. Biostatistics 602 - Statistical Inference Lecture 23

. Interval Estimation

Hyun Min Kang

April 11th, 2013

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 1 / 1 • What is p-value? • What is the advantage of p-value compared to hypothesis testing procedure with size α? • How can one construct a valid p-value? • What is Fisher’s exact p-value? • Is Fisher’s exact p-value uniformly distributed under null hypothesis?

Last Lecture

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 2 / 1 • What is the advantage of p-value compared to hypothesis testing procedure with size α? • How can one construct a valid p-value? • What is Fisher’s exact p-value? • Is Fisher’s exact p-value uniformly distributed under null hypothesis?

Last Lecture

• What is p-value?

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 2 / 1 • How can one construct a valid p-value? • What is Fisher’s exact p-value? • Is Fisher’s exact p-value uniformly distributed under null hypothesis?

Last Lecture

• What is p-value? • What is the advantage of p-value compared to hypothesis testing procedure with size α?

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 2 / 1 • What is Fisher’s exact p-value? • Is Fisher’s exact p-value uniformly distributed under null hypothesis?

Last Lecture

• What is p-value? • What is the advantage of p-value compared to hypothesis testing procedure with size α? • How can one construct a valid p-value?

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 2 / 1 • Is Fisher’s exact p-value uniformly distributed under null hypothesis?

Last Lecture

• What is p-value? • What is the advantage of p-value compared to hypothesis testing procedure with size α? • How can one construct a valid p-value? • What is Fisher’s exact p-value?

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 2 / 1 Last Lecture

• What is p-value? • What is the advantage of p-value compared to hypothesis testing procedure with size α? • How can one construct a valid p-value? • What is Fisher’s exact p-value? • Is Fisher’s exact p-value uniformly distributed under null hypothesis?

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 2 / 1 • If size of the test is (α) small, the decision to reject H0 is convincing. • If α is large, the decision may not be very convincing.

. Definition: p-Value . 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 H1 is true. A p-value is valid if, for every θ ∈ Ω0 and every 0 ≤ α ≤ 1, . Pr(p(X) ≤ α|θ) ≤ α

p-Values

. Conclusions from Hypothesis Testing . • Reject H0 or accept H0.

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 3 / 1 • If α is large, the decision may not be very convincing.

. Definition: p-Value . 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 H1 is true. A p-value is valid if, for every θ ∈ Ω0 and every 0 ≤ α ≤ 1, . Pr(p(X) ≤ α|θ) ≤ α

p-Values

. Conclusions from Hypothesis Testing . • Reject H0 or accept H0.

• If size of the test is (α) small, the decision to reject H0 is convincing.

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 3 / 1 . Definition: p-Value . 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 H1 is true. A p-value is valid if, for every θ ∈ Ω0 and every 0 ≤ α ≤ 1, . Pr(p(X) ≤ α|θ) ≤ α

p-Values

. Conclusions from Hypothesis Testing . • Reject H0 or accept H0.

• If size of the test is (α) small, the decision to reject H0 is convincing. • If α is large, the decision may not be very convincing. .

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 3 / 1 Pr(p(X) ≤ α|θ) ≤ α

p-Values

. Conclusions from Hypothesis Testing . • Reject H0 or accept H0.

• If size of the test is (α) small, the decision to reject H0 is convincing. • If α is large, the decision may not be very convincing. . . Definition: p-Value . 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 H1 is true. A p-value is valid if, for every θ ∈ Ω0 and every 0 ≤ α ≤ 1, .

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 3 / 1 p-Values

. Conclusions from Hypothesis Testing . • Reject H0 or accept H0.

• If size of the test is (α) small, the decision to reject H0 is convincing. • If α is large, the decision may not be very convincing. . . Definition: p-Value . 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 H1 is true. A p-value is valid if, for every θ ∈ Ω0 and every 0 ≤ α ≤ 1, . Pr(p(X) ≤ α|θ) ≤ α

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 3 / 1 p(x) = sup Pr(W(X) ≥ W(x)|θ) θ∈Ω0

Then p(X) is a valid p-value.

Constructing a valid p-value

. Theorem 8.3.27. . Let W(X) be a test statistic such that large values of W give evidence that H1 is true. For each sample point x, define

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 4 / 1 Then p(X) is a valid p-value.

Constructing a valid p-value

. Theorem 8.3.27. . Let W(X) be a test statistic such that large values of W give evidence that H1 is true. For each sample point x, define p(x) = sup Pr(W(X) ≥ W(x)|θ) θ∈Ω0

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 4 / 1 Constructing a valid p-value

. Theorem 8.3.27. . Let W(X) be a test statistic such that large values of W give evidence that H1 is true. For each sample point x, define p(x) = sup Pr(W(X) ≥ W(x)|θ) θ∈Ω0

.Then p(X) is a valid p-value.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 4 / 1 If the null hypothesis is 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 H1 is true. Define

p(x) = Pr(W(X) ≥ W(x)|S = S(x))

If we consider only the conditional distribution, by Theorem 8.3.27, this is a valid p-value, meaning that

Pr(p(X) ≤ α|S = s) ≤ α

p-Values by conditioning on on sufficient statistic

Suppose S(X) is a sufficient statistic for the model {f(x|θ): θ ∈ Ω0}. (not necessarily including alternative hypothesis).

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 5 / 1 Again, let W(X) denote a test statistic where large value give evidence that H1 is true. Define

p(x) = Pr(W(X) ≥ W(x)|S = S(x))

If we consider only the conditional distribution, by Theorem 8.3.27, this is a valid p-value, meaning that

Pr(p(X) ≤ α|S = s) ≤ α

p-Values by conditioning on on sufficient statistic

Suppose S(X) is a sufficient statistic for the model {f(x|θ): θ ∈ Ω0}. (not necessarily including alternative hypothesis). If the null hypothesis is true, the conditional distribution of X given S = s does not depend on θ.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 5 / 1 If we consider only the conditional distribution, by Theorem 8.3.27, this is a valid p-value, meaning that

Pr(p(X) ≤ α|S = s) ≤ α

p-Values by conditioning on on sufficient statistic

Suppose S(X) is a sufficient statistic for the model {f(x|θ): θ ∈ Ω0}. (not necessarily including alternative hypothesis). If the null hypothesis is 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 H1 is true. Define

p(x) = Pr(W(X) ≥ W(x)|S = S(x))

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 5 / 1 p-Values by conditioning on on sufficient statistic

Suppose S(X) is a sufficient statistic for the model {f(x|θ): θ ∈ Ω0}. (not necessarily including alternative hypothesis). If the null hypothesis is 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 H1 is true. Define

p(x) = Pr(W(X) ≥ W(x)|S = S(x))

If we consider only the conditional distribution, by Theorem 8.3.27, this is a valid p-value, meaning that

Pr(p(X) ≤ α|S = s) ≤ α

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 5 / 1 . Solution . Under H0, if we let p denote the common value of p1 = p2. Then the join pmf of (X1, X2) is ( ) ( ) n n 1 x1 n1−x1 2 x2 n2−x2 f(x1, x2|p) = p (1 − p) p (1 − p) (x1)( ) x2 n n − − = 1 2 px1+x2 (1 − p)n1+n2 x1 x2 x1 x2

.Therefore S = X1 + X2 is a sufficient statistic under H0.

Example - Fisher’s Exact Test

. Problem . Let X1 and X2 be independent observations with X1 ∼ Binomial(n1, p1), and X2 ∼ Binomial(n2, p2). Consider testing H0 : p1 = p2 versus .H1 : p1 > p2. Find a valid p-value function.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 6 / 1 ( ) ( ) n n 1 x1 n1−x1 2 x2 n2−x2 f(x1, x2|p) = p (1 − p) p (1 − p) (x1)( ) x2 n n − − = 1 2 px1+x2 (1 − p)n1+n2 x1 x2 x1 x2

Therefore S = X1 + X2 is a sufficient statistic under H0.

Example - Fisher’s Exact Test

. Problem . Let X1 and X2 be independent observations with X1 ∼ Binomial(n1, p1), and X2 ∼ Binomial(n2, p2). Consider testing H0 : p1 = p2 versus .H1 : p1 > p2. Find a valid p-value function. . Solution . Under H0, if we let p denote the common value of p1 = p2. Then the join pmf of (X1, X2) is

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 6 / 1 ( )( ) n n − − = 1 2 px1+x2 (1 − p)n1+n2 x1 x2 x1 x2

Therefore S = X1 + X2 is a sufficient statistic under H0.

Example - Fisher’s Exact Test

. Problem . Let X1 and X2 be independent observations with X1 ∼ Binomial(n1, p1), and X2 ∼ Binomial(n2, p2). Consider testing H0 : p1 = p2 versus .H1 : p1 > p2. Find a valid p-value function. . Solution . Under H0, if we let p denote the common value of p1 = p2. Then the join pmf of (X1, X2) is ( ) ( ) n n 1 x1 n1−x1 2 x2 n2−x2 f(x1, x2|p) = p (1 − p) p (1 − p) x1 x2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 6 / 1 Therefore S = X1 + X2 is a sufficient statistic under H0.

Example - Fisher’s Exact Test

. Problem . Let X1 and X2 be independent observations with X1 ∼ Binomial(n1, p1), and X2 ∼ Binomial(n2, p2). Consider testing H0 : p1 = p2 versus .H1 : p1 > p2. Find a valid p-value function. . Solution . Under H0, if we let p denote the common value of p1 = p2. Then the join pmf of (X1, X2) is ( ) ( ) n n 1 x1 n1−x1 2 x2 n2−x2 f(x1, x2|p) = p (1 − p) p (1 − p) (x1)( ) x2 n n − − = 1 2 px1+x2 (1 − p)n1+n2 x1 x2 x1 x2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 6 / 1 Example - Fisher’s Exact Test

. Problem . Let X1 and X2 be independent observations with X1 ∼ Binomial(n1, p1), and X2 ∼ Binomial(n2, p2). Consider testing H0 : p1 = p2 versus .H1 : p1 > p2. Find a valid p-value function. . Solution . Under H0, if we let p denote the common value of p1 = p2. Then the join pmf of (X1, X2) is ( ) ( ) n n 1 x1 n1−x1 2 x2 n2−x2 f(x1, x2|p) = p (1 − p) p (1 − p) (x1)( ) x2 n n − − = 1 2 px1+x2 (1 − p)n1+n2 x1 x2 x1 x2

.Therefore S = X1 + X2 is a sufficient statistic under H0...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 6 / 1 The conditional distribution of X1 given S = s is a hypergeometric distribution. ( )( ) n1 n2 − f(X = x |s) = x(1 s x)1 1 1 n1+n2 s

Thus, the p-value conditional on the sufficient statistic s = x1 + x2 is

min∑(n1,s) p(x1, x2) = f(j|s)

j=x1

Solution - Fisher’s Exact Test (cont’d)

Given the value of S = s, it is reasonable to use X1 as a test statistic and reject H0 in favor of H1 for large values of X1,because large values of X1 correspond to small values of X2 = s − X1.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 7 / 1 Thus, the p-value conditional on the sufficient statistic s = x1 + x2 is

min∑(n1,s) p(x1, x2) = f(j|s)

j=x1

Solution - Fisher’s Exact Test (cont’d)

Given the value of S = s, it is reasonable to use X1 as a test statistic and reject H0 in favor of H1 for large values of X1,because large values of X1 correspond to small values of X2 = s − X1. The conditional distribution of X1 given S = s is a hypergeometric distribution. ( )( ) n1 n2 − f(X = x |s) = x(1 s x)1 1 1 n1+n2 s

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 7 / 1 Solution - Fisher’s Exact Test (cont’d)

Given the value of S = s, it is reasonable to use X1 as a test statistic and reject H0 in favor of H1 for large values of X1,because large values of X1 correspond to small values of X2 = s − X1. The conditional distribution of X1 given S = s is a hypergeometric distribution. ( )( ) n1 n2 − f(X = x |s) = x(1 s x)1 1 1 n1+n2 s

Thus, the p-value conditional on the sufficient statistic s = x1 + x2 is

min∑(n1,s) p(x1, x2) = f(j|s)

j=x1

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 7 / 1 . Interval 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 that θ ∈ [L(X), U(X)]

.Then we call [L(X), U(X)] an interval estimator of θ. Three types of intervals • Two-sided interval [L(X), U(X)] • One-sided (with lower-bound) interval [L(X), ∞) • One-sided (with upper-bound) interval (−∞, U(X)]

Interval Estimation

θˆ(X) is usually represented as a point estimator

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 8 / 1 θ ∈ [L(X), U(X)]

Then we call [L(X), U(X)] an interval estimator of θ.

Three types of intervals • Two-sided interval [L(X), U(X)] • One-sided (with lower-bound) interval [L(X), ∞) • One-sided (with upper-bound) interval (−∞, U(X)]

Interval Estimation

θˆ(X) is usually represented as a point estimator . Interval 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 that

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 8 / 1 Then we call [L(X), U(X)] an interval estimator of θ.

Three types of intervals • Two-sided interval [L(X), U(X)] • One-sided (with lower-bound) interval [L(X), ∞) • One-sided (with upper-bound) interval (−∞, U(X)]

Interval Estimation

θˆ(X) is usually represented as a point estimator . Interval 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 that θ ∈ [L(X), U(X)]

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 8 / 1 Three types of intervals • Two-sided interval [L(X), U(X)] • One-sided (with lower-bound) interval [L(X), ∞) • One-sided (with upper-bound) interval (−∞, U(X)]

Interval Estimation

θˆ(X) is usually represented as a point estimator . Interval 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 that θ ∈ [L(X), U(X)]

.Then we call [L(X), U(X)] an interval estimator of θ.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 8 / 1 • Two-sided interval [L(X), U(X)] • One-sided (with lower-bound) interval [L(X), ∞) • One-sided (with upper-bound) interval (−∞, U(X)]

Interval Estimation

θˆ(X) is usually represented as a point estimator . Interval 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 that θ ∈ [L(X), U(X)]

.Then we call [L(X), U(X)] an interval estimator of θ. Three types of intervals

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 8 / 1 • One-sided (with lower-bound) interval [L(X), ∞) • One-sided (with upper-bound) interval (−∞, U(X)]

Interval Estimation

θˆ(X) is usually represented as a point estimator . Interval 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 that θ ∈ [L(X), U(X)]

.Then we call [L(X), U(X)] an interval estimator of θ. Three types of intervals • Two-sided interval [L(X), U(X)]

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 8 / 1 • One-sided (with upper-bound) interval (−∞, U(X)]

Interval Estimation

θˆ(X) is usually represented as a point estimator . Interval 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 that θ ∈ [L(X), U(X)]

.Then we call [L(X), U(X)] an interval estimator of θ. Three types of intervals • Two-sided interval [L(X), U(X)] • One-sided (with lower-bound) interval [L(X), ∞)

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 8 / 1 Interval Estimation

θˆ(X) is usually represented as a point estimator . Interval 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 that θ ∈ [L(X), U(X)]

.Then we call [L(X), U(X)] an interval estimator of θ. Three types of intervals • Two-sided interval [L(X), U(X)] • One-sided (with lower-bound) interval [L(X), ∞) • One-sided (with upper-bound) interval (−∞, U(X)]

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 8 / 1 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) →P 1

as n → ∞, where Z ∼ N (0, 1).

Example

i.i.d. Let Xi ∼ N (µ, 1). Define

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 9 / 1 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) →P 1

as n → ∞, where Z ∼ N (0, 1).

Example

i.i.d. Let Xi ∼ N (µ, 1). Define 1. A point estimator of µ : X

Pr(X = µ) = 0

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 9 / 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) →P 1

as n → ∞, where Z ∼ N (0, 1).

Example

i.i.d. Let Xi ∼ 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]

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 9 / 1 = Pr(µ − 1 ≤ X ≤ µ + 1) √ √ √ = Pr(− n ≤ n(X − µ) ≤ n) √ √ = Pr(− n ≤ Z ≤ n) →P 1

as n → ∞, where Z ∼ N (0, 1).

Example

i.i.d. Let Xi ∼ 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)

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 9 / 1 √ √ √ = Pr(− n ≤ n(X − µ) ≤ n) √ √ = Pr(− n ≤ Z ≤ n) →P 1

as n → ∞, where Z ∼ N (0, 1).

Example

i.i.d. Let Xi ∼ 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)

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 9 / 1 √ √ = Pr(− n ≤ Z ≤ n) →P 1

as n → ∞, where Z ∼ N (0, 1).

Example

i.i.d. Let Xi ∼ 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)

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 9 / 1 Example

i.i.d. Let Xi ∼ 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) →P 1

as n → ∞, where Z ∼ N (0, 1).

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 9 / 1 Pr(θ ∈ [L(X), U(X)])

In other words, the probability of a random variable in interval [L(X), U(X)] covers the parameter θ. . Definition: Confidence Coefficient . Confidence coefficient is defined as inf Pr(θ ∈ [L(X), U(X)]) . θ∈Ω

Definitions

. Definition : Coverage Probability . Given an interval estimator [L(X), U(X)] of θ, its coverage probability is defined as

.

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

Definitions

. Definition : Coverage Probability . Given an interval estimator [L(X), U(X)] of θ, its coverage probability is defined as Pr(θ ∈ [L(X), U(X)])

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 10 / 1 . Definition: Confidence Coefficient . Confidence coefficient is defined as inf Pr(θ ∈ [L(X), U(X)]) . θ∈Ω

Definitions

. Definition : Coverage Probability . Given an interval estimator [L(X), U(X)] of θ, its coverage probability is defined as Pr(θ ∈ [L(X), U(X)])

In other words, the probability of a random variable in interval .[L(X), U(X)] covers the parameter θ.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 10 / 1 inf Pr(θ ∈ [L(X), U(X)]) θ∈Ω

Definitions

. Definition : Coverage Probability . Given an interval estimator [L(X), U(X)] of θ, its coverage probability is defined as Pr(θ ∈ [L(X), U(X)])

In other words, the probability of a random variable in interval .[L(X), U(X)] covers the parameter θ. . Definition: Confidence Coefficient . Confidence coefficient is defined as

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 10 / 1 Definitions

. Definition : Coverage Probability . Given an interval estimator [L(X), U(X)] of θ, its coverage probability is defined as Pr(θ ∈ [L(X), U(X)])

In other words, the probability of a random variable in interval .[L(X), U(X)] covers the parameter θ. . Definition: Confidence Coefficient . Confidence coefficient is defined as inf Pr(θ ∈ [L(X), U(X)]) . θ∈Ω

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 10 / 1 . Definition: Expected Length . Given an interval estimator [L(X), U(X)] of θ, its expected length is defined as E[U(X) − L(X)]

where X are random samples from fX(x|θ). In other words, it is the .average length of the interval estimator.

Definitions

. Definition : Confidence Interval . 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

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 11 / 1 E[U(X) − L(X)]

where X are random samples from fX(x|θ). In other words, it is the average length of the interval estimator.

Definitions

. Definition : Confidence Interval . 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 . Definition: Expected Length . Given an interval estimator [L(X), U(X)] of θ, its expected length is defined as

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 11 / 1 where X are random samples from fX(x|θ). In other words, it is the average length of the interval estimator.

Definitions

. Definition : Confidence Interval . 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 . Definition: Expected Length . Given an interval estimator [L(X), U(X)] of θ, its expected length is defined as E[U(X) − L(X)]

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 11 / 1 Definitions

. Definition : Confidence Interval . 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 . Definition: Expected Length . Given an interval estimator [L(X), U(X)] of θ, its expected length is defined as E[U(X) − L(X)]

where X are random samples from fX(x|θ). In other words, it is the average. length of the interval estimator.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 11 / 1 There is a one-to-one correspondence between tests and confidence intervals (or confidence sets).

How to construct confidence interval?

A confidence interval can be obtained by inverting the acceptance region of a test.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 12 / 1 How to construct confidence interval?

A confidence interval can be obtained by inverting the acceptance region of a test. There is a one-to-one correspondence between tests and confidence intervals (or confidence sets).

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 12 / 1

− X √θ0 > zα/2 σ/ n

− X √θ0 ≤ Equivalently, we accept H0 if σ/ n zα/2. Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0.

− − ≤ X √θ0 ≤ zα/2 σ/ n zα/2 σ σ θ − √ z ≤ X ≤ θ + √ z 0 n α/2 0 n α/2 { } − √σ ≤ ≤ √σ Acceptance region is x : θ0 n zα/2 x θ0 + n zα/2

Example

i.i.d. 2 2 Xi ∼ N (θ, σ ) where σ is known. Consider H0 : θ = θ0 vs. H1 : θ ≠ θ0. As previously shown, level α LRT test reject H0 if and only if

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 13 / 1

− X √θ0 ≤ Equivalently, we accept H0 if σ/ n zα/2. Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0.

− − ≤ X √θ0 ≤ zα/2 σ/ n zα/2 σ σ θ − √ z ≤ X ≤ θ + √ z 0 n α/2 0 n α/2 { } − √σ ≤ ≤ √σ Acceptance region is x : θ0 n zα/2 x θ0 + n zα/2

Example

i.i.d. 2 2 Xi ∼ N (θ, σ ) where σ is known. Consider H0 : θ = θ0 vs. H1 : θ ≠ θ0. As previously shown, level α LRT test reject H0 if and only if

− X √θ0 > zα/2 σ/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 13 / 1 Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0.

− − ≤ X √θ0 ≤ zα/2 σ/ n zα/2 σ σ θ − √ z ≤ X ≤ θ + √ z 0 n α/2 0 n α/2 { } − √σ ≤ ≤ √σ Acceptance region is x : θ0 n zα/2 x θ0 + n zα/2

Example

i.i.d. 2 2 Xi ∼ N (θ, σ ) where σ is known. Consider H0 : θ = θ0 vs. H1 : θ ≠ θ0. As previously shown, level α LRT test reject H0 if and only if

− X √θ0 > zα/2 σ/ n

− X √θ0 ≤ Equivalently, we accept H0 if σ/ n zα/2.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 13 / 1 − − ≤ X √θ0 ≤ zα/2 σ/ n zα/2 σ σ θ − √ z ≤ X ≤ θ + √ z 0 n α/2 0 n α/2 { } − √σ ≤ ≤ √σ Acceptance region is x : θ0 n zα/2 x θ0 + n zα/2

Example

i.i.d. 2 2 Xi ∼ N (θ, σ ) where σ is known. Consider H0 : θ = θ0 vs. H1 : θ ≠ θ0. As previously shown, level α LRT test reject H0 if and only if

− X √θ0 > zα/2 σ/ n

− X √θ0 ≤ Equivalently, we accept H0 if σ/ n zα/2. Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 13 / 1 σ σ θ − √ z ≤ X ≤ θ + √ z 0 n α/2 0 n α/2 { } − √σ ≤ ≤ √σ Acceptance region is x : θ0 n zα/2 x θ0 + n zα/2

Example

i.i.d. 2 2 Xi ∼ N (θ, σ ) where σ is known. Consider H0 : θ = θ0 vs. H1 : θ ≠ θ0. As previously shown, level α LRT test reject H0 if and only if

− X √θ0 > zα/2 σ/ n

− X √θ0 ≤ Equivalently, we accept H0 if σ/ n zα/2. Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0.

− − ≤ X √θ0 ≤ zα/2 σ/ n zα/2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 13 / 1 { } − √σ ≤ ≤ √σ Acceptance region is x : θ0 n zα/2 x θ0 + n zα/2

Example

i.i.d. 2 2 Xi ∼ N (θ, σ ) where σ is known. Consider H0 : θ = θ0 vs. H1 : θ ≠ θ0. As previously shown, level α LRT test reject H0 if and only if

− X √θ0 > zα/2 σ/ n

− X √θ0 ≤ Equivalently, we accept H0 if σ/ n zα/2. Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0.

− − ≤ X √θ0 ≤ zα/2 σ/ n zα/2 σ σ θ − √ z ≤ X ≤ θ + √ z 0 n α/2 0 n α/2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 13 / 1 Example

i.i.d. 2 2 Xi ∼ N (θ, σ ) where σ is known. Consider H0 : θ = θ0 vs. H1 : θ ≠ θ0. As previously shown, level α LRT test reject H0 if and only if

− X √θ0 > zα/2 σ/ n

− X √θ0 ≤ Equivalently, we accept H0 if σ/ n zα/2. Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0.

− − ≤ X √θ0 ≤ zα/2 σ/ n zα/2 σ σ θ − √ z ≤ X ≤ θ + √ z 0 n α/2 0 n α/2 { } √σ √σ Acceptance region is x : θ0 − zα/2 ≤ x ≤ θ0 + zα/2 n . . n...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 13 / 1 ( ) σ σ = Pr X − √ z ≤ θ ≤ X + √ z n α/2 0 n α/2

Since θ0 is arbitrary, ( ) σ σ 1 − α = Pr X − √ z ≤ θ ≤ X + √ z n α/2 n α/2

− √σ √σ − Therefore, [X n zα/2, X + n zα/2] is (1 α) confidence interval (CI).

Example (cont’d)

As this is size α test, the probability of accepting H0 is 1 − α. ( ) σ σ 1 − α = Pr θ − √ z ≤ X ≤ θ + √ z 0 n α/2 0 n α/2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 14 / 1 Since θ0 is arbitrary, ( ) σ σ 1 − α = Pr X − √ z ≤ θ ≤ X + √ z n α/2 n α/2

− √σ √σ − Therefore, [X n zα/2, X + n zα/2] is (1 α) confidence interval (CI).

Example (cont’d)

As this is size α test, the probability of accepting H0 is 1 − α. ( ) √σ √σ 1 − α = Pr θ0 − zα/2 ≤ X ≤ θ0 + zα/2 ( n n ) σ σ = Pr X − √ z ≤ θ ≤ X + √ z n α/2 0 n α/2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 14 / 1 − √σ √σ − Therefore, [X n zα/2, X + n zα/2] is (1 α) confidence interval (CI).

Example (cont’d)

As this is size α test, the probability of accepting H0 is 1 − α. ( ) √σ √σ 1 − α = Pr θ0 − zα/2 ≤ X ≤ θ0 + zα/2 ( n n ) σ σ = Pr X − √ z ≤ θ ≤ X + √ z n α/2 0 n α/2

Since θ0 is arbitrary, ( ) σ σ 1 − α = Pr X − √ z ≤ θ ≤ X + √ z n α/2 n α/2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 14 / 1 Example (cont’d)

As this is size α test, the probability of accepting H0 is 1 − α. ( ) √σ √σ 1 − α = Pr θ0 − zα/2 ≤ X ≤ θ0 + zα/2 ( n n ) σ σ = Pr X − √ z ≤ θ ≤ X + √ z n α/2 0 n α/2

Since θ0 is arbitrary, ( ) σ σ 1 − α = Pr X − √ z ≤ θ ≤ X + √ z n α/2 n α/2

− √σ √σ − Therefore, [X n zα/2, X + n zα/2] is (1 α) confidence interval (CI).

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 14 / 1 2. Conversely, if C(X) is a (1 − α) confidence set for θ, for any θ0, define the acceptance region of a test for the hypothesis H0 : θ = θ0 by A(θ0) = {x : θ0 ∈ C(x)}. Then the test has level α.

Confidence intervals and level α test

. Theorem 9.2.2 . 1. For each θ0 ∈ Ω, let A(θ0) be the acceptance region of a level α test of H0 : θ = θ0 vs. H1 : θ ≠ θ0 Define a set C(X) = {θ : x ∈ A(θ)}, then the random set C(X) is a 1 − α confidence set.

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 15 / 1 Confidence intervals and level α test

. Theorem 9.2.2 . 1. For each θ0 ∈ Ω, let A(θ0) be the acceptance region of a level α test of H0 : θ = θ0 vs. H1 : θ ≠ θ0 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 θ0, define the acceptance region of a test for the hypothesis H0 : θ = θ0 by A(θ0) = {x : θ0 ∈ C(x)}. Then the test has level α. .

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 15 / 1 The confidence set C(X) is a subset of the parameter space { } √σ √σ C(X) = θ : θ − zα/2 ≤ X ≤ θ + zα/2 { n n } σ σ = θ : X − √ z ≤ θ ≤ X + √ z n α/2 n α/2

Example

i.i.d. 2 For Xi ∼ N (θ, σ ), the acceptance region A(θ0) is a subset of the sample space { } σ σ A(θ ) = x : θ − √ z ≤ X ≤ θ + √ z 0 0 n α/2 0 n α/2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 16 / 1 { } σ σ = θ : X − √ z ≤ θ ≤ X + √ z n α/2 n α/2

Example

i.i.d. 2 For Xi ∼ N (θ, σ ), the acceptance region A(θ0) is a subset of the sample space { } σ σ A(θ ) = x : θ − √ z ≤ X ≤ θ + √ z 0 0 n α/2 0 n α/2

The confidence set C(X) is a subset of the parameter space { } σ σ C(X) = θ : θ − √ z ≤ X ≤ θ + √ z n α/2 n α/2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 16 / 1 Example

i.i.d. 2 For Xi ∼ N (θ, σ ), the acceptance region A(θ0) is a subset of the sample space { } σ σ A(θ ) = x : θ − √ z ≤ X ≤ θ + √ z 0 0 n α/2 0 n α/2

The confidence set C(X) is a subset of the parameter space { } √σ √σ C(X) = θ : θ − zα/2 ≤ X ≤ θ + zα/2 { n n } σ σ = θ : X − √ z ≤ θ ≤ X + √ z n α/2 n α/2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 16 / 1 1. To obtain (1 − α) two-sided CI [L(X), U(X)], we invert the acceptance region of a level α test for H0 : θ = θ0 vs. H1 : θ ≠ θ0 2. To obtain a lower-bounded CI [L(X), ∞), then we invert the acceptance region of a test for H0 : θ = θ0 vs. H1 : θ > θ0, where Ω = {θ : θ ≥ θ0}. 3. To obtain a upper-bounded CI (−∞, U(X)], then we invert the acceptance region of a test for H0 : θ = θ0 vs. H1 : θ < θ0, where Ω = {θ : θ ≤ θ0}.

Confidence set and confidence interval

There is no guarantee that the confidence set obtained from Theorem 9.2.2 is an interval, but quite often

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 17 / 1 2. To obtain a lower-bounded CI [L(X), ∞), then we invert the acceptance region of a test for H0 : θ = θ0 vs. H1 : θ > θ0, where Ω = {θ : θ ≥ θ0}. 3. To obtain a upper-bounded CI (−∞, U(X)], then we invert the acceptance region of a test for H0 : θ = θ0 vs. H1 : θ < θ0, where Ω = {θ : θ ≤ θ0}.

Confidence set and confidence interval

There is no guarantee that the confidence set obtained from Theorem 9.2.2 is an interval, but quite often 1. To obtain (1 − α) two-sided CI [L(X), U(X)], we invert the acceptance region of a level α test for H0 : θ = θ0 vs. H1 : θ ≠ θ0

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 17 / 1 3. To obtain a upper-bounded CI (−∞, U(X)], then we invert the acceptance region of a test for H0 : θ = θ0 vs. H1 : θ < θ0, where Ω = {θ : θ ≤ θ0}.

Confidence set and confidence interval

There is no guarantee that the confidence set obtained from Theorem 9.2.2 is an interval, but quite often 1. To obtain (1 − α) two-sided CI [L(X), U(X)], we invert the acceptance region of a level α test for H0 : θ = θ0 vs. H1 : θ ≠ θ0 2. To obtain a lower-bounded CI [L(X), ∞), then we invert the acceptance region of a test for H0 : θ = θ0 vs. H1 : θ > θ0, where Ω = {θ : θ ≥ θ0}.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 17 / 1 Confidence set and confidence interval

There is no guarantee that the confidence set obtained from Theorem 9.2.2 is an interval, but quite often 1. To obtain (1 − α) two-sided CI [L(X), U(X)], we invert the acceptance region of a level α test for H0 : θ = θ0 vs. H1 : θ ≠ θ0 2. To obtain a lower-bounded CI [L(X), ∞), then we invert the acceptance region of a test for H0 : θ = θ0 vs. H1 : θ > θ0, where Ω = {θ : θ ≥ θ0}. 3. To obtain a upper-bounded CI (−∞, U(X)], then we invert the acceptance region of a test for H0 : θ = θ0 vs. H1 : θ < θ0, where Ω = {θ : θ ≤ θ0}.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 17 / 1 1. Find 1 − α two-sided CI for µ 2. Find 1 − α upper bound for µ

Example

. Problem . i.i.d. 2 Xi ∼ N (µ, σ ) where both parameters are unknown.

.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 18 / 1 Example

. Problem . i.i.d. 2 Xi ∼ N (µ, σ ) where both parameters are unknown. 1. Find 1 − α two-sided CI for µ 2. Find 1 − α upper bound for µ .

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 18 / 1

− X √µ0 > tn−1,α/2 sX/ n The acceptance region is { } − x √µ0 A(µ0) = x : ≤ tn−1,α/2 sx/ n The confidence set{ is } − x √µ C(x) = µ : ≤ tn−1,α/2 { sx/ n } − x √µ = µ : −tn−1,α/2 ≤ ≤ tn−1,α/2 { sx/ n } sx sx = µ : x − √ t − ≤ µ ≤ x + √ t − n n 1,α/2 n n 1,α/2

Example - two-sided CI - Solution

H0 : µ = µ0 vs H1 : µ ≠ µ0. The LRT test rejects if and only if

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1 The acceptance region is { } − x √µ0 A(µ0) = x : ≤ tn−1,α/2 sx/ n The confidence set{ is } − x √µ C(x) = µ : ≤ tn−1,α/2 { sx/ n } − x √µ = µ : −tn−1,α/2 ≤ ≤ tn−1,α/2 { sx/ n } sx sx = µ : x − √ t − ≤ µ ≤ x + √ t − n n 1,α/2 n n 1,α/2

Example - two-sided CI - Solution

H : µ = µ vs H : µ ≠ µ . The LRT test rejects if and only if 0 0 1 0 − X √µ0 > tn−1,α/2 sX/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1 The confidence set{ is } − x √µ C(x) = µ : ≤ tn−1,α/2 { sx/ n } − x √µ = µ : −tn−1,α/2 ≤ ≤ tn−1,α/2 { sx/ n } sx sx = µ : x − √ t − ≤ µ ≤ x + √ t − n n 1,α/2 n n 1,α/2

Example - two-sided CI - Solution

H : µ = µ vs H : µ ≠ µ . The LRT test rejects if and only if 0 0 1 0 − X √µ0 > tn−1,α/2 sX/ n The acceptance region is { } − x √µ0 A(µ0) = x : ≤ tn−1,α/2 sx/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1 { } − x √µ = µ : −tn−1,α/2 ≤ ≤ tn−1,α/2 { sx/ n } sx sx = µ : x − √ t − ≤ µ ≤ x + √ t − n n 1,α/2 n n 1,α/2

Example - two-sided CI - Solution

H : µ = µ vs H : µ ≠ µ . The LRT test rejects if and only if 0 0 1 0 − X √µ0 > tn−1,α/2 sX/ n The acceptance region is { } − x √µ0 A(µ0) = x : ≤ tn−1,α/2 sx/ n The confidence set{ is } − x √µ C(x) = µ : ≤ tn−1,α/2 sx/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1 { } sx sx = µ : x − √ t − ≤ µ ≤ x + √ t − n n 1,α/2 n n 1,α/2

Example - two-sided CI - Solution

H : µ = µ vs H : µ ≠ µ . The LRT test rejects if and only if 0 0 1 0 − X √µ0 > tn−1,α/2 sX/ n The acceptance region is { } − x √µ0 A(µ0) = x : ≤ tn−1,α/2 sx/ n The confidence set{ is } − x √µ C(x) = µ : ≤ tn−1,α/2 { sx/ n } − x √µ = µ : −tn−1,α/2 ≤ ≤ tn−1,α/2 sx/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1 Example - two-sided CI - Solution

H : µ = µ vs H : µ ≠ µ . The LRT test rejects if and only if 0 0 1 0 − X √µ0 > tn−1,α/2 sX/ n The acceptance region is { } − x √µ0 A(µ0) = x : ≤ tn−1,α/2 sx/ n The confidence set{ is } − x √µ C(x) = µ : ≤ tn−1,α/2 { sx/ n } − x √µ = µ : −tn−1,α/2 ≤ ≤ tn−1,α/2 { sx/ n } sx sx = µ : x − √ t − ≤ µ ≤ x + √ t − n n 1,α/2 n n 1,α/2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1 2 2 Ω0 = {(µ, σ ): µ = µ0, σ > 0} 2 2 Ω = {(µ, σ ): µ ≤ µ0, σ > 0}

LRT statistic is L(ˆµ , σˆ2|x) λ(x) = 0 0 L(ˆµ, σˆ2|x)

2 2 where (ˆµ0, σˆ0) is the MLE restricted to Ω0, and(ˆµ, σˆ∑) is the MLE n − 2 2 i=1(Xi µ0) restricted to Ω, and Within Ω0, µˆ0 = µ0, and σˆ0 = n

Example - upper-bounded CI - Solution

The CI is (−∞, U(X)]. We need to invert a testing procedure for H0 : µ = µ0 vs H1 : µ < µ0.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1 2 2 Ω = {(µ, σ ): µ ≤ µ0, σ > 0}

LRT statistic is L(ˆµ , σˆ2|x) λ(x) = 0 0 L(ˆµ, σˆ2|x)

2 2 where (ˆµ0, σˆ0) is the MLE restricted to Ω0, and(ˆµ, σˆ∑) is the MLE n − 2 2 i=1(Xi µ0) restricted to Ω, and Within Ω0, µˆ0 = µ0, and σˆ0 = n

Example - upper-bounded CI - Solution

The CI is (−∞, U(X)]. We need to invert a testing procedure for H0 : µ = µ0 vs H1 : µ < µ0.

2 2 Ω0 = {(µ, σ ): µ = µ0, σ > 0}

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1 LRT statistic is L(ˆµ , σˆ2|x) λ(x) = 0 0 L(ˆµ, σˆ2|x)

2 2 where (ˆµ0, σˆ0) is the MLE restricted to Ω0, and(ˆµ, σˆ∑) is the MLE n − 2 2 i=1(Xi µ0) restricted to Ω, and Within Ω0, µˆ0 = µ0, and σˆ0 = n

Example - upper-bounded CI - Solution

The CI is (−∞, U(X)]. We need to invert a testing procedure for H0 : µ = µ0 vs H1 : µ < µ0.

2 2 Ω0 = {(µ, σ ): µ = µ0, σ > 0} 2 2 Ω = {(µ, σ ): µ ≤ µ0, σ > 0}

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1 2 2 where (ˆµ0, σˆ0) is the MLE restricted to Ω0, and(ˆµ, σˆ∑) is the MLE n − 2 2 i=1(Xi µ0) restricted to Ω, and Within Ω0, µˆ0 = µ0, and σˆ0 = n

Example - upper-bounded CI - Solution

The CI is (−∞, U(X)]. We need to invert a testing procedure for H0 : µ = µ0 vs H1 : µ < µ0.

2 2 Ω0 = {(µ, σ ): µ = µ0, σ > 0} 2 2 Ω = {(µ, σ ): µ ≤ µ0, σ > 0}

LRT statistic is L(ˆµ , σˆ2|x) λ(x) = 0 0 L(ˆµ, σˆ2|x)

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1 ∑ n − 2 2 i=1(Xi µ0) Within Ω0, µˆ0 = µ0, and σˆ0 = n

Example - upper-bounded CI - Solution

The CI is (−∞, U(X)]. We need to invert a testing procedure for H0 : µ = µ0 vs H1 : µ < µ0.

2 2 Ω0 = {(µ, σ ): µ = µ0, σ > 0} 2 2 Ω = {(µ, σ ): µ ≤ µ0, σ > 0}

LRT statistic is L(ˆµ , σˆ2|x) λ(x) = 0 0 L(ˆµ, σˆ2|x)

2 2 where (ˆµ0, σˆ0) is the MLE restricted to Ω0, and(ˆµ, σˆ ) is the MLE restricted to Ω, and

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1 Example - upper-bounded CI - Solution

The CI is (−∞, U(X)]. We need to invert a testing procedure for H0 : µ = µ0 vs H1 : µ < µ0.

2 2 Ω0 = {(µ, σ ): µ = µ0, σ > 0} 2 2 Ω = {(µ, σ ): µ ≤ µ0, σ > 0}

LRT statistic is L(ˆµ , σˆ2|x) λ(x) = 0 0 L(ˆµ, σˆ2|x)

2 2 where (ˆµ0, σˆ0) is the MLE restricted to Ω0, and(ˆµ, σˆ∑) is the MLE n − 2 2 i=1(Xi µ0) restricted to Ω, and Within Ω0, µˆ0 = µ0, and σˆ0 = n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1 { ∑ n (X −X)2 2 i=1 i ≤ µˆ = X σˆ = ∑ n if X µ0 n − 2 2 i=1(Xi µ0) µˆ = µ0 σˆ = n if X > µ0

  1( ) { } if X > µ0  n ∑ n (X −µ )2 √ 1 exp − i=1 i 0 2 2 λ(x) = ( 2πσ ) { 2ˆσ0 }  n ∑ if X ≤ µ  n (X −X)2 0 √ 1 exp − i=1 i 2 2  2πσ 2ˆσ0  1 if X > µ0 ( ) n = n−1 s2 2  n X ≤ n−1 2 − 2 if X µ0 n sX+(X µ0)

Example - upper bounded CI - Solution (cont’d)

Within Ω, the MLE is

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 21 / 1 ∑ n − 2 2 i=1(Xi µ0) µˆ = µ0 σˆ = n if X > µ0

  1( ) { } if X > µ0  n ∑ n (X −µ )2 √ 1 exp − i=1 i 0 2 2 λ(x) = ( 2πσ ) { 2ˆσ0 }  n ∑ if X ≤ µ  n (X −X)2 0 √ 1 exp − i=1 i 2 2  2πσ 2ˆσ0  1 if X > µ0 ( ) n = n−1 s2 2  n X ≤ n−1 2 − 2 if X µ0 n sX+(X µ0)

Example - upper bounded CI - Solution (cont’d)

Within Ω, the MLE is { ∑ n (X −X)2 2 i=1 i ≤ µˆ = X σˆ = n if X µ0

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 21 / 1   1( ) { } if X > µ0  n ∑ n (X −µ )2 √ 1 exp − i=1 i 0 2 2 λ(x) = ( 2πσ ) { 2ˆσ0 }  n ∑ if X ≤ µ  n (X −X)2 0 √ 1 exp − i=1 i 2 2  2πσ 2ˆσ0  1 if X > µ0 ( ) n = n−1 s2 2  n X ≤ n−1 2 − 2 if X µ0 n sX+(X µ0)

Example - upper bounded CI - Solution (cont’d)

Within Ω, the MLE is { ∑ n (X −X)2 2 i=1 i ≤ µˆ = X σˆ = ∑ n if X µ0 n − 2 2 i=1(Xi µ0) µˆ = µ0 σˆ = n if X > µ0

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 21 / 1   1 if X > µ0 ( ) n = n−1 s2 2  n X ≤ n−1 2 − 2 if X µ0 n sX+(X µ0)

Example - upper bounded CI - Solution (cont’d)

Within Ω, the MLE is { ∑ n (X −X)2 2 i=1 i ≤ µˆ = X σˆ = ∑ n if X µ0 n − 2 2 i=1(Xi µ0) µˆ = µ0 σˆ = n if X > µ0

  1( ) { } if X > µ0  n ∑ n (X −µ )2 √ 1 exp − i=1 i 0 2 2 λ(x) = ( 2πσ ) { 2ˆσ0 }  n ∑ if X ≤ µ  n (X −X)2 0 √ 1 exp − i=1 i 2 2 2πσ 2ˆσ0

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 21 / 1 Example - upper bounded CI - Solution (cont’d)

Within Ω, the MLE is { ∑ n (X −X)2 2 i=1 i ≤ µˆ = X σˆ = ∑ n if X µ0 n − 2 2 i=1(Xi µ0) µˆ = µ0 σˆ = n if X > µ0

  1( ) { } if X > µ0  n ∑ n (X −µ )2 √ 1 exp − i=1 i 0 2 2 λ(x) = ( 2πσ ) { 2ˆσ0 }  n ∑ if X ≤ µ  n (X −X)2 0 √ 1 exp − i=1 i 2 2  2πσ 2ˆσ0  1 if X > µ0 ( ) n = n−1 s2 2  n X ≤ n−1 2 − 2 if X µ0 n sX+(X µ0)

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 21 / 1 ( ) n n−1 s2 2 n X < c n−1 2 − 2 n sX + (X µ0)   n n−1 2  n  2 < c n−1 (X−µ0) + 2 n sx 2 (X − µ0) ∗ 2 > c sX µ − X 0 √ > c∗∗ sX/ n

Example - upper bounded CI - Solution (cont’d)

For 0 < c < 1, LRT test rejects H0 if X ≤ µ0 and

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 22 / 1   n n−1 2  n  2 < c n−1 (X−µ0) + 2 n sx 2 (X − µ0) ∗ 2 > c sX µ − X 0 √ > c∗∗ sX/ n

Example - upper bounded CI - Solution (cont’d)

For 0 < c < 1, LRT test rejects H0 if X ≤ µ0 and

( ) n n−1 s2 2 n X < c n−1 2 − 2 n sX + (X µ0)

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 22 / 1 2 (X − µ0) ∗ 2 > c sX µ − X 0 √ > c∗∗ sX/ n

Example - upper bounded CI - Solution (cont’d)

For 0 < c < 1, LRT test rejects H0 if X ≤ µ0 and

( ) n n−1 s2 2 n X < c n−1 2 − 2 n sX + (X µ0)   n n−1 2  n  2 < c n−1 (X−µ0) + 2 n sx

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 22 / 1 µ − X 0 √ > c∗∗ sX/ n

Example - upper bounded CI - Solution (cont’d)

For 0 < c < 1, LRT test rejects H0 if X ≤ µ0 and

( ) n n−1 s2 2 n X < c n−1 2 − 2 n sX + (X µ0)   n n−1 2  n  2 < c n−1 (X−µ0) + 2 n sx 2 (X − µ0) ∗ 2 > c sX

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 22 / 1 Example - upper bounded CI - Solution (cont’d)

For 0 < c < 1, LRT test rejects H0 if X ≤ µ0 and

( ) n n−1 s2 2 n X < c n−1 2 − 2 n sX + (X µ0)   n n−1 2  n  2 < c n−1 (X−µ0) + 2 n sx 2 (X − µ0) ∗ 2 > c sX µ − X 0 √ > c∗∗ sX/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 22 / 1 ( ) µ − X = Pr 0 √ > c∗∗ s / n ( X ) X − µ = Pr √ 0 < −c∗∗ sX/ n ∗∗ = Pr(Tn−1 < −c ) ∗∗ 1 − α = Pr(Tn−1 > −c ) ∗∗ c = −tn−1,1−α = tn−1,α

Therefore, LRT level α test reject H0 if

X − µ0 √ < −tn−1,α sX/ n

Example - upper bounded CI - Solution (cont’d)

c∗∗ is chosen to satisfy

α = Pr(reject H0|µ0)

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 23 / 1 ( ) X − µ = Pr √ 0 < −c∗∗ sX/ n ∗∗ = Pr(Tn−1 < −c ) ∗∗ 1 − α = Pr(Tn−1 > −c ) ∗∗ c = −tn−1,1−α = tn−1,α

Therefore, LRT level α test reject H0 if

X − µ0 √ < −tn−1,α sX/ n

Example - upper bounded CI - Solution (cont’d)

c∗∗ is chosen to satisfy

α = Pr(reject H |µ ) ( 0 0 ) µ − X = Pr 0 √ > c∗∗ sX/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 23 / 1 ∗∗ = Pr(Tn−1 < −c ) ∗∗ 1 − α = Pr(Tn−1 > −c ) ∗∗ c = −tn−1,1−α = tn−1,α

Therefore, LRT level α test reject H0 if

X − µ0 √ < −tn−1,α sX/ n

Example - upper bounded CI - Solution (cont’d)

c∗∗ is chosen to satisfy

α = Pr(reject H |µ ) ( 0 0 ) µ − X = Pr 0 √ > c∗∗ s / n ( X ) X − µ = Pr √ 0 < −c∗∗ sX/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 23 / 1 ∗∗ 1 − α = Pr(Tn−1 > −c ) ∗∗ c = −tn−1,1−α = tn−1,α

Therefore, LRT level α test reject H0 if

X − µ0 √ < −tn−1,α sX/ n

Example - upper bounded CI - Solution (cont’d)

c∗∗ is chosen to satisfy

α = Pr(reject H |µ ) ( 0 0 ) µ − X = Pr 0 √ > c∗∗ s / n ( X ) X − µ = Pr √ 0 < −c∗∗ sX/ n ∗∗ = Pr(Tn−1 < −c )

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 23 / 1 ∗∗ c = −tn−1,1−α = tn−1,α

Therefore, LRT level α test reject H0 if

X − µ0 √ < −tn−1,α sX/ n

Example - upper bounded CI - Solution (cont’d)

c∗∗ is chosen to satisfy

α = Pr(reject H |µ ) ( 0 0 ) µ − X = Pr 0 √ > c∗∗ s / n ( X ) X − µ = Pr √ 0 < −c∗∗ sX/ n ∗∗ = Pr(Tn−1 < −c ) ∗∗ 1 − α = Pr(Tn−1 > −c )

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 23 / 1 Therefore, LRT level α test reject H0 if

X − µ0 √ < −tn−1,α sX/ n

Example - upper bounded CI - Solution (cont’d)

c∗∗ is chosen to satisfy

α = Pr(reject H |µ ) ( 0 0 ) µ − X = Pr 0 √ > c∗∗ s / n ( X ) X − µ = Pr √ 0 < −c∗∗ sX/ n ∗∗ = Pr(Tn−1 < −c ) ∗∗ 1 − α = Pr(Tn−1 > −c ) ∗∗ c = −tn−1,1−α = tn−1,α

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 23 / 1 X − µ0 √ < −tn−1,α sX/ n

Example - upper bounded CI - Solution (cont’d)

c∗∗ is chosen to satisfy

α = Pr(reject H |µ ) ( 0 0 ) µ − X = Pr 0 √ > c∗∗ s / n ( X ) X − µ = Pr √ 0 < −c∗∗ sX/ n ∗∗ = Pr(Tn−1 < −c ) ∗∗ 1 − α = Pr(Tn−1 > −c ) ∗∗ c = −tn−1,1−α = tn−1,α

Therefore, LRT level α test reject H0 if

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 23 / 1 Example - upper bounded CI - Solution (cont’d)

c∗∗ is chosen to satisfy

α = Pr(reject H |µ ) ( 0 0 ) µ − X = Pr 0 √ > c∗∗ s / n ( X ) X − µ = Pr √ 0 < −c∗∗ sX/ n ∗∗ = Pr(Tn−1 < −c ) ∗∗ 1 − α = Pr(Tn−1 > −c ) ∗∗ c = −tn−1,1−α = tn−1,α

Therefore, LRT level α test reject H0 if

X − µ0 √ < −tn−1,α sX/ n ...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 23 / 1 Inverting the above to get CI C(X) = {µ : X ∈ A(µ)} { } X − µ = µ : √ ≥ −tn−1,α { sX/ n } sX = µ : X − µ ≥ −√ tn−1,α { n } sX = µ : µ ≤ X + √ tn−1,α ( n ] sX = −∞, X + √ t − n n 1,α

Example - upper bounded CI - Solution (cont’d)

Acceptance region is { } X − µ0 A(µ0) = x : √ ≥ −tn−1,α sX/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 24 / 1 { } X − µ = µ : √ ≥ −tn−1,α { sX/ n } sX = µ : X − µ ≥ −√ tn−1,α { n } sX = µ : µ ≤ X + √ tn−1,α ( n ] sX = −∞, X + √ t − n n 1,α

Example - upper bounded CI - Solution (cont’d)

Acceptance region is { } X − µ0 A(µ0) = x : √ ≥ −tn−1,α sX/ n Inverting the above to get CI C(X) = {µ : X ∈ A(µ)}

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 24 / 1 { } sX = µ : X − µ ≥ −√ tn−1,α { n } sX = µ : µ ≤ X + √ tn−1,α ( n ] sX = −∞, X + √ t − n n 1,α

Example - upper bounded CI - Solution (cont’d)

Acceptance region is { } X − µ0 A(µ0) = x : √ ≥ −tn−1,α sX/ n Inverting the above to get CI C(X) = {µ : X ∈ A(µ)} { } X − µ = µ : √ ≥ −tn−1,α sX/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 24 / 1 { } sX = µ : µ ≤ X + √ tn−1,α ( n ] sX = −∞, X + √ t − n n 1,α

Example - upper bounded CI - Solution (cont’d)

Acceptance region is { } X − µ0 A(µ0) = x : √ ≥ −tn−1,α sX/ n Inverting the above to get CI C(X) = {µ : X ∈ A(µ)} { } X − µ = µ : √ ≥ −tn−1,α { sX/ n } sX = µ : X − µ ≥ −√ t − n n 1,α

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 24 / 1 ( ] sX = −∞, X + √ t − n n 1,α

Example - upper bounded CI - Solution (cont’d)

Acceptance region is { } X − µ0 A(µ0) = x : √ ≥ −tn−1,α sX/ n Inverting the above to get CI C(X) = {µ : X ∈ A(µ)} { } X − µ = µ : √ ≥ −tn−1,α { sX/ n } sX = µ : X − µ ≥ −√ tn−1,α { n } sX = µ : µ ≤ X + √ t − n n 1,α

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 24 / 1 Example - upper bounded CI - Solution (cont’d)

Acceptance region is { } X − µ0 A(µ0) = x : √ ≥ −tn−1,α sX/ n Inverting the above to get CI C(X) = {µ : X ∈ A(µ)} { } X − µ = µ : √ ≥ −tn−1,α { sX/ n } sX = µ : X − µ ≥ −√ tn−1,α { n } sX = µ : µ ≤ X + √ tn−1,α ( n ] sX = −∞, X + √ t − n n 1,α ...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 24 / 1 Acceptance region is { } X − µ0 A(µ0) = x : √ ≤ tn−1,α sX/ n Confidence interval is { } X − µ C(X) = {µ : X ∈ A(µ)} = µ : √ ≤ tn−1,α { s}X/ n sX = µ : µ ≥ X − √ tn−1,α [ n ) sX = X − √ t − , ∞ n n 1,α

Example - lower bounded CI - solution

LRT level α test reject H0 if and only if

X − µ0 √ > tn−1,α sX/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 25 / 1 Confidence interval is { } X − µ C(X) = {µ : X ∈ A(µ)} = µ : √ ≤ tn−1,α { s}X/ n sX = µ : µ ≥ X − √ tn−1,α [ n ) sX = X − √ t − , ∞ n n 1,α

Example - lower bounded CI - solution

LRT level α test reject H0 if and only if

X − µ0 √ > tn−1,α sX/ n Acceptance region is { } X − µ0 A(µ0) = x : √ ≤ tn−1,α sX/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 25 / 1 { } sX = µ : µ ≥ X − √ tn−1,α [ n ) sX = X − √ t − , ∞ n n 1,α

Example - lower bounded CI - solution

LRT level α test reject H0 if and only if

X − µ0 √ > tn−1,α sX/ n Acceptance region is { } X − µ0 A(µ0) = x : √ ≤ tn−1,α sX/ n Confidence interval is { } X − µ C(X) = {µ : X ∈ A(µ)} = µ : √ ≤ tn−1,α sX/ n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 25 / 1 [ ) sX = X − √ t − , ∞ n n 1,α

Example - lower bounded CI - solution

LRT level α test reject H0 if and only if

X − µ0 √ > tn−1,α sX/ n Acceptance region is { } X − µ0 A(µ0) = x : √ ≤ tn−1,α sX/ n Confidence interval is { } X − µ C(X) = {µ : X ∈ A(µ)} = µ : √ ≤ tn−1,α { s}X/ n sX = µ : µ ≥ X − √ t − n n 1,α

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 25 / 1 Example - lower bounded CI - solution

LRT level α test reject H0 if and only if

X − µ0 √ > tn−1,α sX/ n Acceptance region is { } X − µ0 A(µ0) = x : √ ≤ tn−1,α sX/ n Confidence interval is { } X − µ C(X) = {µ : X ∈ A(µ)} = µ : √ ≤ tn−1,α { s}X/ n sX = µ : µ ≥ X − √ tn−1,α [ n ) sX = X − √ t − , ∞ n n 1,α ...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 25 / 1 . Solution . Let X be a method of moment estimator for µ. By law of large number, X is consistent( for)µ, and by central limit theorem, σ2 X ∼ AN µ, . n

Example

. Problem . X1, ··· , Xn are iid samples from a distribution with mean µ and finite 2 .variance σ . Construct asymptotic (1 − α) two-sided interval for µ

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 26 / 1 Example

. Problem . X1, ··· , Xn are iid samples from a distribution with mean µ and finite 2 .variance σ . Construct asymptotic (1 − α) two-sided interval for µ . Solution . Let X be a method of moment estimator for µ. By law of large number, X is consistent( for)µ, and by central limit theorem, σ2 X ∼ AN µ, . n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 26 / 1 From previous lectures, we know that 1 ∑n (X − X)2 →P σ2 n − 1 i i=1 √∑ n (X − X)2 i=1 i →P √σ (n − 1)n n The Wald level α test

√ − √(X µ0) n ∑ > zα/2 n − 2 i=1(Xi X) n−1

Example (cont’d)

Consider testing H0 : µ = µ0 vs. H1 : µ ≠ µ0. The Wald statistic

X − µ0 Zn = Sn √ for a consistent estimator of σ/ n.

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 27 / 1 √∑ n (X − X)2 i=1 i →P √σ (n − 1)n n The Wald level α test

√ − √(X µ0) n ∑ > zα/2 n − 2 i=1(Xi X) n−1

Example (cont’d)

Consider testing H0 : µ = µ0 vs. H1 : µ ≠ µ0. The Wald statistic

X − µ0 Zn = Sn √ for a consistent estimator of σ/ n. From previous lectures, we know that 1 ∑n (X − X)2 →P σ2 n − 1 i i=1

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 27 / 1 The Wald level α test

√ − √(X µ0) n ∑ > zα/2 n − 2 i=1(Xi X) n−1

Example (cont’d)

Consider testing H0 : µ = µ0 vs. H1 : µ ≠ µ0. The Wald statistic

X − µ0 Zn = Sn √ for a consistent estimator of σ/ n. From previous lectures, we know that 1 ∑n (X − X)2 →P σ2 n − 1 i i=1 √∑ n (X − X)2 i=1 i →P √σ (n − 1)n n

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 27 / 1 Example (cont’d)

Consider testing H0 : µ = µ0 vs. H1 : µ ≠ µ0. The Wald statistic

X − µ0 Zn = Sn √ for a consistent estimator of σ/ n. From previous lectures, we know that 1 ∑n (X − X)2 →P σ2 n − 1 i i=1 √∑ n (X − X)2 i=1 i →P √σ (n − 1)n n The Wald level α test

√ − √(X µ0) n ∑ > zα/2 n − 2 i=1(Xi X) n−1 ...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 27 / 1 (1 − α) CI is

{ ∈ } C(x) = µ : x A(µ)  √  −  √(x µ) n = µ : ∑ ≤ zα/2  n − 2  i=1(xi x) n−1 [ √ ∑ √ ∑ ] n (x −x)2 n (x −x)2 − √1 i=1 i √1 i=1 i = x n n−1 zα/2, x + n n−1 zα/2

Example (cont’d)

The acceptance region is   √  −  √(x µ0) n A(µ0) = x : ∑ ≤ zα/2  n − 2  i=1(xi x) n−1

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 28 / 1   √  −  √(x µ) n = µ : ∑ ≤ zα/2  n − 2  i=1(xi x) n−1 [ √ ∑ √ ∑ ] n (x −x)2 n (x −x)2 − √1 i=1 i √1 i=1 i = x n n−1 zα/2, x + n n−1 zα/2

Example (cont’d)

The acceptance region is   √  −  √(x µ0) n A(µ0) = x : ∑ ≤ zα/2  n − 2  i=1(xi x) n−1

(1 − α) CI is

C(x) = {µ : x ∈ A(µ)}

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 28 / 1 [ √ ∑ √ ∑ ] n (x −x)2 n (x −x)2 − √1 i=1 i √1 i=1 i = x n n−1 zα/2, x + n n−1 zα/2

Example (cont’d)

The acceptance region is   √  −  √(x µ0) n A(µ0) = x : ∑ ≤ zα/2  n − 2  i=1(xi x) n−1

(1 − α) CI is

{ ∈ } C(x) = µ : x A(µ)  √  −  √(x µ) n = µ : ∑ ≤ zα/2  n − 2  i=1(xi x) n−1

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 28 / 1 Example (cont’d)

The acceptance region is   √  −  √(x µ0) n A(µ0) = x : ∑ ≤ zα/2  n − 2  i=1(xi x) n−1

(1 − α) CI is

{ ∈ } C(x) = µ : x A(µ)  √  −  √(x µ) n = µ : ∑ ≤ zα/2  n − 2  i=1(xi x) n−1 [ √ ∑ √ ∑ ] n (x −x)2 n (x −x)2 − √1 i=1 i √1 i=1 i = x n n−1 zα/2, x + n n−1 zα/2

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 28 / 1 . Next Lectures . • Reviews and Example Problems (every lecture) • E-M algorithm • Non-informative priors • Bayesian Tests .

Summary

. Today . • Interval Estimation • Confidence Interval .

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 29 / 1 Summary

. Today . • Interval Estimation • Confidence Interval . . Next Lectures . • Reviews and Example Problems (every lecture) • E-M algorithm • Non-informative priors • Bayesian Tests .

...... Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 29 / 1