • Rather than just the dichotomous decision ”Accept H ” or ”Reject H ”. . .. . . .. . . . .. .. . . .. . . . . .. . . .. . Advantage to reporting a test result via a p-value he or she considers appropriate to acceptance or rejection to H . Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . .. .. .. .. . . .. . . . . . .. . . .. . . . . .. .. . .. . . .. . . . . . .. . . .. . 6 / 1 • The size α does not need to be predefined

  3. • And then can compare the reported p x to • So that each reader can individually determine whether these data lead • The p-value quantifies the evidence against H . • The smaller the p-value, the stronger, the evidence for rejecting H . • A p-value reports the results of a test on a more continuous scale • Rather than just the dichotomous decision ”Accept H ” or ”Reject H ”. . .. . . .. . . .. . . . .. . . .. .. . . .. . . .. . Advantage to reporting a test result via a p-value to acceptance or rejection to H . Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . .. . . . . .. . . .. . . .. . . .. . . .. . . . .. . . .. . . .. . .. .. . . .. . . 6 / 1 • The size α does not need to be predefined • Each reader can choose the α he or she considers appropriate

  4. • So that each reader can individually determine whether these data lead • The p-value quantifies the evidence against H . • The smaller the p-value, the stronger, the evidence for rejecting H . • A p-value reports the results of a test on a more continuous scale • Rather than just the dichotomous decision ”Accept H ” or ”Reject H ”. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . Advantage to reporting a test result via a p-value to acceptance or rejection to H . Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 .. .. . . . . .. . . .. . . .. . . .. . . .. .. . . . . .. . . .. . .. .. . . .. . . 6 / 1 • The size α does not need to be predefined • Each reader can choose the α he or she considers appropriate • And then can compare the reported p ( x ) to α

  5. • The smaller the p-value, the stronger, the evidence for rejecting H . • A p-value reports the results of a test on a more continuous scale • Rather than just the dichotomous decision ”Accept H ” or ”Reject H ”. . .. . .. . . .. . . .. . . .. . .. . . . . . .. . . .. . Advantage to reporting a test result via a p-value Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . .. .. .. . . .. . . .. . . .. . . .. . . . . . . .. . . .. . 6 / 1 . .. . . .. . . .. • The size α does not need to be predefined • Each reader can choose the α he or she considers appropriate • And then can compare the reported p ( x ) to α • So that each reader can individually determine whether these data lead to acceptance or rejection to H 0 . • The p-value quantifies the evidence against H 0 .

  6. • A p-value reports the results of a test on a more continuous scale • Rather than just the dichotomous decision ”Accept H ” or ”Reject H ”. . . . .. . . .. .. . .. . . .. . . . . .. .. . . .. . . .. . Advantage to reporting a test result via a p-value Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . .. . .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . . 6 / 1 .. . . .. . . .. • The size α does not need to be predefined • Each reader can choose the α he or she considers appropriate • And then can compare the reported p ( x ) to α • So that each reader can individually determine whether these data lead to acceptance or rejection to H 0 . • The p-value quantifies the evidence against H 0 . • The smaller the p-value, the stronger, the evidence for rejecting H 0 .

  7. • Rather than just the dichotomous decision ”Accept H ” or ”Reject H ”. . .. . .. . . .. . . .. . . .. . . . .. . .. . . .. . . .. . Advantage to reporting a test result via a p-value Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . .. . .. . . .. . . .. . . .. . . . . . .. . . .. . . .. . . 6 / 1 .. . . .. . . .. • The size α does not need to be predefined • Each reader can choose the α he or she considers appropriate • And then can compare the reported p ( x ) to α • So that each reader can individually determine whether these data lead to acceptance or rejection to H 0 . • The p-value quantifies the evidence against H 0 . • The smaller the p-value, the stronger, the evidence for rejecting H 0 . • A p-value reports the results of a test on a more continuous scale

  8. . . . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . Advantage to reporting a test result via a p-value Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 .. .. . . . . .. . . .. . . .. . .. .. . . .. .. . . .. . . . 6 / 1 . . .. . . .. . • The size α does not need to be predefined • Each reader can choose the α he or she considers appropriate • And then can compare the reported p ( x ) to α • So that each reader can individually determine whether these data lead to acceptance or rejection to H 0 . • The p-value quantifies the evidence against H 0 . • The smaller the p-value, the stronger, the evidence for rejecting H 0 . • A p-value reports the results of a test on a more continuous scale • Rather than just the dichotomous decision ”Accept H 0 ” or ”Reject H 0 ”.

  9. . .. .. . . .. . . . . . .. . . .. . . .. .. p x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang Then p X is a valid p-value. W x sup Pr W X . . . Theorem 8.3.27. . Constructing a value p-value . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . .. .. . . .. . . .. 7 / 1 . . .. . . .. . . Let W ( X ) be a test statistic such that large values of W give evidence that H 1 is true. For each sample point x , define

  10. . .. . . .. . . .. . . .. . . .. . . . . Theorem 8.3.27. April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang Then p X is a valid p-value. . . . . Constructing a value p-value . .. . . .. .. .. . . .. . .. . . .. . . .. . . .. . . .. . . . . . .. . .. .. . . .. . . .. . . 7 / 1 Let W ( X ) be a test statistic such that large values of W give evidence that H 1 is true. For each sample point x , define Pr ( W ( X ) ≥ W ( x ) | θ ) p ( x ) = sup θ ∈ Ω 0

  11. . . .. . . .. . . .. . . .. . . .. . .. . . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang . . Theorem 8.3.27. Constructing a value p-value . . .. . . .. . . .. .. .. . . .. . . .. . . .. . . .. . . . . . . .. . . . . .. .. . . .. . . .. 7 / 1 Let W ( X ) be a test statistic such that large values of W give evidence that H 1 is true. For each sample point x , define Pr ( W ( X ) ≥ W ( x ) | θ ) p ( x ) = sup θ ∈ Ω 0 Then p ( X ) is a valid p-value.

  12. . . . .. . . .. . .. .. . . .. . . .. . . .. . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang 2 Find a valid p-value, as a function of x . . . . . . . Problem . Example : Two-sided normal p-value . .. . . .. .. .. . . .. . . . . . .. . . .. . . . . .. .. . .. . . .. . . 8 / 1 . . . .. . . .. . Let X = ( X 1 , · · · , X n ) be a random sample from a N ( θ, σ 2 ) population. Consider testing H 0 : θ = θ 0 versus H 1 : θ ̸ = θ 0 . 1 Construct a size α LRT test.

  13. n s X . sup x Solution - Constructing LRT . .. . . .. . x . .. . . .. . . .. L sup .. For the numerator, the MLE of April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang n X i are and n L n X X i X are and For the denominator, the MLE of x . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . . . . . . .. . . .. . . .. . .. . . . .. . . .. 9 / 1 { ( θ, σ 2 ) : θ ∈ R , σ 2 > 0 } Ω =

  14. n s X . sup x Solution - Constructing LRT . .. . . .. . x . .. . . .. . . .. L sup . For the numerator, the MLE of April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang n X i are and n L n X X i X are and For the denominator, the MLE of x . .. .. .. .. . . .. . . .. . . . . . .. . . .. . . .. . . . .. . . . .. . . .. . . .. . .. . . .. . . 9 / 1 { ( θ, σ 2 ) : θ ∈ R , σ 2 > 0 } Ω = { ( θ, σ 2 ) : θ = θ 0 , σ 2 > 0 } Ω 0 =

  15. n s X . . .. . . .. . . .. .. . . .. . .. .. . . Solution - Constructing LRT . For the numerator, the MLE of April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang n X i are and n .. n X X i X are and For the denominator, the MLE of . . . .. . .. . . .. . . . . . .. . . .. . . . 9 / 1 .. .. .. . . .. . . .. . . .. . . . . .. . . { ( θ, σ 2 ) : θ ∈ R , σ 2 > 0 } Ω = { ( θ, σ 2 ) : θ = θ 0 , σ 2 > 0 } Ω 0 = sup { ( θ,σ 2 ): θ = θ 0 ,σ 2 > 0 } L ( θ, σ 2 | x ) λ ( x ) = sup { ( θ,σ 2 ): θ ∈ R ,σ 2 > 0 } L ( θ, σ 2 | x )

  16. n s X . . . .. . . .. . . . .. . . .. . . .. .. . and April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang n X i are For the numerator, the MLE of . n n X X i X Solution - Constructing LRT . .. .. . .. .. .. . . .. . . . . . .. . . .. . . . 9 / 1 .. . . . .. . . .. . . .. . . .. . .. . . { ( θ, σ 2 ) : θ ∈ R , σ 2 > 0 } Ω = { ( θ, σ 2 ) : θ = θ 0 , σ 2 > 0 } Ω 0 = sup { ( θ,σ 2 ): θ = θ 0 ,σ 2 > 0 } L ( θ, σ 2 | x ) λ ( x ) = sup { ( θ,σ 2 ): θ ∈ R ,σ 2 > 0 } L ( θ, σ 2 | x ) For the denominator, the MLE of θ and σ 2 are

  17. . .. .. . . .. . . .. . . .. . . .. . . .. .. and April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang n X i are For the numerator, the MLE of . X n Solution - Constructing LRT . .. . . . . .. . . .. . . . . . . .. . . .. . . .. 9 / 1 . . . . .. . . .. . .. . .. . .. . . .. { ( θ, σ 2 ) : θ ∈ R , σ 2 > 0 } Ω = { ( θ, σ 2 ) : θ = θ 0 , σ 2 > 0 } Ω 0 = sup { ( θ,σ 2 ): θ = θ 0 ,σ 2 > 0 } L ( θ, σ 2 | x ) λ ( x ) = sup { ( θ,σ 2 ): θ ∈ R ,σ 2 > 0 } L ( θ, σ 2 | x ) For the denominator, the MLE of θ and σ 2 are { ˆ θ = X σ 2 = ∑ ( X i − X ) 2 = n − 1 n s 2 ˆ

  18. . . .. . . .. . . .. .. . .. . . .. . .. . n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang n X i X Solution - Constructing LRT . . .. . . .. . . . .. . . .. . . .. . . .. .. . . .. . . . 9 / 1 . . . .. . . .. .. . . . .. . . .. . { ( θ, σ 2 ) : θ ∈ R , σ 2 > 0 } Ω = { ( θ, σ 2 ) : θ = θ 0 , σ 2 > 0 } Ω 0 = sup { ( θ,σ 2 ): θ = θ 0 ,σ 2 > 0 } L ( θ, σ 2 | x ) λ ( x ) = sup { ( θ,σ 2 ): θ ∈ R ,σ 2 > 0 } L ( θ, σ 2 | x ) For the denominator, the MLE of θ and σ 2 are { ˆ θ = X σ 2 = ∑ ( X i − X ) 2 = n − 1 n s 2 ˆ For the numerator, the MLE of θ and σ 2 are

  19. . . .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . Solution - Constructing LRT n X n Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . .. .. . . .. . . .. . . .. .. . . .. . . . 9 / 1 . . .. . . .. . . .. . .. . . . .. . { ( θ, σ 2 ) : θ ∈ R , σ 2 > 0 } Ω = { ( θ, σ 2 ) : θ = θ 0 , σ 2 > 0 } Ω 0 = sup { ( θ,σ 2 ): θ = θ 0 ,σ 2 > 0 } L ( θ, σ 2 | x ) λ ( x ) = sup { ( θ,σ 2 ): θ ∈ R ,σ 2 > 0 } L ( θ, σ 2 | x ) For the denominator, the MLE of θ and σ 2 are { ˆ θ = X σ 2 = ∑ ( X i − X ) 2 = n − 1 n s 2 ˆ For the numerator, the MLE of θ and σ 2 are { ˆ θ 0 = θ 0 ∑ ( X i − θ 0 ) 2 σ 2 ˆ 0 =

  20. . . .. . . .. . . .. . Solution - Constructing and Simplifying the Test .. . . .. . . .. . Combining the results together . c April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang c x i x x i n LRT test rejects H if and only if n x i n x x i c n . .. .. . . . .. . . .. . .. .. . . .. . . .. . . . 10 / 1 . .. . .. . . .. . . .. . . . . .. . . .. . ( ˆ ) n /2 σ 2 λ ( x ) = σ 2 ˆ 0

  21. . . . . .. . . .. . .. . . . .. . . .. . .. Solution - Constructing and Simplifying the Test .. c April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang c x i x x i n Combining the results together n x i n x x i c n . .. . .. . .. . . .. . . . . . .. . . .. . . . 10 / 1 .. . .. . . .. . . .. . . .. . .. . . . .. . ( ˆ ) n /2 σ 2 λ ( x ) = σ 2 ˆ 0 LRT test rejects H 0 if and only if

  22. . . . .. . . .. . .. .. . . .. . . .. . . . .. c April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang c x i x x i n Solution - Constructing and Simplifying the Test n x i n x x i c Combining the results together . .. . .. . .. . . .. . . . . . .. . . .. . . . 10 / 1 .. . . .. . . . . . .. . .. .. . . .. .. . . ( ˆ ) n /2 σ 2 λ ( x ) = σ 2 ˆ 0 LRT test rejects H 0 if and only if ( ˆ ) n /2 σ 2 ≤ σ 2 ˆ 0

  23. . .. .. . . .. . . .. . . .. . . .. . . .. .. x i April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang c x i x c . c Combining the results together Solution - Constructing and Simplifying the Test . .. . . . . .. . . .. . . . . . . .. . . .. . . .. 10 / 1 . . . .. . . .. . .. .. .. . . .. . . . ( ˆ ) n /2 σ 2 λ ( x ) = σ 2 ˆ 0 LRT test rejects H 0 if and only if ( ˆ ) n /2 σ 2 ≤ σ 2 ˆ 0 ) n /2 ( ∑ ( x i − x ) 2 / n ≤ ∑ ( x i − θ 0 ) 2 / n

  24. . . .. . . .. . . .. . .. .. . . .. . .. . . . .. . . .. . Solution - Constructing and Simplifying the Test Combining the results together c c Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . . .. . . .. . . .. . .. .. . . . .. . . . 10 / 1 . . . . .. . . . .. . . .. . . .. .. ( ˆ ) n /2 σ 2 λ ( x ) = σ 2 ˆ 0 LRT test rejects H 0 if and only if ( ˆ ) n /2 σ 2 ≤ σ 2 ˆ 0 ) n /2 ( ∑ ( x i − x ) 2 / n ≤ ∑ ( x i − θ 0 ) 2 / n ∑ ( x i − x ) 2 c ∗ ≤ ∑ ( x i − θ 0 ) 2

  25. n x . x i n x Solution - Simplifying the LRT . .. . . .. . x .. . . .. . . .. . . c .. n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang test. size . The next step is specify c to get c s X x i x LRT test rejects H if c n s X x c x .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . . . . .. .. . . .. . . .. . . . . . .. . . .. 11 / 1 ∑ ( x i − x ) 2 c ∗ ≤ ∑ ( x i − s ) 2 + n ( x − θ 0 ) 2

  26. . . .. . . .. . . .. . Solution - Simplifying the LRT .. . . .. .. . .. . n x . n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang test. size . The next step is specify c to get c s X x i x LRT test rejects H if c n s X x c x . . .. . .. . . .. . . .. . .. . . . .. . . .. . . . 11 / 1 . . .. . . .. . .. . . .. . .. . . .. . . ∑ ( x i − x ) 2 c ∗ ≤ ∑ ( x i − s ) 2 + n ( x − θ 0 ) 2 1 c ∗ ≤ 1 + n ( x − θ 0 ) 2 ∑ ( x i − x ) 2

  27. . .. .. . . .. . . . . . .. .. . .. . . . .. . n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang test. size . The next step is specify c to get c s X . x LRT test rejects H if c n s X x Solution - Simplifying the LRT .. . . .. .. .. . . .. . . . . . .. . . .. . . . 11 / 1 .. . . . .. . . .. . . .. . . .. . .. . . ∑ ( x i − x ) 2 c ∗ ≤ ∑ ( x i − s ) 2 + n ( x − θ 0 ) 2 1 c ∗ ≤ 1 + n ( x − θ 0 ) 2 ∑ ( x i − x ) 2 n ( x − θ 0 ) 2 c ∗∗ ≥ ∑ ( x i − x ) 2

  28. . .. .. . . .. . . . . . .. . . .. . . .. .. c April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang test. size . The next step is specify c to get n . s X x LRT test rejects H if Solution - Simplifying the LRT . .. . . .. . .. . . .. . . . . . . .. . . .. . . .. 11 / 1 . .. . . .. .. . . .. . . .. . . .. . . ∑ ( x i − x ) 2 c ∗ ≤ ∑ ( x i − s ) 2 + n ( x − θ 0 ) 2 1 c ∗ ≤ 1 + n ( x − θ 0 ) 2 ∑ ( x i − x ) 2 n ( x − θ 0 ) 2 c ∗∗ ≥ ∑ ( x i − x ) 2 � � x − θ 0 c ∗∗∗ � � s X / √ n ≥ � � � �

  29. . .. .. . . .. . . . . . .. . . .. . . .. .. c April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang test. size . The next step is specify c to get n . s X x LRT test rejects H if Solution - Simplifying the LRT . .. . . .. . .. . . .. . . . . . . .. . . .. . . .. 11 / 1 . .. . . .. .. . . .. . . .. . . .. . . ∑ ( x i − x ) 2 c ∗ ≤ ∑ ( x i − s ) 2 + n ( x − θ 0 ) 2 1 c ∗ ≤ 1 + n ( x − θ 0 ) 2 ∑ ( x i − x ) 2 n ( x − θ 0 ) 2 c ∗∗ ≥ ∑ ( x i − x ) 2 � � x − θ 0 c ∗∗∗ � � s X / √ n ≥ � � � �

  30. . . . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . Solution - Simplifying the LRT Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 .. .. . . . . .. .. . . .. . . .. . . .. . 11 / 1 .. .. . . . . . .. . . .. . . .. . ∑ ( x i − x ) 2 c ∗ ≤ ∑ ( x i − s ) 2 + n ( x − θ 0 ) 2 1 c ∗ ≤ 1 + n ( x − θ 0 ) 2 ∑ ( x i − x ) 2 n ( x − θ 0 ) 2 c ∗∗ ≥ ∑ ( x i − x ) 2 � � x − θ 0 c ∗∗∗ � � s X / √ n ≥ � � � � � � � x − θ 0 � ≥ c ∗∗∗ . The next step is specify c to get s X / √ n � � LRT test rejects H 0 if size α test.

  31. . . .. . . .. . . .. . Pr .. . . .. . . .. . X . LRT test rejects H if and only if April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang t n n s X x Therefore, size s X t n c c T n Pr c n . .. .. . .. . . .. . . .. . .. . . . .. . . .. . . . 12 / 1 .. . .. . . .. . . . . .. . . .. . .. . . Solution - Obtaining size α test Under H 0 X − θ 0 s X / √ n ∼ T n − 1

  32. . .. .. . . .. . . . . .. .. . . .. . . . .. . x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang t n n s X LRT test rejects H if and only if . Therefore, size t n c c T n Pr Pr .. . . .. .. . . .. . . . . . .. . . .. . . . .. .. . . . .. . . .. . . .. . . .. . . 12 / 1 .. . Solution - Obtaining size α test Under H 0 X − θ 0 s X / √ n ∼ T n − 1 (� � X − θ 0 ) � ≥ c ∗∗∗ � � s X / √ n = α � � �

  33. . .. .. . . .. . .. . . . .. . . .. . . .. .. x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang t n n s X LRT test rejects H if and only if . Therefore, size t n c Pr . .. . . . . . . . . . . .. . . .. . . .. . . .. .. . .. .. . . .. . . .. . . .. . 12 / 1 . .. . . Solution - Obtaining size α test Under H 0 X − θ 0 s X / √ n ∼ T n − 1 (� � X − θ 0 ) � ≥ c ∗∗∗ � � s X / √ n = α � � � Pr ( | T n − 1 | ≥ c ∗∗∗ ) = α

  34. . . . . .. .. . .. . . .. . . .. . . .. . . x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang t n n s X LRT test rejects H if and only if .. Therefore, size Pr . .. . . .. . . . . . .. . .. .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. 12 / 1 . . . .. . Solution - Obtaining size α test Under H 0 X − θ 0 s X / √ n ∼ T n − 1 (� � X − θ 0 ) � ≥ c ∗∗∗ � � s X / √ n = α � � � Pr ( | T n − 1 | ≥ c ∗∗∗ ) = α c ∗∗∗ = t n − 1 ,α /2

  35. . . . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . Pr Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 .. .. . . . . .. . .. . .. . . .. . . .. . 12 / 1 .. .. . . . . . .. . . .. . . .. . Solution - Obtaining size α test Under H 0 X − θ 0 s X / √ n ∼ T n − 1 (� � X − θ 0 ) � ≥ c ∗∗∗ � � s X / √ n = α � � � Pr ( | T n − 1 | ≥ c ∗∗∗ ) = α c ∗∗∗ = t n − 1 ,α /2 � � � x − θ 0 Therefore, size α LRT test rejects H 0 if and only if � ≥ t n − 1 ,α /2 s X / √ n � �

  36. . .. . .. . . .. . . . under H , regardless of the value of . .. . . .. .. . Solution - p-value from two-sided test , W X . W x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang freedom. degrees of is the inverse CDF of t-distribution with n where F T n F T n T n W x Pr T n W x Pr W X W x sup Pr W X p x . Then, a valid p-value can be defined by .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . . 13 / 1 .. . . . .. . . . . .. . . . .. .. . . .. � � � X − θ 0 For a test statistic W ( X ) = s X / √ n � � � ,

  37. . . . .. . . .. . .. .. . . .. . . .. . . . .. W x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang freedom. degrees of is the inverse CDF of t-distribution with n where F T n F T n Solution - p-value from two-sided test W x Pr T n W x Pr W X W x sup Pr W X p x . .. . .. . .. . . .. . . . . . .. . . .. . . . 13 / 1 .. . .. . . .. . . .. . . .. . .. . . . .. . � � � X − θ 0 For a test statistic W ( X ) = s X / √ n � � � , under H 0 , regardless of the value of σ 2 , W ( X ) ∼ T n − 1 . Then, a valid p-value can be defined by

  38. . .. .. . . .. . . . . . .. . .. .. . . . .. . where F T n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang freedom. degrees of is the inverse CDF of t-distribution with n W x . F T n W x Pr T n W x Pr W X sup Solution - p-value from two-sided test .. . . .. .. .. . . .. . . . . . .. . . .. . . . 13 / 1 .. . .. . . .. . .. . . . .. . . . .. . . � � � X − θ 0 For a test statistic W ( X ) = s X / √ n � � � , under H 0 , regardless of the value of σ 2 , W ( X ) ∼ T n − 1 . Then, a valid p-value can be defined by Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) p ( x ) = θ ∈ Ω 0

  39. . . . .. . . .. . .. .. . .. .. . . .. . . . where F T n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang freedom. degrees of is the inverse CDF of t-distribution with n W x . F T n W x Pr T n sup Solution - p-value from two-sided test . .. . . .. .. . . . .. . . . . . .. . . .. . . .. 13 / 1 . .. . .. . . .. . . .. . . .. . . . .. . � � � X − θ 0 For a test statistic W ( X ) = s X / √ n � � � , under H 0 , regardless of the value of σ 2 , W ( X ) ∼ T n − 1 . Then, a valid p-value can be defined by Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) p ( x ) = θ ∈ Ω 0 Pr ( W ( X ) ≥ W ( x ) | θ 0 , σ 2 ) =

  40. . .. .. . . .. . . . . .. .. . . .. . . .. .. where F T n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang freedom. degrees of is the inverse CDF of t-distribution with n W x . F T n sup Solution - p-value from two-sided test . .. . . . . .. . . .. . . . . . . .. . . .. . . .. 13 / 1 . .. .. . . .. . . .. . . .. . . .. . . � � � X − θ 0 For a test statistic W ( X ) = s X / √ n � � � , under H 0 , regardless of the value of σ 2 , W ( X ) ∼ T n − 1 . Then, a valid p-value can be defined by Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) p ( x ) = θ ∈ Ω 0 Pr ( W ( X ) ≥ W ( x ) | θ 0 , σ 2 ) = = 2 Pr ( T n − 1 ≥ W ( x ))

  41. . .. . . .. . . .. . . .. . . .. . . . . where F T n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang freedom. degrees of is the inverse CDF of t-distribution with n sup . Solution - p-value from two-sided test . .. . . .. .. .. . . . .. . . .. .. . .. .. . . .. . . . 13 / 1 . . . .. . . .. .. . . . . . .. . . .. � � � X − θ 0 For a test statistic W ( X ) = s X / √ n � � � , under H 0 , regardless of the value of σ 2 , W ( X ) ∼ T n − 1 . Then, a valid p-value can be defined by Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) p ( x ) = θ ∈ Ω 0 Pr ( W ( X ) ≥ W ( x ) | θ 0 , σ 2 ) = = 2 Pr ( T n − 1 ≥ W ( x )) [ ] 1 − F − 1 T n − 1 { W ( x ) } = 2

  42. . . . .. . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . Solution - p-value from two-sided test sup freedom. Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . .. . . . . .. . . . .. . . .. . . .. . 13 / 1 .. .. .. . . .. . . . . . .. . . .. . � � � X − θ 0 For a test statistic W ( X ) = s X / √ n � � � , under H 0 , regardless of the value of σ 2 , W ( X ) ∼ T n − 1 . Then, a valid p-value can be defined by Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) p ( x ) = θ ∈ Ω 0 Pr ( W ( X ) ≥ W ( x ) | θ 0 , σ 2 ) = = 2 Pr ( T n − 1 ≥ W ( x )) [ ] 1 − F − 1 T n − 1 { W ( x ) } = 2 where F − 1 T n − 1 ( · ) is the inverse CDF of t-distribution with n − 1 degrees of

  43. . . . .. . . .. . .. .. . . .. . . .. . . .. . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang 2 Find a valid p-value, as a function of x . . . . . . . Problem . Example : One-sided normal p-value . .. . . .. .. .. . . .. . . . . . .. . . .. . . . . .. .. . .. . . .. . . 14 / 1 . . . .. . . .. . Let X = ( X 1 , · · · , X n ) be a random sample from a N ( θ, σ 2 ) population. Consider testing H 0 : θ ≤ θ 0 versus H 1 : θ > θ 0 . 1 Construct a size α LRT test.

  44. . . . . .. . . .. . .. . . . .. . . .. . .. Constructing LRT test .. sup Pr W X April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang does not matter. , and the value of always occurs at when W x p x W x want to show that the supreme in the definition of p-value , we first Because the null hypothesis contains multiple possible t n n s X x . .. . .. . .. . . .. . . . . . .. . . .. . . . .. . . .. . . .. . . .. . .. . . .. . . .. . 15 / 1 As shown in previous lectures, the LRT size α test rejects H 0 if

  45. . . . .. . . .. . .. .. . . .. . . .. . . . W x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang does not matter. , and the value of always occurs at when sup Pr W X . p x want to show that the supreme in the definition of p-value , we first Because the null hypothesis contains multiple possible Constructing LRT test . .. .. . .. .. . . . .. . . . . . .. . . .. . . . .. .. . . .. . . .. . . .. . .. . . . .. . 15 / 1 As shown in previous lectures, the LRT size α test rejects H 0 if x − θ 0 s X / √ n ≥ t n − 1 ,α W ( x ) =

  46. . .. .. . . .. . . . . . .. . . .. .. . .. .. W x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang does not matter. , and the value of always occurs at when sup Pr W X . p x want to show that the supreme in the definition of p-value Constructing LRT test . .. . . . . . . . .. . . .. . . .. . . .. . . .. . . .. .. . . .. . . .. . . 15 / 1 . .. . .. . . As shown in previous lectures, the LRT size α test rejects H 0 if x − θ 0 s X / √ n ≥ t n − 1 ,α W ( x ) = Because the null hypothesis contains multiple possible θ ≤ θ 0 , we first

  47. . . .. . .. .. . . .. . . .. . . .. . .. . want to show that the supreme in the definition of p-value April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang does not matter. , and the value of always occurs at when Constructing LRT test . . .. . . .. . . . .. .. . . .. . . .. . . .. . . . . . .. . . . .. . . .. . . 15 / 1 .. . . .. . . .. As shown in previous lectures, the LRT size α test rejects H 0 if x − θ 0 s X / √ n ≥ t n − 1 ,α W ( x ) = Because the null hypothesis contains multiple possible θ ≤ θ 0 , we first Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) p ( x ) = sup θ ∈ Ω 0

  48. . .. . .. .. . .. . . .. . . .. . . . .. . .. . . .. . . .. . Constructing LRT test want to show that the supreme in the definition of p-value Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . . . .. . . .. . . .. . . .. . . .. . . . . . .. . . .. . . 15 / 1 .. . .. . . .. . As shown in previous lectures, the LRT size α test rejects H 0 if x − θ 0 s X / √ n ≥ t n − 1 ,α W ( x ) = Because the null hypothesis contains multiple possible θ ≤ θ 0 , we first Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) p ( x ) = sup θ ∈ Ω 0 always occurs at when θ = θ 0 , and the value of σ does not matter.

  49. . .. . .. . . .. . . . Pr . .. . . .. . .. Obtaining one-sided p-value Pr . n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang W x Pr W X W x Pr T n s X X W x T n Pr n s X W x n s X .. . . . .. . . .. . . .. . .. .. . . .. . . .. . . . . 16 / 1 . . . .. . . .. . . .. . .. . . .. . . .. Consider any θ ≤ θ 0 and any σ . ( X − θ 0 ) Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) s X / √ n ≥ W ( x ) | θ, σ 2 =

  50. . .. .. . . .. . . . . . .. . . .. . . . .. . Pr T n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang W x Pr W X W x n . s X W x T n Pr Pr Pr Obtaining one-sided p-value .. .. . .. .. .. . . .. . . . . . .. . . .. . . . 16 / 1 .. . .. . . .. .. . . . .. . . . . .. . . Consider any θ ≤ θ 0 and any σ . ( X − θ 0 ) Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) s X / √ n ≥ W ( x ) | θ, σ 2 = ( X − θ s X / √ n ≥ W ( x ) + θ 0 − θ ) s X / √ n | θ, σ 2 =

  51. . .. .. . . .. . . . . .. .. . . .. . . .. .. Pr T n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang W x Pr W X W x Pr . Pr Pr Obtaining one-sided p-value . .. . . . . .. . . .. . . . . . . .. . . .. . . .. 16 / 1 . . . . .. .. . . . .. .. .. . . .. . . Consider any θ ≤ θ 0 and any σ . ( X − θ 0 ) Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) s X / √ n ≥ W ( x ) | θ, σ 2 = ( X − θ s X / √ n ≥ W ( x ) + θ 0 − θ ) s X / √ n | θ, σ 2 = ( T n − 1 ≥ W ( x ) + θ 0 − θ ) s X / √ n | θ, σ 2 =

  52. . .. . . .. . . .. .. . .. . . .. . . . . Pr April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang W x Pr W X Pr Pr . Obtaining one-sided p-value . .. . . .. .. . . . . .. . . .. .. . .. .. . . .. . . . 16 / 1 . . .. . . .. . . .. . . . .. . . .. . Consider any θ ≤ θ 0 and any σ . ( X − θ 0 ) Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) s X / √ n ≥ W ( x ) | θ, σ 2 = ( X − θ s X / √ n ≥ W ( x ) + θ 0 − θ ) s X / √ n | θ, σ 2 = ( T n − 1 ≥ W ( x ) + θ 0 − θ ) s X / √ n | θ, σ 2 = ≤ Pr ( T n − 1 ≥ W ( x ))

  53. . . .. . . .. . . .. . .. .. . . .. . .. . Pr April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang Pr Pr Pr Obtaining one-sided p-value . . .. . . .. . . . .. . . .. . . .. . . .. .. . . .. . . . 16 / 1 . .. . . . .. . . .. . . .. . . .. . Consider any θ ≤ θ 0 and any σ . ( X − θ 0 ) Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) s X / √ n ≥ W ( x ) | θ, σ 2 = ( X − θ s X / √ n ≥ W ( x ) + θ 0 − θ ) s X / √ n | θ, σ 2 = ( T n − 1 ≥ W ( x ) + θ 0 − θ ) s X / √ n | θ, σ 2 = ≤ Pr ( T n − 1 ≥ W ( x )) W ( X ) ≥ W ( x ) | θ 0 , σ 2 ) ( =

  54. . . . . .. . . .. . . .. . . .. . . .. .. . . Pr T n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang W x F T n W x W x .. Pr W X W x sup Pr W X p x Thus, the p-value for this one-side test is Obtaining one-sided p-value (cont’d) . . .. .. .. .. . . .. . . . . . .. . . .. . . . .. . . . .. . . .. . .. . . . .. . . .. . 17 / 1

  55. . .. .. . . .. . . . . . .. . . .. . . .. .. Pr T n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang W x F T n W x W x . Pr W X sup Thus, the p-value for this one-side test is Obtaining one-sided p-value (cont’d) . .. . .. . . . . . .. . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . 17 / 1 .. .. . . .. . . Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) p ( x ) = θ ∈ Ω 0

  56. . . . . .. . . .. . . .. . . .. . .. .. . . Pr T n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang W x F T n W x sup .. Thus, the p-value for this one-side test is Obtaining one-sided p-value (cont’d) . .. . . .. . . . .. . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . 17 / 1 .. . . . .. . . Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) p ( x ) = θ ∈ Ω 0 Pr ( W ( X ) ≥ W ( x ) | θ 0 , σ 2 ) =

  57. . . . .. .. . .. . . .. . . .. . . .. . .. .. . . .. . . .. . Obtaining one-sided p-value (cont’d) Thus, the p-value for this one-side test is sup Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . . . .. . . .. . . .. . . .. . . .. . . . . . .. . . .. . . 17 / 1 .. . . .. . . .. Pr ( W ( X ) ≥ W ( x ) | θ, σ 2 ) p ( x ) = θ ∈ Ω 0 Pr ( W ( X ) ≥ W ( x ) | θ 0 , σ 2 ) = Pr ( T n − 1 ≥ W ( x )) = 1 − F − 1 = T n − 1 [ W ( x )]

  58. . . .. . . .. . . .. . p-Values by conditioning on on sufficient statistic .. . . .. . . .. . (not necessarily including alternative hypothesis). .. If we consider only the conditional distribution, by Theorem 8.3.27, this is April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang s S Pr p X a valid p-value, meaning that S x If the null hypothesis is W x S Pr W X p x that H is true. Define Again, let W X denote a test statistic where large value give evidence . s does not depend on true, the conditional distribution of X given S . . .. . . . .. . . .. . .. . . . .. . . .. . . .. 18 / 1 . .. . .. . . .. . . . . . .. . . .. . . .. Suppose S ( X ) is a sufficient statistic for the model { f ( x | θ ) : θ ∈ Ω 0 } .

  59. . .. .. . . .. . . . . . .. . . .. . . . .. .. If we consider only the conditional distribution, by Theorem 8.3.27, this is April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang s S Pr p X a valid p-value, meaning that S x . W x S Pr W X p x that H is true. Define Again, let W X denote a test statistic where large value give evidence (not necessarily including alternative hypothesis). If the null hypothesis is p-Values by conditioning on on sufficient statistic .. . . .. .. . . .. . . . .. . .. . . .. . . . . .. .. . . .. . . .. . . . . . .. . . .. . 18 / 1 Suppose S ( X ) is a sufficient statistic for the model { f ( x | θ ) : θ ∈ Ω 0 } . true, the conditional distribution of X given S = s does not depend on θ .

  60. . . . . .. . . .. . . .. . . .. .. . .. . . a valid p-value, meaning that April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang s S Pr p X If we consider only the conditional distribution, by Theorem 8.3.27, this is .. (not necessarily including alternative hypothesis). If the null hypothesis is p-Values by conditioning on on sufficient statistic . .. . . .. . . . . .. .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . 18 / 1 .. . . .. . . Suppose S ( X ) is a sufficient statistic for the model { f ( x | θ ) : θ ∈ Ω 0 } . true, the conditional distribution of X given S = s does not depend on θ . Again, let W ( X ) denote a test statistic where large value give evidence that H 1 is true. Define p ( x ) = Pr ( W ( X ) ≥ W ( x ) | S = S ( x ))

  61. . . .. .. . .. . . .. . . .. . . .. . .. . . . .. . . .. . p-Values by conditioning on on sufficient statistic (not necessarily including alternative hypothesis). If the null hypothesis is If we consider only the conditional distribution, by Theorem 8.3.27, this is a valid p-value, meaning that Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . . .. .. . . .. . . .. . . .. . . . . . .. . . . .. . . .. . 18 / 1 . .. . . .. . . .. Suppose S ( X ) is a sufficient statistic for the model { f ( x | θ ) : θ ∈ Ω 0 } . true, the conditional distribution of X given S = s does not depend on θ . Again, let W ( X ) denote a test statistic where large value give evidence that H 1 is true. Define p ( x ) = Pr ( W ( X ) ≥ W ( x ) | S = S ( x )) Pr ( p ( X ) ≤ α | S = s ) ≤ α

  62. . . . . .. . . .. . . .. . . .. . . .. .. . .. s April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang Thus, p X is a valid p-value. s Pr S s .. s Pr S S Pr p X s Pr p X p-Values by conditioning on sufficient statistic (cont’d) . . . .. .. .. . . .. . . . . . .. . . .. . . . . .. .. . .. . . .. . . . . . .. . . .. . 19 / 1 Then for any θ ∈ Ω 0 , unconditionally we have

  63. . .. . . .. . . .. . . .. . . .. . . . . s April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang Thus, p X is a valid p-value. s Pr S s . p-Values by conditioning on sufficient statistic (cont’d) . .. . . .. .. .. . . .. . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . 19 / 1 .. . . .. . . Then for any θ ∈ Ω 0 , unconditionally we have ∑ Pr ( p ( X ) ≤ α | θ ) = Pr ( p ( X ) ≤ α | S = s ) Pr ( S = s | θ )

  64. . . .. .. . .. . . .. . . .. . . .. . .. . . . .. . . .. . p-Values by conditioning on sufficient statistic (cont’d) s s Thus, p X is a valid p-value. Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . . .. .. . . .. . . .. . . .. . . . . . .. . . . .. . . .. . 19 / 1 . .. . . .. . . .. Then for any θ ∈ Ω 0 , unconditionally we have ∑ Pr ( p ( X ) ≤ α | θ ) = Pr ( p ( X ) ≤ α | S = s ) Pr ( S = s | θ ) ∑ ≤ α Pr ( S = s | θ ) = α

  65. . . . .. .. . .. . . .. . . .. . . .. . .. .. . . .. . . .. . p-Values by conditioning on sufficient statistic (cont’d) s s Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . . . .. . . .. . . .. . . .. . . .. . . . . . .. . . .. . . 19 / 1 .. . . .. . . .. Then for any θ ∈ Ω 0 , unconditionally we have ∑ Pr ( p ( X ) ≤ α | θ ) = Pr ( p ( X ) ≤ α | S = s ) Pr ( S = s | θ ) ∑ ≤ α Pr ( S = s | θ ) = α Thus, p ( X ) is a valid p-value.

  66. p n p n p n . . X pmf of X p . Then the join Under H , if we let p denote the common value of p . . . . . . f x . Solution . . . Problem . Example - Fisher’s Exact Test . .. . is n x p p x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang X is a sufficient statistic under H . X Therefore S x x n x x .. n x n x p x x n x p x x .. . . . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . . . . .. . . .. . . .. . . . .. . .. . . .. . . .. . . .. 20 / 1 Let X 1 and X 2 be independent observations with X 1 ∼ Binomial ( n 1 , p 1 ) , and X 2 ∼ Binomial ( n 2 , p 2 ) . Consider testing H 0 : p 1 = p 2 versus H 1 : p 1 > p 2 . Find a valid p-value function.

  67. p n p n p n Problem x p f x . . Solution . . . . x Example - Fisher’s Exact Test . .. . . .. . . .. n . p x x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang X is a sufficient statistic under H . X Therefore S x x n p x . x n x n x p x x n x .. . .. . . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . . .. . .. .. . . .. . . 20 / 1 Let X 1 and X 2 be independent observations with X 1 ∼ Binomial ( n 1 , p 1 ) , and X 2 ∼ Binomial ( n 2 , p 2 ) . Consider testing H 0 : p 1 = p 2 versus H 1 : p 1 > p 2 . Find a valid p-value function. Under H 0 , if we let p denote the common value of p 1 = p 2 . Then the join pmf of ( X 1 , X 2 ) is

  68. p n . Problem . Example - Fisher’s Exact Test . .. . . .. . . . .. . .. .. . . .. . . . x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang X is a sufficient statistic under H . X Therefore S x n Solution x p x x n x n . . . . .. . .. . . . .. . . .. . . .. . . .. . . .. . . . 20 / 1 . . . .. . . . . .. . .. .. . . .. . .. . . .. Let X 1 and X 2 be independent observations with X 1 ∼ Binomial ( n 1 , p 1 ) , and X 2 ∼ Binomial ( n 2 , p 2 ) . Consider testing H 0 : p 1 = p 2 versus H 1 : p 1 > p 2 . Find a valid p-value function. Under H 0 , if we let p denote the common value of p 1 = p 2 . Then the join pmf of ( X 1 , X 2 ) is ( n 1 ) ( n 2 ) p x 1 (1 − p ) n 1 − x 1 p x 2 (1 − p ) n 2 − x 2 f ( x 1 , x 2 | p ) = x 1 x 2

  69. . .. .. . . .. . . . . . .. . . .. . . . .. . . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang X is a sufficient statistic under H . X Therefore S . . Solution . . . Problem . Example - Fisher’s Exact Test .. .. . .. .. . . .. . .. . . . .. . . .. . . . . .. .. . . .. . . .. . . . 20 / 1 . .. . . . .. . Let X 1 and X 2 be independent observations with X 1 ∼ Binomial ( n 1 , p 1 ) , and X 2 ∼ Binomial ( n 2 , p 2 ) . Consider testing H 0 : p 1 = p 2 versus H 1 : p 1 > p 2 . Find a valid p-value function. Under H 0 , if we let p denote the common value of p 1 = p 2 . Then the join pmf of ( X 1 , X 2 ) is ( n 1 ) ( n 2 ) p x 1 (1 − p ) n 1 − x 1 p x 2 (1 − p ) n 2 − x 2 f ( x 1 , x 2 | p ) = x 1 x 2 ( n 1 )( n 2 ) p x 1 + x 2 (1 − p ) n 1 + n 2 − x 1 − x 2 = x 1 x 2

  70. . .. .. . . .. .. . . . . .. . . .. . . .. .. . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang . . Solution . . . Problem . Example - Fisher’s Exact Test . .. . . . . . . . .. . . .. . . . .. . . .. . . .. 20 / 1 . . . .. . . .. .. . .. . .. . . .. . . Let X 1 and X 2 be independent observations with X 1 ∼ Binomial ( n 1 , p 1 ) , and X 2 ∼ Binomial ( n 2 , p 2 ) . Consider testing H 0 : p 1 = p 2 versus H 1 : p 1 > p 2 . Find a valid p-value function. Under H 0 , if we let p denote the common value of p 1 = p 2 . Then the join pmf of ( X 1 , X 2 ) is ( n 1 ) ( n 2 ) p x 1 (1 − p ) n 1 − x 1 p x 2 (1 − p ) n 2 − x 2 f ( x 1 , x 2 | p ) = x 1 x 2 ( n 1 )( n 2 ) p x 1 + x 2 (1 − p ) n 1 + n 2 − x 1 − x 2 = x 1 x 2 Therefore S = X 1 + X 2 is a sufficient statistic under H 0 .

  71. . . f X s is a hypergeometric distribution. X given S The conditional distribution of Solution - Fisher’s Exact Test (cont’d) . .. . .. n . . .. . . .. . . x s x .. x April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang f j s x j s min n p x n x is x Thus, the p-value conditional on the sufficient statistic s s n n x s .. . . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . .. .. . . .. . . 21 / 1 Given the value of S = s , it is reasonable to use X 1 as a test statistic and reject H 0 in favor of H 1 for large values of X 1 ,because large values of X 1 correspond to small values of X 2 = s − X 1 .

  72. . .. .. . . .. . . . . . .. . .. .. . . . .. . s April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang f j s x j min n . x p x x is x Thus, the p-value conditional on the sufficient statistic s s Solution - Fisher’s Exact Test (cont’d) .. . . .. .. .. . . .. . . . . . .. . . .. . . . 21 / 1 .. . . . .. . . .. . . .. . . .. . . .. . Given the value of S = s , it is reasonable to use X 1 as a test statistic and reject H 0 in favor of H 1 for large values of X 1 ,because large values of X 1 correspond to small values of X 2 = s − X 1 . The conditional distribution of X 1 given S = s is a hypergeometric distribution. ( n 1 )( n 2 ) s − x 1 f ( X 1 = x 1 | s ) = x 1 ( n 1 + n 2 )

  73. . .. . .. . . .. . .. .. . . .. . . . .. . .. . . .. . . .. . Solution - Fisher’s Exact Test (cont’d) s Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 . . . .. . . .. . . . . . .. . . .. . . .. . . .. . . .. . . 21 / 1 .. . .. . . . .. Given the value of S = s , it is reasonable to use X 1 as a test statistic and reject H 0 in favor of H 1 for large values of X 1 ,because large values of X 1 correspond to small values of X 2 = s − X 1 . The conditional distribution of X 1 given S = s is a hypergeometric distribution. ( n 1 )( n 2 ) s − x 1 f ( X 1 = x 1 | s ) = x 1 ( n 1 + n 2 ) Thus, the p-value conditional on the sufficient statistic s = x 1 + x 2 is min ( n 1 , s ) ∑ p ( x 1 , x 2 ) = f ( j | s ) j = x 1

  74. . . . . Problem . Exercise 8.1 . .. . .. to assume that the coin is fair? Justify your answer. . . .. . . .. . . In 1,000 tosses of a coin, 560 heads and 440 tails appear. Is it reasonable . . . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang 2 H . . 1 H . be the probability of head. Hypothesis Let . . . . . . . . .. . .. . .. . . .. . . .. . .. . . . .. . . .. . . . .. .. .. . . .. . . .. . . . . . .. . . .. . . .. . 22 / 1

  75. . . .. . . .. . . .. . Exercise 8.1 .. . . .. . . .. . . . . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang 2 H . . 1 H . Problem . Hypothesis . to assume that the coin is fair? Justify your answer. In 1,000 tosses of a coin, 560 heads and 440 tails appear. Is it reasonable . . . .. .. .. . .. . . .. . . . .. . .. . . .. . . . . . .. . .. . . .. . . . . . .. . . .. . . .. 22 / 1 Let θ ∈ (0 , 1) be the probability of head.

  76. . . . . .. . . .. . .. . . . .. . . .. . .. Exercise 8.1 .. . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang . . . . . . Hypothesis . to assume that the coin is fair? Justify your answer. In 1,000 tosses of a coin, 560 heads and 440 tails appear. Is it reasonable . . Problem .. . . .. . .. . . .. . . . .. . .. . . .. . . . . . . .. . . .. . . .. 22 / 1 . . .. . . .. . . .. Let θ ∈ (0 , 1) be the probability of head. 1 H 0 : θ = 1/2 2 H 1 : θ ̸ = 1/2

  77. . . . . . . Two possible strategies . .. . .. . . . .. . . .. . . . 2 Test whether the observation rejects H or not. .. . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang 2 Compute p-value as a quantitative support for the null hypothesis. . . 1 Obtain a p-value function p X . . . . . . . . . . Obtaining p-value . 3 Conclude that H is true or false at level .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . . . .. . . . .. . . .. . . .. . . .. . . .. . . .. 23 / 1 Performing size α Hypothesis Testing 1 Define a level α test for a reasonably small α .

  78. . . . . . . Two possible strategies . .. . .. . . . .. . . .. . . . . . . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang 2 Compute p-value as a quantitative support for the null hypothesis. . . 1 Obtain a p-value function p X . . . . . . . . . . Obtaining p-value . 3 Conclude that H is true or false at level .. .. . . . .. . . .. . . .. . .. .. . . .. . . .. . . . . .. . . . .. . . .. . . .. . .. . . .. . . .. . 23 / 1 Performing size α Hypothesis Testing 1 Define a level α test for a reasonably small α . 2 Test whether the observation rejects H 0 or not.

  79. . . .. . . .. . . .. . Two possible strategies .. . . .. . . .. . . . . April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang 2 Compute p-value as a quantitative support for the null hypothesis. . . . . . Obtaining p-value . . . . . . . . .. .. .. . .. . . .. . . . . . .. . . .. . . . .. . . . .. . . .. . . .. . .. . . .. . . .. . 23 / 1 Performing size α Hypothesis Testing 1 Define a level α test for a reasonably small α . 2 Test whether the observation rejects H 0 or not. 3 Conclude that H 0 is true or false at level α 1 Obtain a p-value function p ( X ) .

  80. . . . .. . . .. . .. .. . . .. . . .. .. . . .. At level April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang z z Z x , the H is rejected if and only if n 1,000 tosses are large enough to approximate using CLT. X X X Z X A two-sided Wald test statistic can be defined by n X . . . .. . .. . . .. . . . .. . .. . . .. . . . . . . .. . . .. . . .. . . .. . .. . . .. . 24 / 1 Asymptotic size α test ( θ, θ (1 − θ ) ) ∼ AN

  81. . . . .. . . .. . .. .. . . .. .. . .. . . . , the H is rejected if and only if April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang z z Z x At level . n A two-sided Wald test statistic can be defined by n X 1,000 tosses are large enough to approximate using CLT. . .. . . .. .. .. . . .. . . . . . .. . . .. . . . . .. . . .. . . .. . . .. . . 24 / 1 . .. . .. . Asymptotic size α test ( θ, θ (1 − θ ) ) ∼ AN X − θ 0 Z ( X ) = √ X (1 − X )

  82. . .. . . .. . .. .. . . .. . . .. . . . . n April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang z n A two-sided Wald test statistic can be defined by X . 1,000 tosses are large enough to approximate using CLT. . .. . . .. .. . . . . .. . . .. .. . .. .. . . .. . . . 24 / 1 . . . . .. . . . .. . . .. . . .. .. . Asymptotic size α test ( θ, θ (1 − θ ) ) ∼ AN X − θ 0 Z ( X ) = √ X (1 − X ) At level α , the H 0 is rejected if and only if | Z ( x ) | > z α /2

  83. . . .. . . .. . . .. . .. .. . . .. . .. . X April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang n A two-sided Wald test statistic can be defined by n 1,000 tosses are large enough to approximate using CLT. . . .. . . .. . . . .. . . .. . . .. . . .. .. . . .. . . . 24 / 1 . .. . .. .. . . . . . . .. . . .. . Asymptotic size α test ( θ, θ (1 − θ ) ) ∼ AN X − θ 0 Z ( X ) = √ X (1 − X ) At level α , the H 0 is rejected if and only if | Z ( x ) | > z α /2 0 . 56 − 0 . 5 = 3 . 822 > z α /2 √ 0 . 56 × 0 . 44 1000

  84. . . . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . Hypothesis Testing Hyun Min Kang Biostatistics 602 - Lecture 22 April 9th, 2013 .. .. . .. . . .. . . .. . . .. . . .. . . .. . . . .. . . .. . . .. . . .. . 25 / 1 Since z α /2 is 1.96, 2.57, and 4.42 for α = 0 . 05 , 0 . 01 , and 10 − 5 , respectively, we can conclude that the coin is biased at level 0 . 05 and 0 . 01 . However, at the level of 10 − 5 , the coin can be assumed to be fair.

  85. . .. .. . . .. . . . . . .. . . .. . . .. .. Pr Z X April 9th, 2013 Biostatistics 602 - Lecture 22 Hyun Min Kang suggesting a strong evidence for rejecting H . , So, under the null hypothesis, the size of test is less than Z x . Pr Z X If the normal approximation is used, the p-value can be obtained as Using p-value function . .. . . . .. . . . .. . . .. . . .. . . .. . . .. . . .. .. . . .. . . . .. . .. . . .. . . 26 / 1

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