[PPT] - Lecture 23 April 11th, 2013 Biostatistics 602 - Lecture 23 Hyun PowerPoint Presentation

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. .

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

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Last Lecture

What is p-value?
What is the advantage of p-value compared to hypothesis testing

procedure with size ?

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

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Last Lecture

What is p-value?
What is the advantage of p-value compared to hypothesis testing

procedure with size ?

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

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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

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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

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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

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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

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p-Values

.

Conclusions from Hypothesis Testing

. .

Reject H0 or accept H0.
If size of the test is ( ) small, the decision to reject H is convincing.
If

is large, the decision may not be very convincing. .

Definition: p-Value

. . . . . . . . A p-value p X is a test statistic satisfying p x for every sample point x. Small values of p X given evidence that H is true. A p-value is valid if, for every and every , Pr p X

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

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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

is large, the decision may not be very convincing. .

Definition: p-Value

. . . . . . . . A p-value p X is a test statistic satisfying p x for every sample point x. Small values of p X given evidence that H is true. A p-value is valid if, for every and every , Pr p X

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

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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 p x for every sample point x. Small values of p X given evidence that H is true. A p-value is valid if, for every and every , Pr p X

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

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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

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

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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

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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 Then p X is a valid p-value.

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 4 / 1

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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

θ∈Ω0

Pr(W(X) ≥ W(x)|θ) Then p X is a valid p-value.

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 4 / 1

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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

θ∈Ω0

Pr(W(X) ≥ W(x)|θ) Then p(X) is a valid p-value.

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 4 / 1

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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 . Again, let W X denote a test statistic where large value give evidence that H 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 S s

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 5 / 1

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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 H 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 S s

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 5 / 1

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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 S s

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 5 / 1

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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

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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 H , if we let p denote the common value of p p . Then the join pmf of X X is f x x p n x px p n

x

n x px p n

x

n x n x px

x

p n

n x x

Therefore S X X is a sufficient statistic under H .

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 6 / 1

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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 f x x p n x px p n

x

n x px p n

x

n x n x px

x

p n

n x x

Therefore S X X is a sufficient statistic under H .

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 6 / 1

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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 f(x1, x2|p) = (n1 x1 ) px1(1 − p)n1−x1 (n2 x2 ) px2(1 − p)n2−x2 n x n x px

x

p n

n x x

Therefore S X X is a sufficient statistic under H .

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 6 / 1

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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 f(x1, x2|p) = (n1 x1 ) px1(1 − p)n1−x1 (n2 x2 ) px2(1 − p)n2−x2 = (n1 x1 )(n2 x2 ) px1+x2(1 − p)n1+n2−x1−x2 Therefore S X X is a sufficient statistic under H .

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 6 / 1

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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 f(x1, x2|p) = (n1 x1 ) px1(1 − p)n1−x1 (n2 x2 ) px2(1 − p)n2−x2 = (n1 x1 )(n2 x2 ) px1+x2(1 − p)n1+n2−x1−x2 Therefore S = X1 + X2 is a sufficient statistic under H0.

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 6 / 1

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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 X given S s is a hypergeometric distribution. f X x s

n x n s x n n s

Thus, the p-value conditional on the sufficient statistic s x x is p x x

min n s j x

f j s

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 7 / 1

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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. f(X1 = x1|s) = (n1

x1

)( n2

s−x1

) (n1+n2

s

) Thus, the p-value conditional on the sufficient statistic s x x is p x x

min n s j x

f j s

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 7 / 1

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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. f(X1 = x1|s) = (n1

x1

)( n2

s−x1

) (n1+n2

s

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

min(n1,s)

∑

j=x1

f(j|s)

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 7 / 1

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Interval Estimation

ˆ θ(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 L X U X Then we call L X U X an interval estimator of . 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

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Interval Estimation

ˆ θ(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 L X U X Then we call L X U X an interval estimator of . 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

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Interval Estimation

ˆ θ(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 . 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

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Interval Estimation

ˆ θ(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

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Interval Estimation

ˆ θ(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

SLIDE 33

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Interval Estimation

ˆ θ(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

SLIDE 34

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Interval Estimation

ˆ θ(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

SLIDE 35

. .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. . . .. .

Interval Estimation

ˆ θ(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

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Example

Let Xi

i.i.d.

∼ N(µ, 1). Define

1. A point estimator of

: X Pr X

2. An interval estimator of

: X X Pr X X Pr X X Pr X Pr n n X n Pr n Z n

P

as n , where Z .

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

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Example

Let Xi

i.i.d.

∼ N(µ, 1). Define

1. A point estimator of µ : X

Pr(X = µ) = 0

2. An interval estimator of

: X X Pr X X Pr X X Pr X Pr n n X n Pr n Z n

P

as n , where Z .

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

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Example

Let Xi

i.i.d.

∼ 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 X X Pr X X Pr X Pr n n X n Pr n Z n

P

as n , where Z .

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

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Example

Let Xi

i.i.d.

∼ 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 X Pr n n X n Pr n Z n

P

as n , where Z .

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

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Example

Let Xi

i.i.d.

∼ 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 n Pr n Z n

P

as n , where Z .

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

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Example

Let Xi

i.i.d.

∼ 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

as n , where Z .

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

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Example

Let Xi

i.i.d.

∼ 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

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Definitions

.

Definition : Coverage Probability

. . Given an interval estimator [L(X), U(X)] of θ, its coverage probability is defined as Pr L X U X In other words, the probability of a random variable in interval L X U X covers the parameter . .

Definition: Confidence Coefficient

. . . . . . . . Confidence coefficient is defined as inf Pr L X U X

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

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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 . .

Definition: Confidence Coefficient

. . . . . . . . Confidence coefficient is defined as inf Pr L X U X

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

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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 L X U X

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

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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 L X U X

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

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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

θ∈Ω Pr(θ ∈ [L(X), U(X)])

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

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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

f

, its expected length is defined as E U X L X where X are random samples from fX x . In other words, it is the average length of the interval estimator.

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

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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 . In other words, it is the average length of the interval estimator.

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

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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 . In other words, it is the average length of the interval estimator.

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

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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

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How to construct confidence interval?

A confidence interval can be obtained by inverting the acceptance region

f 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

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How to construct confidence interval?

A confidence interval can be obtained by inverting the acceptance region

f 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

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Example

Xi

i.i.d.

∼ N(θ, σ2) where σ2 is known. Consider H0 : θ = θ0 vs. H1 : θ ̸= θ0.

As previously shown, level α LRT test reject H0 if and only if X n z Equivalently, we accept H if

X n

z . Accepting H because we believe our data ”agrees with” the hypothesis . z

X n

z nz X nz Acceptance region is x

nz

x

nz

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

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Example

Xi

i.i.d.

∼ N(θ, σ2) where σ2 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

σ/√n

> zα/2

Equivalently, we accept H if

X n

z . Accepting H because we believe our data ”agrees with” the hypothesis . z

X n

z nz X nz Acceptance region is x

nz

x

nz

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

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Example

Xi

i.i.d.

∼ N(θ, σ2) where σ2 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

σ/√n

> zα/2

Equivalently, we accept H0 if

X−θ0

σ/√n

≤ zα/2.

Accepting H because we believe our data ”agrees with” the hypothesis . z

X n

z nz X nz Acceptance region is x

nz

x

nz

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

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Example

Xi

i.i.d.

∼ N(θ, σ2) where σ2 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

σ/√n

> zα/2

Equivalently, we accept H0 if

X−θ0

σ/√n

≤ zα/2.

Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0. z

X n

z nz X nz Acceptance region is x

nz

x

nz

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

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Example

Xi

i.i.d.

∼ N(θ, σ2) where σ2 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

σ/√n

> zα/2

Equivalently, we accept H0 if

X−θ0

σ/√n

≤ zα/2.

Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0. −zα/2 ≤

X−θ0 σ/√n

≤ zα/2 nz X nz Acceptance region is x

nz

x

nz

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

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Example

Xi

i.i.d.

∼ N(θ, σ2) where σ2 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

σ/√n

> zα/2

Equivalently, we accept H0 if

X−θ0

σ/√n

≤ zα/2.

Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0. −zα/2 ≤

X−θ0 σ/√n

≤ zα/2 θ0 − σ √nzα/2 ≤ X ≤ θ0 + σ √nzα/2 Acceptance region is x

nz

x

nz

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

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Example

Xi

i.i.d.

∼ N(θ, σ2) where σ2 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

σ/√n

> zα/2

Equivalently, we accept H0 if

X−θ0

σ/√n

≤ zα/2.

Accepting H0 : θ = θ0 because we believe our data ”agrees with” the hypothesis θ = θ0. −zα/2 ≤

X−θ0 σ/√n

≤ zα/2 θ0 − σ √nzα/2 ≤ X ≤ θ0 + σ √nzα/2 Acceptance region is { x : θ0 −

σ √nzα/2 ≤ x ≤ θ0 + σ √nzα/2

}

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

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Example (cont’d)

As this is size α test, the probability of accepting H0 is 1 − α. 1 − α = Pr ( θ0 − σ √nzα/2 ≤ X ≤ θ0 + σ √nzα/2 ) Pr X nz X nz Since is arbitrary, Pr X nz X nz Therefore, X

nz

X

nz

is confidence interval (CI).

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 14 / 1

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Example (cont’d)

As this is size α test, the probability of accepting H0 is 1 − α. 1 − α = Pr ( θ0 − σ √nzα/2 ≤ X ≤ θ0 + σ √nzα/2 ) = Pr ( X − σ √nzα/2 ≤ θ0 ≤ X + σ √nzα/2 ) Since is arbitrary, Pr X nz X nz Therefore, X

nz

X

nz

is confidence interval (CI).

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 14 / 1

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Example (cont’d)

As this is size α test, the probability of accepting H0 is 1 − α. 1 − α = Pr ( θ0 − σ √nzα/2 ≤ X ≤ θ0 + σ √nzα/2 ) = Pr ( X − σ √nzα/2 ≤ θ0 ≤ X + σ √nzα/2 ) Since θ0 is arbitrary, 1 − α = Pr ( X − σ √nzα/2 ≤ θ ≤ X + σ √nzα/2 ) Therefore, X

nz

X

nz

is confidence interval (CI).

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 14 / 1

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Example (cont’d)

As this is size α test, the probability of accepting H0 is 1 − α. 1 − α = Pr ( θ0 − σ √nzα/2 ≤ X ≤ θ0 + σ √nzα/2 ) = Pr ( X − σ √nzα/2 ≤ θ0 ≤ X + σ √nzα/2 ) Since θ0 is arbitrary, 1 − α = Pr ( X − σ √nzα/2 ≤ θ ≤ X + σ √nzα/2 ) Therefore, [X −

σ √nzα/2, X + σ √nzα/2] is (1 − α) confidence interval (CI).

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 14 / 1

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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

f 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

confidence set for , for any , define the acceptance region of a test for the hypothesis H by A x C x . Then the test has level .

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 15 / 1

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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

f 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

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Example

For Xi

i.i.d.

∼ N(θ, σ2), the acceptance region A(θ0) is a subset of the

sample space A(θ0) = { x : θ0 − σ √nzα/2 ≤ X ≤ θ0 + σ √nzα/2 } The confidence set C X is a subset of the parameter space C X nz X nz X nz X nz

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Example

For Xi

i.i.d.

∼ N(θ, σ2), the acceptance region A(θ0) is a subset of the

sample space A(θ0) = { x : θ0 − σ √nzα/2 ≤ X ≤ θ0 + σ √nzα/2 } The confidence set C(X) is a subset of the parameter space C(X) = { θ : θ − σ √nzα/2 ≤ X ≤ θ + σ √nzα/2 } X nz X nz

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Example

For Xi

i.i.d.

∼ N(θ, σ2), the acceptance region A(θ0) is a subset of the

sample space A(θ0) = { x : θ0 − σ √nzα/2 ≤ X ≤ θ0 + σ √nzα/2 } The confidence set C(X) is a subset of the parameter space C(X) = { θ : θ − σ √nzα/2 ≤ X ≤ θ + σ √nzα/2 } = { θ : X − σ √nzα/2 ≤ θ ≤ X + σ √nzα/2 }

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

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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

two-sided CI L X U X , we invert the acceptance region of a level test for H

vs. H

. . 2 To obtain a lower-bounded CI L X

, then we invert the acceptance region of a test for H

vs. H

, where .

. . 3 To obtain a upper-bounded CI

U X , then we invert the acceptance region of a test for H

vs. H

, where .

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 17 / 1

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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 H

vs. H

, where .

. . 3 To obtain a upper-bounded CI

U X , then we invert the acceptance region of a test for H

vs. H

, where .

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 17 / 1

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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 H

vs. H

, where .

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 17 / 1

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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

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Example

.

Problem

. . Xi

i.i.d.

∼ N(µ, σ2) where both parameters are unknown.

. . 1 Find

two-sided CI for

. . 2 Find

upper bound for

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

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Example

.

Problem

. . Xi

i.i.d.

∼ N(µ, σ2) 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

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Example - two-sided CI - Solution

H0 : µ = µ0 vs H1 : µ ̸= µ0. The LRT test rejects if and only if X sX n tn The acceptance region is A x x sx n tn The confidence set is C x x sx n tn tn x sx n tn x sx ntn x sx ntn

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1

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Example - two-sided CI - Solution

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

X − µ0

sX/√n

> tn−1,α/2

The acceptance region is A x x sx n tn The confidence set is C x x sx n tn tn x sx n tn x sx ntn x sx ntn

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1

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Example - two-sided CI - Solution

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

X − µ0

sX/√n

> tn−1,α/2

The acceptance region is A(µ0) = { x :

x − µ0

sx/√n

≤ tn−1,α/2

} The confidence set is C x x sx n tn tn x sx n tn x sx ntn x sx ntn

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1

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Example - two-sided CI - Solution

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

X − µ0

sX/√n

> tn−1,α/2

The acceptance region is A(µ0) = { x :

x − µ0

sx/√n

≤ tn−1,α/2

} The confidence set is C(x) = { µ :

x − µ

sx/√n

≤ tn−1,α/2

} tn x sx n tn x sx ntn x sx ntn

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1

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Example - two-sided CI - Solution

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

X − µ0

sX/√n

> tn−1,α/2

The acceptance region is A(µ0) = { x :

x − µ0

sx/√n

≤ tn−1,α/2

} The confidence set is C(x) = { µ :

x − µ

sx/√n

≤ tn−1,α/2

} = { µ : −tn−1,α/2 ≤ x − µ sx/√n ≤ tn−1,α/2 } x sx ntn x sx ntn

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1

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Example - two-sided CI - Solution

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

X − µ0

sX/√n

> tn−1,α/2

The acceptance region is A(µ0) = { x :

x − µ0

sx/√n

≤ tn−1,α/2

} The confidence set is C(x) = { µ :

x − µ

sx/√n

≤ tn−1,α/2

} = { µ : −tn−1,α/2 ≤ x − µ sx/√n ≤ tn−1,α/2 } = { µ : x − sx √ntn−1,α/2 ≤ µ ≤ x + sx √ntn−1,α/2 }

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 19 / 1

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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. LRT statistic is x L x L x where is the MLE restricted to , and is the MLE restricted to , and Within , , and

n i

Xi n

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1

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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. Ω0 = {(µ, σ2) : µ = µ0, σ2 > 0} LRT statistic is x L x L x where is the MLE restricted to , and is the MLE restricted to , and Within , , and

n i

Xi n

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1

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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. Ω0 = {(µ, σ2) : µ = µ0, σ2 > 0} Ω = {(µ, σ2) : µ ≤ µ0, σ2 > 0} LRT statistic is x L x L x where is the MLE restricted to , and is the MLE restricted to , and Within , , and

n i

Xi n

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1

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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. Ω0 = {(µ, σ2) : µ = µ0, σ2 > 0} Ω = {(µ, σ2) : µ ≤ µ0, σ2 > 0} LRT statistic is λ(x) = L(ˆ µ0, ˆ σ2

0|x)

L(ˆ µ, ˆ σ2|x) where is the MLE restricted to , and is the MLE restricted to , and Within , , and

n i

Xi n

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1

SLIDE 86

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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. Ω0 = {(µ, σ2) : µ = µ0, σ2 > 0} Ω = {(µ, σ2) : µ ≤ µ0, σ2 > 0} LRT statistic is λ(x) = L(ˆ µ0, ˆ σ2

0|x)

L(ˆ µ, ˆ σ2|x) where (ˆ µ0, ˆ σ2

0) is the MLE restricted to Ω0, and(ˆ

µ, ˆ σ2) is the MLE restricted to Ω, and Within , , and

n i

Xi n

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1

SLIDE 87

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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. Ω0 = {(µ, σ2) : µ = µ0, σ2 > 0} Ω = {(µ, σ2) : µ ≤ µ0, σ2 > 0} LRT statistic is λ(x) = L(ˆ µ0, ˆ σ2

0|x)

L(ˆ µ, ˆ σ2|x) where (ˆ µ0, ˆ σ2

0) is the MLE restricted to Ω0, and(ˆ

µ, ˆ σ2) is the MLE restricted to Ω, and Within Ω0, ˆ µ0 = µ0, and ˆ σ2

0 = ∑n

i=1(Xi−µ0)2

n

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 20 / 1

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

Within Ω, the MLE is X

n i

Xi X n

if X

n i

Xi n

if X x if X

n

exp

n i Xi n

exp

n i Xi X

if X if X

n n sX n n sX

X

n

if X

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 21 / 1

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

Within Ω, the MLE is { ˆ µ = X ˆ σ2 =

∑n

i=1(Xi−X)2

n

if X ≤ µ0

n i

Xi n

if X x if X

n

exp

n i Xi n

exp

n i Xi X

if X if X

n n sX n n sX

X

n

if X

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 21 / 1

SLIDE 90

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

Within Ω, the MLE is { ˆ µ = X ˆ σ2 =

∑n

i=1(Xi−X)2

n

if X ≤ µ0 ˆ µ = µ0 ˆ σ2 =

∑n

i=1(Xi−µ0)2

n

if X > µ0 x if X

n

exp

n i Xi n

exp

n i Xi X

if X if X

n n sX n n sX

X

n

if X

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 21 / 1

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

Within Ω, the MLE is { ˆ µ = X ˆ σ2 =

∑n

i=1(Xi−X)2

n

if X ≤ µ0 ˆ µ = µ0 ˆ σ2 =

∑n

i=1(Xi−µ0)2

n

if X > µ0 λ(x) =        1 if X > µ0

(

1

√

2πσ2

)n exp { −

∑n i=1(Xi−µ0)2 2ˆ σ2

} (

1

√

2πσ2

)n exp { −

∑n i=1(Xi−X)2 2ˆ σ2

}

if X ≤ µ0 if X

n n sX n n sX

X

n

if X

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 21 / 1

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

Within Ω, the MLE is { ˆ µ = X ˆ σ2 =

∑n

i=1(Xi−X)2

n

if X ≤ µ0 ˆ µ = µ0 ˆ σ2 =

∑n

i=1(Xi−µ0)2

n

if X > µ0 λ(x) =        1 if X > µ0

(

1

√

2πσ2

)n exp { −

∑n i=1(Xi−µ0)2 2ˆ σ2

} (

1

√

2πσ2

)n exp { −

∑n i=1(Xi−X)2 2ˆ σ2

}

if X ≤ µ0 =    1 if X > µ0 (

n−1 n s2 X n−1 n s2 X+(X−µ0)2

) n

2

if X ≤ µ0

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 21 / 1

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

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

n n sX n n sX

X

n

c

n n n n X sx

n

c X sX c X sX n c

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 22 / 1

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

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

n−1 n s2 X n−1 n s2 X + (X − µ0)2

) n

2

< c

n n n n X sx

n

c X sX c X sX n c

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 22 / 1

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

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

n−1 n s2 X n−1 n s2 X + (X − µ0)2

) n

2

< c  

n−1 n n−1 n

+ (X−µ0)2

s2

x

 

n 2

< c X sX c X sX n c

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 22 / 1

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

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

n−1 n s2 X n−1 n s2 X + (X − µ0)2

) n

2

< c  

n−1 n n−1 n

+ (X−µ0)2

s2

x

 

n 2

< c (X − µ0)2 s2

X

> c∗ X sX n c

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 22 / 1

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

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

n−1 n s2 X n−1 n s2 X + (X − µ0)2

) n

2

< c  

n−1 n n−1 n

+ (X−µ0)2

s2

x

 

n 2

< c (X − µ0)2 s2

X

> c∗ µ0 − X sX/√n > c∗∗

Hyun Min Kang Biostatistics 602 - Lecture 23 April 11th, 2013 22 / 1

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