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. . . . .. . . .. . . .. . . .. . . .. . . Lecture 23 April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang April 11th, 2013 Hyun Min Kang Interval Estimation Biostatistics 602 - Statistical Inference .. . . .

  1. . . .. . . .. . . .. . 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 11th, 2013 Biostatistics 602 - Lecture 23 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 . . .. . . . .. . . .. . .. . . . .. . . .. . . .. 5 / 1 . .. . .. . . .. . . . . . .. . . .. . . .. Suppose S ( X ) is a sufficient statistic for the model { f ( x | θ ) : θ ∈ Ω 0 } .

  2. . .. .. . . .. . . . . . .. . . .. . . . .. .. If we consider only the conditional distribution, by Theorem 8.3.27, this is April 11th, 2013 Biostatistics 602 - Lecture 23 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 .. . . .. .. . . .. . . . .. . .. . . .. . . . . .. .. . . .. . . .. . . . . . .. . . .. . 5 / 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 .

  3. . . . . .. . . .. . . .. . . .. .. . .. . . a valid p-value, meaning that April 11th, 2013 Biostatistics 602 - Lecture 23 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 . .. . . .. . . . . .. .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . 5 / 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 ))

  4. . . .. .. . .. . . .. . . .. . . .. . .. . . . .. . . .. . 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 23 April 11th, 2013 . . .. .. . . .. . . .. . . .. . . . . . .. . . . .. . . .. . 5 / 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

  5. 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 11th, 2013 Biostatistics 602 - Lecture 23 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 .. . . . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . . . . .. . . .. . . .. . . . .. . .. . . .. . . .. . . .. 6 / 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.

  6. p n p n p n Problem x p f x . . Solution . . . . x Example - Fisher’s Exact Test . .. . . .. . . .. n . p x x April 11th, 2013 Biostatistics 602 - Lecture 23 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 .. . .. . . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . . .. . .. .. . . .. . . 6 / 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

  7. p n . Problem . Example - Fisher’s Exact Test . .. . . .. . . . .. . .. .. . . .. . . . x April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang X is a sufficient statistic under H . X Therefore S x n Solution x p x x n x n . . . . .. . .. . . . .. . . .. . . .. . . .. . . .. . . . 6 / 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 ) f ( x 1 , x 2 | p ) = p x 1 (1 − p ) n 1 − x 1 p x 2 (1 − p ) n 2 − x 2 x 1 x 2

  8. . .. .. . . .. . . . . . .. . . .. . . . .. . . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang X is a sufficient statistic under H . X Therefore S . . Solution . . . Problem . Example - Fisher’s Exact Test .. .. . .. .. . . .. . .. . . . .. . . .. . . . . .. .. . . .. . . .. . . . 6 / 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 ) f ( x 1 , x 2 | p ) = p x 1 (1 − p ) n 1 − x 1 p x 2 (1 − p ) n 2 − x 2 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

  9. . .. .. . . .. .. . . . . .. . . .. . . .. .. . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . . Solution . . . Problem . Example - Fisher’s Exact Test . .. . . . . . . . .. . . .. . . . .. . . .. . . .. 6 / 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 ) f ( x 1 , x 2 | p ) = p x 1 (1 − p ) n 1 − x 1 p x 2 (1 − p ) n 2 − x 2 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 .

  10. . . 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 11th, 2013 Biostatistics 602 - Lecture 23 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 .. . . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . .. .. . . .. . . 7 / 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 .

  11. . .. .. . . .. . . . . . .. . .. .. . . . .. . s April 11th, 2013 Biostatistics 602 - Lecture 23 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) .. . . .. .. .. . . .. . . . . . .. . . .. . . . 7 / 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 )

  12. . .. . .. . . .. . .. .. . . .. . . . .. . .. . . .. . . .. . Solution - Fisher’s Exact Test (cont’d) s Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 . . . .. . . .. . . . . . .. . . .. . . .. . . .. . . .. . . 7 / 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

  13. • Two-sided interval L X • One-sided (with lower-bound) interval L X • One-sided (with upper-bound) interval . .. . . . Interval Estimator . Interval Estimation . .. . . . . . .. . . .. . . . U X April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X U X Three types of intervals . an interval estimator of Then we call L X . U X L X that U X . Based on the observed sample x , we can make an inference L X U X , where L X and U X are functions of sample X and Let L X .. . . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . 8 / 1 ˆ θ ( X ) is usually represented as a point estimator

  14. • Two-sided interval L X • One-sided (with lower-bound) interval L X • One-sided (with upper-bound) interval . .. .. . . .. . . .. . . Interval Estimation . . .. . . . Interval Estimator . . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X U X Three types of intervals an interval estimator of . U X Then we call L X U X L X that . . .. .. . . .. . . .. . . .. . .. .. . . .. . . .. . . . . 8 / 1 . . . .. . . .. . . .. . .. . . .. . . .. ˆ θ ( X ) is usually represented as a point estimator Let [ L ( X ) , U ( X )] , where L ( X ) and U ( X ) are functions of sample X and L ( X ) ≤ U ( X ) . Based on the observed sample x , we can make an inference

  15. • Two-sided interval L X • One-sided (with lower-bound) interval L X • One-sided (with upper-bound) interval . . . .. . . .. . . .. .. . .. . . .. . . . . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X U X Three types of intervals an interval estimator of Interval Estimation U X Then we call L X that . . Interval Estimator . .. . .. . . . .. . . .. . .. . . . .. . . .. . . .. 8 / 1 .. . . . .. . . . . . . .. .. . . . .. . .. ˆ θ ( X ) is usually represented as a point estimator Let [ L ( X ) , U ( X )] , where L ( X ) and U ( X ) are functions of sample X and L ( X ) ≤ U ( X ) . Based on the observed sample x , we can make an inference θ ∈ [ L ( X ) , U ( X )]

  16. • Two-sided interval L X • One-sided (with lower-bound) interval L X • One-sided (with upper-bound) interval . . .. . . .. . . .. .. . . .. . .. . .. . that April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X U X Three types of intervals . . . Interval Estimator . Interval Estimation . . .. . . .. .. . . .. . . . . . .. . . .. . . . .. .. . . . .. . . .. . . .. . . .. . . .. . 8 / 1 ˆ θ ( X ) is usually represented as a point estimator Let [ L ( X ) , U ( X )] , where L ( X ) and U ( X ) are functions of sample X and L ( X ) ≤ U ( X ) . Based on the observed sample x , we can make an inference θ ∈ [ L ( X ) , U ( X )] Then we call [ L ( X ) , U ( X )] an interval estimator of θ .

  17. • Two-sided interval L X • One-sided (with lower-bound) interval L X • One-sided (with upper-bound) interval . . .. . . .. . . .. .. . . .. . .. . .. . that April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X U X Three types of intervals . . . Interval Estimator . Interval Estimation . . .. . . .. .. . . .. . . . . . .. . . .. . . . .. .. . . . .. . . .. . . .. . . .. . . .. . 8 / 1 ˆ θ ( X ) is usually represented as a point estimator Let [ L ( X ) , U ( X )] , where L ( X ) and U ( X ) are functions of sample X and L ( X ) ≤ U ( X ) . Based on the observed sample x , we can make an inference θ ∈ [ L ( X ) , U ( X )] Then we call [ L ( X ) , U ( X )] an interval estimator of θ .

  18. • One-sided (with lower-bound) interval L X • One-sided (with upper-bound) interval . . .. . . .. . . .. . . .. . . .. . . .. . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X Three types of intervals that . . Interval Estimator . Interval Estimation . .. . . .. .. .. .. . . .. . . . . . .. . . .. . . . . .. . . .. . . .. . . .. 8 / 1 . . .. . . .. . ˆ θ ( X ) is usually represented as a point estimator Let [ L ( X ) , U ( X )] , where L ( X ) and U ( X ) are functions of sample X and L ( X ) ≤ U ( X ) . Based on the observed sample x , we can make an inference θ ∈ [ L ( X ) , U ( X )] Then we call [ L ( X ) , U ( X )] an interval estimator of θ . • Two-sided interval [ L ( X ) , U ( X )]

  19. • One-sided (with upper-bound) interval . .. .. . . .. . . . . . .. . . .. . . .. .. . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X Three types of intervals that . . Interval Estimator . Interval Estimation . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . . .. . . .. . . .. .. 8 / 1 . .. . . .. . . ˆ θ ( X ) is usually represented as a point estimator Let [ L ( X ) , U ( X )] , where L ( X ) and U ( X ) are functions of sample X and L ( X ) ≤ U ( X ) . Based on the observed sample x , we can make an inference θ ∈ [ L ( X ) , U ( X )] Then we call [ L ( X ) , U ( X )] an interval estimator of θ . • Two-sided interval [ L ( X ) , U ( X )] • One-sided (with lower-bound) interval [ L ( X ) , ∞ )

  20. . . . . .. . . .. . . .. . . .. . . .. . . . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang Three types of intervals that . Interval Estimator .. . Interval Estimation . .. . . .. .. . . . . .. . . .. . . .. . . .. . . .. .. . . . . .. . . .. .. 8 / 1 . .. . . .. . . ˆ θ ( X ) is usually represented as a point estimator Let [ L ( X ) , U ( X )] , where L ( X ) and U ( X ) are functions of sample X and L ( X ) ≤ U ( X ) . Based on the observed sample x , we can make an inference θ ∈ [ L ( X ) , U ( X )] Then we call [ L ( X ) , U ( X )] an interval estimator of θ . • Two-sided interval [ L ( X ) , U ( X )] • One-sided (with lower-bound) interval [ L ( X ) , ∞ ) • One-sided (with upper-bound) interval ( −∞ , U ( X )]

  21. Pr X : X . . 2. An interval estimator of : X 1. A point estimator of i.i.d. Let X i Example .. Pr . . .. . . .. . . .. X X . Z April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . , where Z as n P n n X Pr n n X n Pr X Pr X Pr X . .. .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . 9 / 1 ∼ N ( µ, 1) . Define

  22. : X . . 2. An interval estimator of i.i.d. Let X i Example . .. . .. Pr . . .. . . .. . . X X . Z April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . , where Z as n P n n X Pr n n X n Pr X Pr X Pr X .. .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . .. .. . . .. . . 9 / 1 ∼ N ( µ, 1) . Define 1. A point estimator of µ : X Pr ( X = µ ) = 0

  23. . .. i.i.d. Let X i Example . .. . . . X . .. . . .. . . .. Pr X . Z April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . , where Z as n P n n Pr X Pr n n X n Pr X Pr X .. . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. 9 / 1 . . . .. . . .. . . .. . .. . . .. . . .. . ∼ N ( µ, 1) . Define 1. A point estimator of µ : X Pr ( X = µ ) = 0 2. An interval estimator of µ : [ X − 1 , X + 1]

  24. . . .. . . .. . . .. . Example .. . . .. . .. .. . Let X i . n April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . , where Z as n P Z i.i.d. n Pr n n X n Pr X Pr . . .. . . . .. . . .. . .. .. . . .. . . .. . . . 9 / 1 . .. . .. . . .. . . .. . . . . .. . . .. . ∼ N ( µ, 1) . Define 1. A point estimator of µ : X Pr ( X = µ ) = 0 2. An interval estimator of µ : [ X − 1 , X + 1] Pr ( µ ∈ [ X − 1 , X + 1]) = Pr ( X − 1 ≤ µ ≤ X + 1)

  25. . . . . .. . . .. . .. . . . .. . .. .. . .. Example .. n April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . , where Z as n P Z Let X i n Pr n n X n Pr i.i.d. . . . .. . .. . . .. . . . . . .. . . .. . . . 9 / 1 .. . .. . . .. . . .. . . .. . .. . . . .. . ∼ N ( µ, 1) . Define 1. A point estimator of µ : X Pr ( X = µ ) = 0 2. An interval estimator of µ : [ X − 1 , X + 1] Pr ( µ ∈ [ X − 1 , X + 1]) = Pr ( X − 1 ≤ µ ≤ X + 1) = Pr ( µ − 1 ≤ X ≤ µ + 1)

  26. . . . . .. . . .. . . .. . . .. . . .. .. . . P April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . , where Z as n n .. Z n Pr i.i.d. Let X i Example . . .. .. .. .. . . . . . . . . .. . . .. . . . .. .. . . .. . . .. . . .. . . .. 9 / 1 . . .. . ∼ N ( µ, 1) . Define 1. A point estimator of µ : X Pr ( X = µ ) = 0 2. An interval estimator of µ : [ X − 1 , X + 1] Pr ( µ ∈ [ X − 1 , X + 1]) = Pr ( X − 1 ≤ µ ≤ X + 1) = Pr ( µ − 1 ≤ X ≤ µ + 1) Pr ( −√ n ≤ √ n ( X − µ ) ≤ √ n ) =

  27. . . .. . . .. .. . .. . . .. . . .. . .. . . . .. . . .. . Example Let X i i.i.d. P Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 . . .. .. . . .. . . .. . . . . . .. . . .. . . . .. . . . .. 9 / 1 . .. . . .. . . .. ∼ N ( µ, 1) . Define 1. A point estimator of µ : X Pr ( X = µ ) = 0 2. An interval estimator of µ : [ X − 1 , X + 1] Pr ( µ ∈ [ X − 1 , X + 1]) = Pr ( X − 1 ≤ µ ≤ X + 1) = Pr ( µ − 1 ≤ X ≤ µ + 1) Pr ( −√ n ≤ √ n ( X − µ ) ≤ √ n ) = Pr ( −√ n ≤ Z ≤ √ n ) = → 1 as n → ∞ , where Z ∼ N (0 , 1) .

  28. . . L X Pr defined as . . Definition : Coverage Probability . Definitions .. In other words, the probability of a random variable in interval . . .. . . .. . . .. U X L X . . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X L X inf Pr Confidence coefficient is defined as . . U X . . . . . Definition: Confidence Coefficient . . covers the parameter . .. .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . 10 / 1 Given an interval estimator [ L ( X ) , U ( X )] of θ , its coverage probability is

  29. . . . . Definition : Coverage Probability . Definitions . .. . .. In other words, the probability of a random variable in interval . . .. . . .. . . defined as L X .. . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X L X inf Pr Confidence coefficient is defined as . . U X . . . . . Definition: Confidence Coefficient . . covers the parameter .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . . . .. . . . .. . . .. . . .. . .. . . .. . . .. . 10 / 1 Given an interval estimator [ L ( X ) , U ( X )] of θ , its coverage probability is Pr ( θ ∈ [ L ( X ) , U ( X )])

  30. . . . Definitions . .. . . .. . . .. . . .. . . .. . Definition : Coverage Probability . .. . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X L X inf Pr Confidence coefficient is defined as . . defined as . . . . . Definition: Confidence Coefficient . In other words, the probability of a random variable in interval .. . . . .. . . .. . . .. . .. . . . .. . . .. . . . 10 / 1 . .. .. . . .. . . .. . . . .. . .. . . .. . . Given an interval estimator [ L ( X ) , U ( X )] of θ , its coverage probability is Pr ( θ ∈ [ L ( X ) , U ( X )]) [ L ( X ) , U ( X )] covers the parameter θ .

  31. . . . . .. . . .. . .. . . . .. . . .. . .. Definitions .. . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang U X L X inf Pr Confidence coefficient is defined as . . Definition: Confidence Coefficient . In other words, the probability of a random variable in interval defined as . . Definition : Coverage Probability .. . . .. . .. . . .. . . . .. . .. . . .. . . . . . . .. . . .. . . .. . .. . . . .. . . .. 10 / 1 Given an interval estimator [ L ( X ) , U ( X )] of θ , its coverage probability is Pr ( θ ∈ [ L ( X ) , U ( X )]) [ L ( X ) , U ( X )] covers the parameter θ .

  32. . .. .. . . .. . . . . . .. . . .. . . . .. . Definition: Confidence Coefficient April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang inf Confidence coefficient is defined as . . . . In other words, the probability of a random variable in interval defined as . . Definition : Coverage Probability . Definitions .. .. . .. .. . . .. . . . . . .. . . .. . . .. . .. .. . . .. . . .. . . 10 / 1 . . . .. . . .. . Given an interval estimator [ L ( X ) , U ( X )] of θ , its coverage probability is Pr ( θ ∈ [ L ( X ) , U ( X )]) [ L ( X ) , U ( X )] covers the parameter θ . θ ∈ Ω Pr ( θ ∈ [ L ( X ) , U ( X )])

  33. where X are random samples from f X x . . . . Definition : Confidence Interval . Definitions . .. . .. Definition: Expected Length . . .. . . .. . . . . .. , its expected length is April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang average length of the interval estimator. . In other words, it is the L X E U X defined as of . U X Given an interval estimator L X . . . . . . .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . . . .. . . . .. . . .. . . .. . . .. . . .. . . .. 11 / 1 Given an interval estimator [ L ( X ) , U ( X )] of θ , if its confidence coefficient is 1 − α , we call it a (1 − α ) confidence interval

  34. where X are random samples from f X x . . . .. . . .. . . .. .. . .. . . .. . . Definitions . defined as April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang average length of the interval estimator. . In other words, it is the L X E U X . .. . Definition: Expected Length . . . Definition : Confidence Interval . .. . . .. . .. . . .. . . . .. . .. . . .. . . . . . . .. . . .. . . .. . 11 / 1 . .. . . .. . . .. Given an interval estimator [ L ( X ) , U ( X )] of θ , if its confidence coefficient is 1 − α , we call it a (1 − α ) confidence interval Given an interval estimator [ L ( X ) , U ( X )] of θ , its expected length is

  35. where X are random samples from f X x . .. .. . . .. . . . . . .. . . .. . . . .. . . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang average length of the interval estimator. . In other words, it is the defined as . . Definition: Expected Length . . . Definition : Confidence Interval . Definitions .. .. . .. .. . . .. . . . . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . .. . . .. . 11 / 1 Given an interval estimator [ L ( X ) , U ( X )] of θ , if its confidence coefficient is 1 − α , we call it a (1 − α ) confidence interval Given an interval estimator [ L ( X ) , U ( X )] of θ , its expected length is E [ U ( X ) − L ( X )]

  36. . . . . .. . . .. . . .. . . .. . . .. .. . . . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang average length of the interval estimator. defined as . Definition: Expected Length .. . . . Definition : Confidence Interval . Definitions . . .. .. .. . . . .. . . . . . .. . . .. . . . .. .. . . .. . . .. . . .. . . .. . . .. . 11 / 1 Given an interval estimator [ L ( X ) , U ( X )] of θ , if its confidence coefficient is 1 − α , we call it a (1 − α ) confidence interval Given an interval estimator [ L ( X ) , U ( X )] of θ , its expected length is E [ U ( X ) − L ( X )] where X are random samples from f X ( x | θ ) . In other words, it is the

  37. . . .. . . .. . . .. . . .. . . .. . .. . A confidence interval can be obtained by inverting the acceptance region April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang intervals (or confidence sets). There is a one-to-one correspondence between tests and confidence of a test. How to construct confidence interval? . . .. . . .. . . .. .. .. . . .. . . .. . . .. . . .. . . . . . . .. . . .. . .. . . . .. . . .. 12 / 1

  38. . . .. . . .. . . .. . . .. . . .. . .. . A confidence interval can be obtained by inverting the acceptance region April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang intervals (or confidence sets). There is a one-to-one correspondence between tests and confidence of a test. How to construct confidence interval? . . .. . . .. . . .. .. .. . . .. . . .. . . .. . . .. . . . . . . .. . . .. . .. . . . .. . . .. 12 / 1

  39. . .. Equivalently, we accept H if z n X i.i.d. X i Example . . n . .. . . .. . . .. . X z .. nz April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n z x n z x Acceptance region is X . nz z n X z . hypothesis because we believe our data ”agrees with” the Accepting H . .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. .. . . .. . . .. . . . .. . .. . . .. . . 13 / 1 ∼ N ( θ, σ 2 ) where σ 2 is known. Consider H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 . As previously shown, level α LRT test reject H 0 if and only if

  40. . . X Equivalently, we accept H if i.i.d. X i Example . .. . .. z . . .. . .. .. . . n . . nz April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n z x n z x Acceptance region is X Accepting H nz z n X z . hypothesis because we believe our data ”agrees with” the .. . . . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . .. 13 / 1 . . . . .. . . .. . .. . .. . . .. . . ∼ N ( θ, σ 2 ) where σ 2 is known. Consider H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 . As previously shown, level α LRT test reject H 0 if and only if � � X − θ 0 � � σ / √ n � > z α /2 � � �

  41. . . Example . .. . . .. . .. i.i.d. .. . .. . . .. . . X i Accepting H . Acceptance region is April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n z x n z x nz because we believe our data ”agrees with” the X nz z n X z . hypothesis .. . . . .. . . .. . . .. . .. .. . . .. . . .. . . . 13 / 1 . .. . .. . . .. . . . .. . .. . . .. . . . ∼ N ( θ, σ 2 ) where σ 2 is known. Consider H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 . As previously shown, level α LRT test reject H 0 if and only if � � X − θ 0 � � σ / √ n � > z α /2 � � � � � � X − θ 0 � ≤ z α /2 . � σ / √ n � Equivalently, we accept H 0 if

  42. . .. . . .. . . .. . .. . . . .. . . .. . .. Example .. Acceptance region is April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n z x n z x nz X i X nz z n X z i.i.d. . . . .. . .. . . .. . . . . . .. . . .. . . . 13 / 1 .. .. .. . . .. . . . .. . . . . .. . .. . . ∼ N ( θ, σ 2 ) where σ 2 is known. Consider H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 . As previously shown, level α LRT test reject H 0 if and only if � � X − θ 0 � � σ / √ n � > z α /2 � � � � � � X − θ 0 � ≤ z α /2 . � σ / √ n � Equivalently, we accept H 0 if Accepting H 0 : θ = θ 0 because we believe our data ”agrees with” the hypothesis θ = θ 0 .

  43. . . . . .. . .. .. . . .. . . .. . . .. .. . . x April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n z x n z Acceptance region is .. nz X nz i.i.d. X i Example . . . .. .. . . .. . . . . . . .. . . .. . . .. 13 / 1 . .. . . .. . . .. . . . .. . . .. . . .. ∼ N ( θ, σ 2 ) where σ 2 is known. Consider H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 . As previously shown, level α LRT test reject H 0 if and only if � � X − θ 0 � � σ / √ n � > z α /2 � � � � � � X − θ 0 � ≤ z α /2 . � σ / √ n � Equivalently, we accept H 0 if Accepting H 0 : θ = θ 0 because we believe our data ”agrees with” the hypothesis θ = θ 0 . X − θ 0 − z α /2 ≤ ≤ z α /2 σ / √ n

  44. . .. .. .. . .. . . . . . .. . . .. . . .. .. x April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n z x n z Acceptance region is . X i.i.d. X i Example . .. . . . . .. . . .. . . . . . . .. . . .. . . .. 13 / 1 . .. . .. . . .. . .. . . . . .. . . .. ∼ N ( θ, σ 2 ) where σ 2 is known. Consider H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 . As previously shown, level α LRT test reject H 0 if and only if � � X − θ 0 � � σ / √ n � > z α /2 � � � � � � X − θ 0 � ≤ z α /2 . � σ / √ n � Equivalently, we accept H 0 if Accepting H 0 : θ = θ 0 because we believe our data ”agrees with” the hypothesis θ = θ 0 . X − θ 0 − z α /2 ≤ ≤ z α /2 σ / √ n θ 0 − σ ≤ θ 0 + σ √ nz α /2 ≤ √ nz α /2

  45. . . .. . . .. . . .. . . .. . . .. . .. . X i April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang Acceptance region is X i.i.d. Example .. . .. . . .. . . . .. . . .. . . .. . .. .. . . . .. . . . 13 / 1 . . . . . .. . . .. .. . . .. . . .. ∼ N ( θ, σ 2 ) where σ 2 is known. Consider H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 . As previously shown, level α LRT test reject H 0 if and only if � � X − θ 0 � � σ / √ n � > z α /2 � � � � � � X − θ 0 � ≤ z α /2 . � σ / √ n � Equivalently, we accept H 0 if Accepting H 0 : θ = θ 0 because we believe our data ”agrees with” the hypothesis θ = θ 0 . X − θ 0 − z α /2 ≤ ≤ z α /2 σ / √ n θ 0 − σ ≤ θ 0 + σ √ nz α /2 ≤ √ nz α /2 { } σ σ x : θ 0 − √ n z α /2 ≤ x ≤ θ 0 + √ n z α /2

  46. Therefore, X . Example (cont’d) . .. . . .. . . Pr .. . . .. . . .. . Pr nz X n z April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang confidence interval (CI). is n z X nz .. X nz X Pr is arbitrary, Since nz X . .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. 14 / 1 . . . .. . . .. . .. . .. .. . . . . .. . As this is size α test, the probability of accepting H 0 is 1 − α . ( ) θ 0 − σ √ nz α /2 ≤ X ≤ θ 0 + σ 1 − α √ nz α /2 =

  47. Therefore, X . . . .. . . .. . . . .. . .. .. . . .. .. Pr Example (cont’d) n z April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang confidence interval (CI). is n z X nz . X nz X Pr is arbitrary, Since Pr . . .. . . . .. . . .. . .. .. . . .. . . .. . . . 14 / 1 . .. . . .. . . .. . . . .. . . . .. . . .. As this is size α test, the probability of accepting H 0 is 1 − α . ( ) θ 0 − σ √ nz α /2 ≤ X ≤ θ 0 + σ 1 − α √ nz α /2 = ( ) X − σ √ nz α /2 ≤ θ 0 ≤ X + σ √ nz α /2 =

  48. Therefore, X . . .. . . .. .. . .. .. . . .. . . .. . . . X April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang confidence interval (CI). is n z n z . Pr Pr Pr Example (cont’d) . .. . . .. .. . . .. . . . . . . .. . . .. . . .. 14 / 1 . . . .. . . .. .. . . . . .. . .. . . .. As this is size α test, the probability of accepting H 0 is 1 − α . ( ) θ 0 − σ √ nz α /2 ≤ X ≤ θ 0 + σ 1 − α √ nz α /2 = ( ) X − σ √ nz α /2 ≤ θ 0 ≤ X + σ √ nz α /2 = Since θ 0 is arbitrary, ( ) X − σ √ nz α /2 ≤ θ ≤ X + σ √ nz α /2 1 − α =

  49. . . .. . . .. . . .. . . .. .. . .. . .. . . . .. . . .. . Example (cont’d) Pr Pr Pr Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 . . .. .. . . .. . . . . . .. . . .. . . .. . . . .. . . . .. 14 / 1 . .. . . .. . . .. As this is size α test, the probability of accepting H 0 is 1 − α . ( ) θ 0 − σ √ nz α /2 ≤ X ≤ θ 0 + σ 1 − α √ nz α /2 = ( ) X − σ √ nz α /2 ≤ θ 0 ≤ X + σ √ nz α /2 = Since θ 0 is arbitrary, ( ) X − σ √ nz α /2 ≤ θ ≤ X + σ √ nz α /2 1 − α = σ σ Therefore, [ X − √ n z α /2 , X + √ n z α /2 ] is (1 − α ) confidence interval (CI).

  50. . . .. . . .. . . .. . . .. . . .. . . .. . Theorem 9.2.2 . by A April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . . Then the test has level C x x define the acceptance region of a test for the hypothesis H . , , for any confidence set for 2 Conversely, if C X is a . . . . .. . .. . . . .. . . .. . .. . . . .. . . .. . . . .. . . . .. . . .. . . .. . .. . . .. . . .. 15 / 1 Confidence intervals and level α test 1 For each θ 0 ∈ Ω , let A ( θ 0 ) be the acceptance region of a level α test of H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 Define a set C ( X ) = { θ : x ∈ A ( θ ) } , then the random set C ( X ) is a 1 − α confidence set.

  51. . . . . .. . . .. . . .. . . .. .. . .. . . . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . . . . .. Theorem 9.2.2 . . .. . . .. . . . . .. .. . . .. . . .. . . .. . . .. . . . . . .. . . .. .. 15 / 1 . .. . . .. . . Confidence intervals and level α test 1 For each θ 0 ∈ Ω , let A ( θ 0 ) be the acceptance region of a level α test of H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 Define a set C ( X ) = { θ : x ∈ A ( θ ) } , then the random set C ( X ) is a 1 − α confidence set. 2 Conversely, if C ( X ) is a (1 − α ) confidence set for θ , for any θ 0 , define the acceptance region of a test for the hypothesis H 0 : θ = θ 0 by A ( θ 0 ) = { x : θ 0 ∈ C ( x ) } . Then the test has level α .

  52. . .. .. . . .. . . . . . .. . . .. .. . . .. . nz April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang nz X nz X X . nz C X The confidence set C X is a subset of the parameter space sample space i.i.d. For X i Example .. . . .. .. . . .. . . . . . .. . . .. . . . .. .. . . . .. . . .. . . .. . .. . . . .. . 16 / 1 ∼ N ( θ, σ 2 ) , the acceptance region A ( θ 0 ) is a subset of the { } x : θ 0 − σ √ nz α /2 ≤ X ≤ θ 0 + σ √ nz α /2 A ( θ 0 ) =

  53. . .. .. . . .. . . . . . .. . . .. . . .. .. X April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang nz X nz sample space . i.i.d. For X i Example . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . .. .. . . .. . . .. 16 / 1 . . . . .. . . .. ∼ N ( θ, σ 2 ) , the acceptance region A ( θ 0 ) is a subset of the { } x : θ 0 − σ √ nz α /2 ≤ X ≤ θ 0 + σ √ nz α /2 A ( θ 0 ) = The confidence set C ( X ) is a subset of the parameter space { } θ : θ − σ √ nz α /2 ≤ X ≤ θ + σ √ nz α /2 C ( X ) =

  54. . . .. . . .. . . .. .. . .. . . .. . .. . . . .. . . .. . Example For X i i.i.d. sample space Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 . . .. .. . . .. . . . . . .. . . .. . . .. . . . .. . . . .. 16 / 1 . .. . . .. . . .. ∼ N ( θ, σ 2 ) , the acceptance region A ( θ 0 ) is a subset of the { } x : θ 0 − σ √ nz α /2 ≤ X ≤ θ 0 + σ √ nz α /2 A ( θ 0 ) = The confidence set C ( X ) is a subset of the parameter space { } θ : θ − σ √ nz α /2 ≤ X ≤ θ + σ √ nz α /2 C ( X ) = { } θ : X − σ √ nz α /2 ≤ θ ≤ X + σ √ nz α /2 =

  55. . .. two-sided CI L X 1 To obtain . . 9.2.2 is an interval, but quite often There is no guarantee that the confidence set obtained from Theorem Confidence set and confidence interval . . acceptance region of a level . .. . . .. . . .. . U X , we invert the test for H .. 3 To obtain a upper-bounded CI April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . , where vs. H acceptance region of a test for H U X , then we invert the . vs. H . . , where vs. H acceptance region of a test for H , then we invert the 2 To obtain a lower-bounded CI L X . . . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . . . . . .. . . .. . . .. . .. . . . .. . . .. . . .. 17 / 1

  56. . . There is no guarantee that the confidence set obtained from Theorem Confidence set and confidence interval . .. . . .. . . .. . . .. . . .. . 9.2.2 is an interval, but quite often . .. 3 To obtain a upper-bounded CI April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . , where vs. H acceptance region of a test for H U X , then we invert the . . . . , where vs. H acceptance region of a test for H , then we invert the 2 To obtain a lower-bounded CI L X . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . . 17 / 1 . . .. . . .. . . .. . .. . . .. . . .. . . .. 1 To obtain (1 − α ) two-sided CI [ L ( X ) , U ( X )] , we invert the acceptance region of a level α test for H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0

  57. . . . . .. . . .. . .. . . . .. . . .. .. .. Confidence set and confidence interval .. U X , then we invert the April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . , where vs. H acceptance region of a test for H 3 To obtain a upper-bounded CI There is no guarantee that the confidence set obtained from Theorem . . . . . . 9.2.2 is an interval, but quite often . . . .. . .. . . .. . . . .. . .. . . .. . . . . . .. .. . . .. . . .. . . . . . .. . . .. 17 / 1 1 To obtain (1 − α ) two-sided CI [ L ( X ) , U ( X )] , we invert the acceptance region of a level α test for H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 2 To obtain a lower-bounded CI [ L ( X ) , ∞ ) , then we invert the acceptance region of a test for H 0 : θ = θ 0 vs. H 1 : θ > θ 0 , where Ω = { θ : θ ≥ θ 0 } .

  58. . .. .. . . .. . . . . . .. . . .. . . .. .. . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . . . . . . 9.2.2 is an interval, but quite often There is no guarantee that the confidence set obtained from Theorem Confidence set and confidence interval . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . .. .. . . .. . .. . . . . . .. . . .. 17 / 1 1 To obtain (1 − α ) two-sided CI [ L ( X ) , U ( X )] , we invert the acceptance region of a level α test for H 0 : θ = θ 0 vs. H 1 : θ ̸ = θ 0 2 To obtain a lower-bounded CI [ L ( X ) , ∞ ) , then we invert the acceptance region of a test for H 0 : θ = θ 0 vs. H 1 : θ > θ 0 , where Ω = { θ : θ ≥ θ 0 } . 3 To obtain a upper-bounded CI ( −∞ , U ( X )] , then we invert the acceptance region of a test for H 0 : θ = θ 0 vs. H 1 : θ < θ 0 , where Ω = { θ : θ ≤ θ 0 } .

  59. . . . . .. . . .. . .. . . . .. . . .. . .. Example .. two-sided CI for April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang upper bound for 2 Find . . 1 Find . . . i.i.d. X i . . Problem . .. . .. . .. . . .. . . . . . .. . . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. 18 / 1 ∼ N ( µ, σ 2 ) where both parameters are unknown.

  60. . . . . .. . . .. . . .. . . .. . . .. .. . . . April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang . . . i.i.d. .. X i . . Problem . Example . .. . .. .. .. . . .. . . . . . .. . . .. . . . . .. . . .. . . .. . . 18 / 1 .. . . .. . . .. . ∼ N ( µ, σ 2 ) where both parameters are unknown. 1 Find 1 − α two-sided CI for µ 2 Find 1 − α upper bound for µ

  61. . . x A The acceptance region is t n n s X X Example - two-sided CI - Solution .. s x . . .. . . .. . . .. x n .. t n April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang nt n s x x nt n s x x n t n s x x t n t n n s x x C x The confidence set is . . .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . 19 / 1 H 0 : µ = µ 0 vs H 1 : µ ̸ = µ 0 . The LRT test rejects if and only if

  62. . . x x A The acceptance region is Example - two-sided CI - Solution . .. . .. n . . .. . .. .. . . s x t n . x April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang nt n s x x nt n s x t n The confidence set is n s x x t n t n n s x x C x .. . . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . . .. .. .. . . .. . . .. . . .. . . . 19 / 1 . H 0 : µ = µ 0 vs H 1 : µ ̸ = µ 0 . The LRT test rejects if and only if � � X − µ 0 � � s X / √ n � > t n − 1 ,α /2 � � �

  63. . . Example - two-sided CI - Solution . .. . . .. .. .. The confidence set is . . .. . . .. . . The acceptance region is C x . s x April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang nt n s x x nt n x x t n n s x x t n t n n s x .. . . . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. 19 / 1 . . . .. . . .. . .. .. . . .. . . . . H 0 : µ = µ 0 vs H 1 : µ ̸ = µ 0 . The LRT test rejects if and only if � � X − µ 0 � � s X / √ n � > t n − 1 ,α /2 � � � � � { x − µ 0 } � � A ( µ 0 ) = x : s x / √ n � ≤ t n − 1 ,α /2 � � �

  64. . . . .. . . .. . .. .. . . .. . . .. . . .. .. s x April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang nt n s x x nt n x Example - two-sided CI - Solution t n n s x x t n The confidence set is The acceptance region is . . . .. . .. . . .. . . . . . .. . . .. . . . 19 / 1 .. .. . . .. . . .. . . .. . . . . . .. .. . H 0 : µ = µ 0 vs H 1 : µ ̸ = µ 0 . The LRT test rejects if and only if � � X − µ 0 � � s X / √ n � > t n − 1 ,α /2 � � � � � { x − µ 0 } � � A ( µ 0 ) = x : s x / √ n � ≤ t n − 1 ,α /2 � � � � � { x − µ } � � C ( x ) = µ : s x / √ n � ≤ t n − 1 ,α /2 � � �

  65. . .. .. . . .. . . . . . .. . . .. . . .. .. nt n April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang nt n s x x s x . x The confidence set is The acceptance region is Example - two-sided CI - Solution .. .. . . . . .. . . .. . . . . . . .. . . .. . . .. 19 / 1 . . .. . . .. . . .. . . .. . .. . .. . H 0 : µ = µ 0 vs H 1 : µ ̸ = µ 0 . The LRT test rejects if and only if � � X − µ 0 � � s X / √ n � > t n − 1 ,α /2 � � � � � { x − µ 0 } � � A ( µ 0 ) = x : s x / √ n � ≤ t n − 1 ,α /2 � � � � � { x − µ } � � C ( x ) = µ : s x / √ n � ≤ t n − 1 ,α /2 � � � { µ : − t n − 1 ,α /2 ≤ x − µ } = s x / √ n ≤ t n − 1 ,α /2

  66. . . . .. . . .. . . .. . . .. . . .. . .. .. . . .. . . .. .. Example - two-sided CI - Solution The acceptance region is The confidence set is Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 . . . . . . .. . . . .. . . .. . . .. . 19 / 1 .. .. . . . .. . . .. . .. . . .. . . H 0 : µ = µ 0 vs H 1 : µ ̸ = µ 0 . The LRT test rejects if and only if � � X − µ 0 � � s X / √ n � > t n − 1 ,α /2 � � � � � { x − µ 0 } � � A ( µ 0 ) = x : s x / √ n � ≤ t n − 1 ,α /2 � � � � � { x − µ } � � C ( x ) = µ : s x / √ n � ≤ t n − 1 ,α /2 � � � { µ : − t n − 1 ,α /2 ≤ x − µ } = s x / √ n ≤ t n − 1 ,α /2 { } µ : x − s x √ nt n − 1 ,α /2 ≤ µ ≤ x + s x √ nt n − 1 ,α /2 =

  67. . . Example - upper-bounded CI - Solution . .. . . .. . .. x . . .. . . .. . . LRT statistic is L .. , and April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n X i i n , x , and Within restricted to is the MLE , and is the MLE restricted to where x L .. . . . .. . . .. . . .. . .. . . . .. . . .. . . . .. .. .. . . .. . . .. . . . . .. . . .. . . 20 / 1 The CI is ( −∞ , U ( X )] . We need to invert a testing procedure for H 0 : µ = µ 0 vs H 1 : µ < µ 0 .

  68. . . Example - upper-bounded CI - Solution . .. . . .. . .. x . . .. . . .. . .. LRT statistic is L . , and April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n X i i n , x , and Within restricted to is the MLE , and is the MLE restricted to where x L .. . . . . .. . . .. . . .. . .. .. . . .. . . .. . . . 20 / 1 .. . . . .. . . .. . . .. . .. . . .. . . The CI is ( −∞ , U ( X )] . We need to invert a testing procedure for H 0 : µ = µ 0 vs H 1 : µ < µ 0 . { ( µ, σ 2 ) : µ = µ 0 , σ 2 > 0 } Ω 0 =

  69. . . Example - upper-bounded CI - Solution . .. . . .. . .. x . . .. . . .. .. . LRT statistic is L . , and April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n X i i n , x , and Within restricted to is the MLE , and is the MLE restricted to where x L .. . . . . . .. . . .. . . .. . .. .. . . .. . . .. . . .. 20 / 1 .. . . . .. . . . . .. . . . .. . . .. The CI is ( −∞ , U ( X )] . We need to invert a testing procedure for H 0 : µ = µ 0 vs H 1 : µ < µ 0 . { ( µ, σ 2 ) : µ = µ 0 , σ 2 > 0 } Ω 0 = { ( µ, σ 2 ) : µ ≤ µ 0 , σ 2 > 0 } Ω =

  70. . . . .. . . .. . .. .. . .. .. . . .. . . . .. , and April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n X i i n , Example - upper-bounded CI - Solution , and Within restricted to is the MLE , and is the MLE restricted to where LRT statistic is . . . .. . .. . . .. . . . . . .. . . .. . . . 20 / 1 .. .. . . .. . . . . .. . . . . . .. .. . .. The CI is ( −∞ , U ( X )] . We need to invert a testing procedure for H 0 : µ = µ 0 vs H 1 : µ < µ 0 . { ( µ, σ 2 ) : µ = µ 0 , σ 2 > 0 } Ω 0 = { ( µ, σ 2 ) : µ ≤ µ 0 , σ 2 > 0 } Ω = σ 2 0 | x ) L (ˆ µ 0 , ˆ λ ( x ) = σ 2 | x ) L (ˆ µ, ˆ

  71. . .. .. . . .. .. . . . . .. . . .. . . .. .. n April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang n X i i , and . , Within LRT statistic is Example - upper-bounded CI - Solution . .. . . . . .. . . .. . . . . . . .. . . .. . . .. 20 / 1 . . . .. .. . . .. . . .. . .. . .. . . The CI is ( −∞ , U ( X )] . We need to invert a testing procedure for H 0 : µ = µ 0 vs H 1 : µ < µ 0 . { ( µ, σ 2 ) : µ = µ 0 , σ 2 > 0 } Ω 0 = { ( µ, σ 2 ) : µ ≤ µ 0 , σ 2 > 0 } Ω = σ 2 0 | x ) L (ˆ µ 0 , ˆ λ ( x ) = σ 2 | x ) L (ˆ µ, ˆ σ 2 σ 2 ) is the MLE where (ˆ µ 0 , ˆ 0 ) is the MLE restricted to Ω 0 , and (ˆ µ, ˆ restricted to Ω , and

  72. . . . .. . . .. . . .. . . .. . .. .. . .. .. . . .. . . .. . Example - upper-bounded CI - Solution LRT statistic is n Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 . . . . . . .. . . . .. . . .. . . .. . 20 / 1 .. . .. . . . .. .. . . .. . . .. . . The CI is ( −∞ , U ( X )] . We need to invert a testing procedure for H 0 : µ = µ 0 vs H 1 : µ < µ 0 . { ( µ, σ 2 ) : µ = µ 0 , σ 2 > 0 } Ω 0 = { ( µ, σ 2 ) : µ ≤ µ 0 , σ 2 > 0 } Ω = σ 2 0 | x ) L (ˆ µ 0 , ˆ λ ( x ) = σ 2 | x ) L (ˆ µ, ˆ σ 2 σ 2 ) is the MLE where (ˆ µ 0 , ˆ 0 ) is the MLE restricted to Ω 0 , and (ˆ µ, ˆ i =1 ( X i − µ 0 ) 2 ∑ n σ 2 restricted to Ω , and Within Ω 0 , ˆ µ 0 = µ 0 , and ˆ 0 =

  73. if X if X if X n s X n s X n n X i i n n X X i i . X if X . .. . . .. . . .. Example - upper bounded CI - Solution (cont’d) n x if X April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang if X n X n n X .. Xi i n exp n Xi i n exp . . .. .. .. . . .. . . .. . . . . . .. . . .. . . .. . . . .. . . . .. . . .. . . .. . .. . . . .. . . .. . . .. . 21 / 1 Within Ω , the MLE is

  74. if X if X n s X n s X .. if X n X i i n n Example - upper bounded CI - Solution (cont’d) . . . . .. . . .. .. . .. x exp n if X April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang if X n X n n X . Xi i n exp n Xi i n . . .. . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . . . .. . . . .. . . .. . . .. . . .. 21 / 1 . . .. . Within Ω , the MLE is { i =1 ( X i − X ) 2 σ 2 = ∑ n if X ≤ µ 0 µ = X ˆ ˆ

  75. if X if X n s X n s X x n n Example - upper bounded CI - Solution (cont’d) . .. . . . .. n . .. . . .. . . . n exp n April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang if X n X n if X . X Xi i n exp n Xi i .. .. . . . .. . . .. . . .. . . .. . .. .. . . .. . . .. . . .. . . . 21 / 1 .. . . . . .. . .. .. . . .. . . . Within Ω , the MLE is { i =1 ( X i − X ) 2 σ 2 = ∑ n if X ≤ µ 0 µ = X ˆ ˆ i =1 ( X i − µ 0 ) 2 σ 2 = ∑ n µ = µ 0 ˆ ˆ if X > µ 0

  76. n s X n s X . .. . . .. . . .. . . . .. . . .. . . . .. n April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang if X n X n .. if X exp exp n n .. Example - upper bounded CI - Solution (cont’d) . 21 / 1 . . . . .. . . .. . .. .. . . . .. . . .. . . . . .. . . .. . . .. . . .. . . .. . . .. Within Ω , the MLE is { i =1 ( X i − X ) 2 σ 2 = ∑ n if X ≤ µ 0 µ = X ˆ ˆ i =1 ( X i − µ 0 ) 2 σ 2 = ∑ n µ = µ 0 ˆ ˆ if X > µ 0  1 if X > µ 0   i =1( Xi − µ 0)2 ( ) n { } ∑ n  1 √ λ ( x ) = − σ 2 2 πσ 2 2ˆ 0 if X ≤ µ 0 i =1( Xi − X )2 ( ) n { }  ∑ n 1  √ −  σ 2 2 πσ 2 2ˆ 0

  77. . .. . . .. . . .. . . .. . . .. . . . . n April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang X exp exp .. . Example - upper bounded CI - Solution (cont’d) . .. . . .. .. n . . . .. . . .. . .. .. .. . . .. . . . 21 / 1 . .. . . . . .. . . . .. .. . . .. . . Within Ω , the MLE is { i =1 ( X i − X ) 2 σ 2 = ∑ n if X ≤ µ 0 µ = X ˆ ˆ i =1 ( X i − µ 0 ) 2 σ 2 = ∑ n µ = µ 0 ˆ ˆ if X > µ 0  1 if X > µ 0   i =1( Xi − µ 0)2 ( ) n { } ∑ n  1 √ λ ( x ) = − σ 2 2 πσ 2 2ˆ 0 if X ≤ µ 0 i =1( Xi − X )2 ( ) n { }  ∑ n 1  √ −  σ 2 2 πσ 2 2ˆ 0  1 if X > µ 0  = ( ) n n − 1 n s 2 2 if X ≤ µ 0 n − 1  n s 2 X +( X − µ 0 ) 2

  78. n s X n s X .. n n Example - upper bounded CI - Solution (cont’d) . .. . . . X . .. . . .. . . .. . n .. s X April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang c n s X X c X c c n s x X n n n n . . .. . .. . . .. . . .. . .. . . . .. . . .. . . . .. . . . .. . . .. . . .. . . .. . . .. . . .. . 22 / 1 For 0 < c < 1 , LRT test rejects H 0 if X ≤ µ 0 and

  79. . .. . .. . . .. . . . X . .. . . .. . . Example - upper bounded CI - Solution (cont’d) c . c April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang c n s X X s X n X c n s x X n n n .. .. . . . .. . . .. . . .. . . .. . . .. . . .. . . .. 22 / 1 . . . . .. . . .. . .. . . .. . .. . . .. For 0 < c < 1 , LRT test rejects H 0 if X ≤ µ 0 and ( ) n n − 1 n s 2 2 < n − 1 n s 2 X + ( X − µ 0 ) 2

  80. . . . . .. . . .. . .. . . . .. . . .. . .. Example - upper bounded CI - Solution (cont’d) .. c April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang c n s X X s X X X c n x n n c . .. . .. . .. . . .. . . . . . .. . . .. . . . 22 / 1 .. .. . . . .. . . .. . .. . . .. . . . .. . For 0 < c < 1 , LRT test rejects H 0 if X ≤ µ 0 and ( ) n n − 1 n s 2 2 < n − 1 n s 2 X + ( X − µ 0 ) 2   2 n − 1 <   + ( X − µ 0 ) 2 n − 1 s 2

  81. . .. .. . . .. . .. . . . .. . . .. . . . .. . X April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang c n s X X c . n x n n c X Example - upper bounded CI - Solution (cont’d) .. . . .. .. .. . . .. . . . . . .. . . .. . . . 22 / 1 .. . . .. .. . . . .. . . .. . .. . . . . For 0 < c < 1 , LRT test rejects H 0 if X ≤ µ 0 and ( ) n n − 1 n s 2 2 < n − 1 n s 2 X + ( X − µ 0 ) 2   2 n − 1 <   + ( X − µ 0 ) 2 n − 1 s 2 ( X − µ 0 ) 2 c ∗ > s 2

  82. . .. .. . .. .. . . . . . .. . . .. . . .. .. x April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang X c n n . n c X Example - upper bounded CI - Solution (cont’d) . .. . . . . .. . . .. . . . . . . .. . . .. . . .. 22 / 1 . . .. . . .. . . .. .. . .. . . . .. . For 0 < c < 1 , LRT test rejects H 0 if X ≤ µ 0 and ( ) n n − 1 n s 2 2 < n − 1 n s 2 X + ( X − µ 0 ) 2   2 n − 1 <   + ( X − µ 0 ) 2 n − 1 s 2 ( X − µ 0 ) 2 c ∗ > s 2 µ 0 − X c ∗∗ s X / √ n >

  83. . . n s X X Pr Example - upper bounded CI - Solution (cont’d) . .. . .. Pr . . .. . . .. . . c X . Therefore, LRT level April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang t n n s X X test reject H if t n s X t n c c Pr T n c Pr T n c n .. .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. . .. . . . .. . . .. . . .. . .. .. . . .. . . 23 / 1 c ∗∗ is chosen to satisfy α = Pr ( reject H 0 | µ 0 )

  84. . . Pr Example - upper bounded CI - Solution (cont’d) . .. . . .. . X .. . . .. . . .. . Pr s X .. test reject H if April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang t n n s X X Therefore, LRT level n t n t n c c Pr T n c Pr T n c . .. . . . . .. . . .. . . .. . . .. . . .. . . .. . . .. 23 / 1 . . .. . . .. . . .. . . .. . .. . . .. . c ∗∗ is chosen to satisfy α = Pr ( reject H 0 | µ 0 ) ( µ 0 − X ) s X / √ n > c ∗∗ =

  85. . . .. . . .. . . .. . Example - upper bounded CI - Solution (cont’d) .. . . .. . . .. . Pr . test reject H if April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun Min Kang t n n s X X Therefore, LRT level Pr t n t n c c Pr T n c Pr T n . .. .. .. . .. . . .. . . . . . .. . . .. . . . .. . .. . .. . . .. . . .. . . .. . . . . .. . 23 / 1 c ∗∗ is chosen to satisfy α = Pr ( reject H 0 | µ 0 ) ( µ 0 − X ) s X / √ n > c ∗∗ = ( X − µ 0 ) s X / √ n < − c ∗∗ =

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