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# 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|

= v1v1/2y1−(v1/2)v2v2/2y(v1+v2)/2 B(v21,v22)(v2y + v1)(v1+v2)/2

1 y2

= v1v1/2v2v2/2y(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)

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X2

Y2

## from two ind. Normal

Given SX2, SY2 are unbiased estimates of σ2X, σY2 derived from samples of size n and m , respectively, from two independent normal populations. Find a 100(1 − α)% CI for σX22Y.

(n − 1)SX22X ∼ χ2(n − 1) , (m − 1)SY2Y2 ∼ χ2(m − 1)

(m−1)SY2

σ2Y /(m − 1)

(n−1)S12

σ12 /(n − 1)

= SY22Y SX22X

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

SY2Y2

SX2X2 ∼ F (m − 1, n − 1)

31

X2

Y2

## from two ind. Normal

Given SX2, SY2 are unbiased estimates of σ2X, σY2 derived from samples of size n and m , respectively, from two independent normal populations. Find a 100(1 − α)% CI for σX22Y. (n − 1)SX22X ∼ χ2(n − 1) , (m − 1)SY2Y2 ∼ χ2(m − 1)

(m−1)SY2

σ2Y /(m − 1)

(n−1)S12

σ12 /(n − 1)

= SY22Y SX22X

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

SY2Y2

SX2X2 ∼ F (m − 1, n − 1)

31

SY22Y

SX22X ∼ F (m − 1, n − 1)

So we select c = F1−α/2(m − 1, n − 1) and d = Fα/2(m − 1, n − 1), and

P(c ≤ SY2Y2

SX2X2 ≤ d ) = 1 − α That is

P c SX2 SY2 ≤ σX2

σY2 ≤ d SX2

SY2 = 1 − α

Often from table we have

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

d = Fα/2(m − 1, n − 1), let sx2 and sy2 be the realization of SX2 and SY2, then a 100(1 − α)% CIs for σX22Y is

[ 1

Fα/2(n − 1, m − 1) sX2

sY2 , Fα/2(m − 1, n − 1)sX2 sY2 ]

32

SY22Y

SX22X ∼ F (m − 1, n − 1)

So we select c = F1−α/2(m − 1, n − 1) and d = Fα/2(m − 1, n − 1), and

P(c ≤ SY2Y2

SX2X2 ≤ d ) = 1 − α That is

P c SX2 SY2 ≤ σX2

σY2 ≤ d SX2

SY2 = 1 − α Often from table we have

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

d = Fα/2(m − 1, n − 1), let sx2 and sy2 be the realization of SX2 and SY2, then a 100(1 − α)% CIs for σX22Y is

[ 1

Fα/2(n − 1, m − 1) sX2

sY2, Fα/2(m − 1, n − 1)sX2 sY2 ]

32

## Example

From two ind Normal with unknown means and variances, we have (12)sX2 = 128.4 from a random sample of size 13 and (8)sY2 = 36.72 from a random sample of size 9. Find a 98%

CIs for σX2Y2.

SY2Y2

SX2X2 ∼ F (8, 12) From F -table we have F0.01(12, 8) = 5.67 and F0.01(8, 12) = 4.50, so a 98% CIs for σ2XY2 is

[ 1

5.67

 128.4/12

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

33

## Example

From two ind Normal with unknown means and variances, we have (12)sX2 = 128.4 from a random sample of size 13 and (8)sY2 = 36.72 from a random sample of size 9. Find a 98%

CIs for σX2Y2.

SY2Y2

SX2X2 ∼ F (8, 12) From F -table we have F0.01(12, 8) = 5.67 and F0.01(8, 12) = 4.50, so a 98% CIs for σ2XY2 is

[ 1

5.67

 128.4/12

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

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## 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 = Yn.

E (ˆp) = E (Y n) = p and

Var (ˆp) = 1

n2Var(Y ) = 1

n2np(1 − p) = p(1 − p) n

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## 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(−zα/2< (Y /n) − p

pp(1 − p)/n < zα/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 ± zα/2

rˆp(1 − ˆp) n ] = [Y

n ± zα/2

r(Y /n)(1 − Y /n)

n ]

35

## 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(−zα/2< (Y /n) − p

pp(1 − p)/n < zα/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 ± zα/2

rp(1 − ˆˆ p) n ] = [Y

n ± zα/2

r(Y /n)(1 − Y /n)

n ]

35

## Example

Assume Y ∼ Bin(n, p), we have n = 36 and y /n = 0.222, find an approximate 90% CIs for p

[ 0.222 ± 1.645

r(0.222)(1 − 0.222)

36 ]

### Example

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?

36

## Example

Assume Y ∼ Bin(n, p), we have n = 36 and y /n = 0.222, find an approximate 90% CIs for p

[ 0.222 ± 1.645

r(0.222)(1 − 0.222)

36 ]

### Example

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?

36

## Example

Assume Y ∼ Bin(n, p), we have n = 36 and y /n = 0.222, find an approximate 90% CIs for p

[ 0.222 ± 1.645

r(0.222)(1 − 0.222)

36 ]

### Example

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?

36

37

## 民 民調 調 調的 的 的解 解 解讀 讀 讀

「....施政滿意度4成4。本次調查是以台灣地區住宅電話簿 為抽樣清冊， 並以電話的後四碼進行隨機抽樣。共成功訪 問1056位台灣地區20歲以上民眾。在95%的信心水準下，

1. 這項民調的母體是什麼？樣本數為多少？

2. 受訪民眾中對施政滿意約有多少人？

3. 算出這次調查的信賴區間？

38

## 民 民調 調 調的 的 的解 解 解讀 讀 讀

1. 在本次調查中，母體是台灣地區20歲以上的民眾，樣

2. 區間[0.44 − 0.03, 0.44 + 0.03] = [0.41, 0.47]，稱為信賴 區間 (信賴區間：[估計值-最大誤差 , 估計值+最大誤 差] )

3. 95%的信心水準: p是不可知的，而抽樣都會有誤差，

4. 而我們有95%的信心說，真正的滿意度會落在我們所

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## Sample Size for proportion

z0.025 = 1.96 and (1.96)

q(0.63)(1−0.63)

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

40

## Sample Size for proportion

z0.025 = 1.96 and (1.96)

q(0.63)(1−0.63)

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

40

## Sample Size for proportion

z0.025 = 1.96 and (1.96)

q(0.63)(1−0.63)

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

40

## Sample Size for proportion

z0.025 = 1.96 and (1.96)

q(0.63)(1−0.63)

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

40

## 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

41

## 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]

42

## Sample Size for mean

100(1 − α)% CI for µ is [x ± zα/2(σ/√

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

n) the maximum error of the estimate

n = (zα/2)2(σ)2

2 where it is assumed that σ2 is known.

43

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 z0.1 = 1.282 and thus n = 93

44

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 z0.1 = 1.282 and thus n = 93

44

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 z0.1 = 1.282 and thus n = 93

44

## 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, · · · , Xn is a random sample from a N (µ, σ2) distribution.

45