distribution

Reciprocal of an F

Let the r.v. F ∼ F (v₁, v₂) and let Y = 1/F . Then Y has a pdf.

f (y )^∗ = g (F )|dF dy|

= v₁^v¹^/2y^1−(v¹^/2)v₂^v²^/2y^(v¹^+v²^)/2 B(^v₂¹,^v₂²)(v₂y + v₁)^(v¹^+v²^)/2

1 y²

= v₁^v¹^/2v₂^v²^/2y^(v²^/2)−1

B(^v₂²,^v₂¹)(v₁+ v₂y )^(v¹^+v²^)/2 y ∈ [0, ∞)

That is if F ∼ F (v₁, v₂) and Y = 1/F , then Y ∼ F (v₂, v₁)

CI for σ

_X²

/σ

_Y²

from two ind. Normal

Given S_X², S_Y² are unbiased estimates of σ²_X, σ_Y² derived from samples of size n and m , respectively, from two independent normal populations. Find a 100(1 − α)% CI for σ_X²/σ²_Y.

(n − 1)S_X²/σ²_X ∼ χ²(n − 1) , (m − 1)S_Y²/σ_Y² ∼ χ²(m − 1)

(m−1)S_Y²

σ²_Y /(m − 1)

(n−1)S₁²

σ₁² /(n − 1)

= S_Y²/σ²_Y S_X²/σ²_X

follow a F distribution with degrees of freedom (m − 1) and (n − 1) i.e.,

S_Y²/σ_Y²

S_X²/σ_X² ∼ F (m − 1, n − 1)

CI for σ

_X²

/σ

_Y²

from two ind. Normal

Given S_X², S_Y² are unbiased estimates of σ²_X, σ_Y² derived from samples of size n and m , respectively, from two independent normal populations. Find a 100(1 − α)% CI for σ_X²/σ²_Y. (n − 1)S_X²/σ²_X ∼ χ²(n − 1) , (m − 1)S_Y²/σ_Y² ∼ χ²(m − 1)

(m−1)S_Y²

σ²_Y /(m − 1)

(n−1)S₁²

σ₁² /(n − 1)

= S_Y²/σ²_Y S_X²/σ²_X

follow a F distribution with degrees of freedom (m − 1) and (n − 1) i.e.,

S_Y²/σ_Y²

S_X²/σ_X² ∼ F (m − 1, n − 1)

S_Y²/σ²_Y

S_X²/σ²_X ∼ F (m − 1, n − 1)

So we select c = F_1−α/2(m − 1, n − 1) and d = Fα/2(m − 1, n − 1), and

P(c ≤ S_Y²/σ_Y²

S_X²/σ_X² ≤ d ) = 1 − α That is

P c S_X² S_Y² ≤ σ_X²

σ_Y² ≤ d S_X²

S_Y² = 1 − α

Often from table we have

c = F1−α/2(m − 1, n − 1) = 1/Fα/2(n − 1, m − 1) and

d = F_α/2(m − 1, n − 1), let s_x² and s_y² be the realization of S_X² and S_Y², then a 100(1 − α)% CIs for σ_X²/σ²_Y is

[ 1

F_α/2(n − 1, m − 1) s_X²

s_Y² , F_α/2(m − 1, n − 1)s_X² s_Y² ]

S_Y²/σ²_Y

S_X²/σ²_X ∼ F (m − 1, n − 1)

So we select c = F_1−α/2(m − 1, n − 1) and d = Fα/2(m − 1, n − 1), and

P(c ≤ S_Y²/σ_Y²

S_X²/σ_X² ≤ d ) = 1 − α That is

P c S_X² S_Y² ≤ σ_X²

σ_Y² ≤ d S_X²

S_Y² = 1 − α Often from table we have

c = F_1−α/2(m − 1, n − 1) = 1/F_α/2(n − 1, m − 1) and

d = F_α/2(m − 1, n − 1), let s_x² and s_y² be the realization of S_X² and S_Y², then a 100(1 − α)% CIs for σ_X²/σ²_Y is

[ 1

F_α/2(n − 1, m − 1) s_X²

s_Y², F_α/2(m − 1, n − 1)s_X² s_Y² ]

Example

From two ind Normal with unknown means and variances, we have (12)s_X² = 128.4 from a random sample of size 13 and (8)s_Y² = 36.72 from a random sample of size 9. Find a 98%

CIs for σ_X²/σ_Y².

S_Y²/σ_Y²

S_X²/σ_X² ∼ F (8, 12) From F -table we have F_0.01(12, 8) = 5.67 and F_0.01(8, 12) = 4.50, so a 98% CIs for σ²_X/σ_Y² is

[ 1

5.67

128.4/12

36.72/8 , 4.50 128.4/12 36.72/8 ]

Example

From two ind Normal with unknown means and variances, we have (12)s_X² = 128.4 from a random sample of size 13 and (8)s_Y² = 36.72 from a random sample of size 9. Find a 98%

CIs for σ_X²/σ_Y².

S_Y²/σ_Y²

S_X²/σ_X² ∼ F (8, 12) From F -table we have F0.01(12, 8) = 5.67 and F_0.01(8, 12) = 4.50, so a 98% CIs for σ²_X/σ_Y² is

[ 1

5.67

128.4/12

36.72/8 , 4.50 128.4/12 36.72/8 ]

Confidence intervals for proportions (p)

Estimate proportions. Construct a CI for p in the Bin(n, p) distribution.

Assume that sampling is from a binomial population and hence that the problem is to estimate p in the Bin(n, p) distribution where p is unknown.

recall:

Given Y is distributed as Bin(n, p), an unbiased estimate of p is ˆp = ^Y_n.

E (ˆp) = E (Y n) = p and

Var (ˆp) = 1

n²Var(Y ) = 1

n²np(1 − p) = p(1 − p) n

Confidence intervals for proportions (p)

For large n,

Y − np

pnp(1 − p) = (Y /n) − p pp(1 − p)/n

can be approximated by the standard normal N (0, 1).

Thus an approximate 100(1 − α)% CI for p is obtained by considering

P(−z_α/2< (Y /n) − p

pp(1 − p)/n < z_α/2) = 1 − α

Replace the variance of ˆp = Y /n by its estimate ˆp(1 − ˆp)/n, giving a simple expression for the CI for p is

[ˆp ± z_α/2

rˆp(1 − ˆp) n ] = [Y

n ± z_α/2

r(Y /n)(1 − Y /n)

n ]

Confidence intervals for proportions (p)

For large n,

Y − np

pnp(1 − p) = (Y /n) − p pp(1 − p)/n

can be approximated by the standard normal N (0, 1).

Thus an approximate 100(1 − α)% CI for p is obtained by considering

P(−z_α/2< (Y /n) − p

pp(1 − p)/n < z_α/2) = 1 − α

Replace the variance of ˆp = Y /n by its estimate ˆp(1 − ˆp)/n, giving a simple expression for the CI for p is

[ˆp ± z_α/2

rp(1 − ˆˆ p) n ] = [Y

n ± z_α/2

Poll n = 100 and y = 51 say yes, find 95% CI for p .41, 0.61 Poll n = 351 and y = 185 say yes, find 95% CI for p?

CI for difference of two proportions

民

民民調調調的的的解解解讀讀讀

n = 0.032 we have n = (0.63)(0.37)(1.96)²/0.032² = 864

The maximum error of the estimate for 98% confidence coefficient is 0.01 for ˆp = 0.08, find the n

z0.01= 2.326 and (2.326)

q(0.08)(0.92)

n = 0.01 we have n = (0.08)(0.92)(2.326)²/0.01² = 3982

unknown ˆ p

For estimating p, we have p^∗(1 − p^∗) ≤ 1/4. Hence we need n = 2.326²/(4 ∗ 0.01²) = 13530

for the maximum error of the estimate for 98% confidence coefficient is 0.01.

95% confidence coefficient for = 0.01, we have n = 9604 95% confidence coefficient for = 0.03, we have n = 1067

Sample Size and CIs for given ˆ p

The 95% CI for the proportion of people of supporting A when there is 51% people support A in polls of 100, 400 or 10,000 sample.

[0.41, 0.61], [0.46, 0.56], [0.50, 0.52]

[0.51 ± 0.1], [0.51 ± 0.05], [0.51 ± 0.01]

Sample Size for mean

100(1 − α)% CI for µ is [x ± z_α/2(σ/√

n)]. Denote such interval as x ± . we sometime call = z_α/2(σ/√

n) the maximum error of the estimate

n = (z_α/2)²(σ)²

² where it is assumed that σ² is known.

Example

we want the 95% CIs for µ to be x ± 1 for a normal population with standard deviation σ = 15, find the sample size.

1.96 15

√n = 1 we have n ≈ 864.35 = 865.

The 80% CIs for µ is x ± 2, then we have

1.282 15

√n = 2 where z_0.1 = 1.282 and thus n = 93

Example

we want the 95% CIs for µ to be x ± 1 for a normal population with standard deviation σ = 15, find the sample size.

1.96 15

√n = 1 we have n ≈ 864.35 = 865.

The 80% CIs for µ is x ± 2, then we have

1.282 15

√n = 2 where z_0.1 = 1.282 and thus n = 93

Example

we want the 95% CIs for µ to be x ± 1 for a normal population with standard deviation σ = 15, find the sample size.

1.96 15

√n = 1 we have n ≈ 864.35 = 865.

The 80% CIs for µ is x ± 2, then we have

1.282 15

√n = 2 where z_0.1 = 1.282 and thus n = 93

Pivotal quantity I

A very useful method for finding confidence intervals uses a pivotal quantity.

What is a pivotal quantity? A pivotal quantity is a function of data and the unknown parameter, say g (X, θ), and the distribution of g (X, θ) does not depend on the unknown parameter.

Example

Given X₁, · · · , X_n is a random sample from a N (µ, σ²) distribution.

在文檔中 StatisticalInference:ConﬁdenceIntervals 101 學年統計學 (I) 授課老師：蔡碧紋 (頁 34-61)

CI for σ

/σ

from two ind. Normal

CI for σ

/σ

from two ind. Normal

Example

Example

Confidence intervals for proportions (p)

Confidence intervals for proportions (p)

Confidence intervals for proportions (p)

Example

Example

Example

Example

Example

Example

CI for difference of two proportions

民

民 民調 調 調的 的 的解 解 解讀 讀 讀

民

民 民調 調 調的 的 的解 解 解讀 讀 讀

Sample Size for proportion

Sample Size for proportion

Sample Size for proportion

Sample Size for proportion

unknown ˆ p

Sample Size and CIs for given ˆ p

Sample Size for mean

Pivotal quantity I

Example

民民調調調的的的解解解讀讀讀

民民調調調的的的解解解讀讀讀