2011 c f - ustcstaff.ustc.edu.cn/~zwp/teach/math-stat/lec9.pdf · 2011-04-11 · i=...

17
Lec9: bu ² 2011 c 4 11 F bu=ƒ^Ø/%b?1. buK'üa: 1. ºŒ.bu: =oN'/(X!Œ!'), oN'6 uºŒ(ºŒ) θ, u·kºŒb. XX N (a, σ 2 ),a, u H 0 : a = a 0 ←→ H 1 : a 6= a 0 H 0 : a a 0 ←→ H 1 : a>a 0 2. ºŒ.bu: XJoN'/, dIk«oN'xNŒ ˘/ˆO{, ¡ºŒ{. Xu1Œ·˜5g,oN, Æuø aK. 1 bueZ˜Vg !uKJ{ ²buKJ{, e¡~f. ~5.1.1 ,)1, æßA. U5g˙pL0.01,83¥ ˜100, ²uk3g, ø1˜? uøK, 3¡c3ü«U5: :0 <p 0.01; :0.01 <p< 1. ˇLlø1¥˜5ß, ü«U5¥=/Æ. øK~eªJ: ?bH 0 :0 <p 0.01 §"b (null hypothesis) b, k{¡b. ,U· H 1 :0.01 <p< 1 ØÆbJb(alternative hypothesis). 8·ˇLßH 0 , ·ÆH 0 . //rK/ H 0 :0 <p 0.01 ←→ H 1 :0.01 <p< 1

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Page 1: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

Lec9: b�u�

Ü�²

2011c 4� 11F

b�u�=¦^��é¤'%�b�?1íä. b�u�¯K��©�ü�a:

1. ëê.b�u�: =oN�©Ù/ª®�(X��!�ê!��©Ù�) , oN©Ù�6

u��ëê(½ëê�þ) θ, �u��´k'��ëê�b�. XX ∼ N(a, σ2), a��, u�

H0 : a = a0 ←→ H1 : a 6= a0 ½ H0 : a ≤ a0 ←→ H1 : a > a0

2. �ëê.b�u�: XJoN©Ù/ª��, d�ÒI�k�«�oN©Ùx�äNê

Æ/ªÃ'�ÚO�{, ¡��ëê�{. Xu��1êâ´Ä5g,�®��oN, Òáuù

a¯K.

1 b�u��eZ�Vg

�!u�¯K�J{

�`²b�u�¯K�J{, �e¡�~f.

~~~5.1.1 ,ó�)����1�¬, �ñ�ûA. U5½g¬ÇpØ��L0.01,83Ù¥

Ä�100�, ²u�k3�g¬, ¯ù1�¬�ÄÑ�?

'uù�¯K, 3·�¡c�3ü«�U5:

`: 0 < p ≤ 0.01; ¯: 0.01 < p < 1.

·��ÏLlù1�¬¥Ä�5û½`, ¯ü«�U5¥=�¤á.

ù�¯K~±eã�ªJÑ: Ú?��“b�”

H0 : 0 < p ≤ 0.01

§��"b� (null hypothesis) ½�b�, k��{¡�b�. ,���U´

H1 : 0.01 < p < 1

��éáb�½�Jb�(alternative hypothesis).

·��8�´�ÏL��û½�ÉH0, �´áýH0. �±/�/r¯K�¤

H0 : 0 < p ≤ 0.01←→ H1 : 0.01 < p < 1

Page 2: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

5¿ù�J{¥òH0�3¥% �, §´u��é�. H0ÚH1� �Ø�6�. lù�~

f�òb�u�¯K��z, J{Xe:

�këê©Ùx{Fθ, θ ∈ Θ},d?Θ�ëê�m. X1, · · · , Xn´lþã©Ùx¥Ä�

�{ü�Å��. 3ëêb�u�¯K¥, ·�a,��´θ´Äáuëê�mΘ�

,�ýf8Θ0, K·KH0 : θ ∈ Θ¡�"b�½�b�, Ù(�¹Â´: �3�

�θ0 ∈ Θ0¦�X�©Ù�Fθ0 . PΘ1 = Θ−Θ0,K·KH1 : θ ∈ Θ1¡�H0�éáb

�½�Jb�(3~5.1.1¥,Θ = (0, 1), Θ0 = (0, 0.01], Θ1 = (0.01, 1)) . Kb�u�

¯KL�

H0 : θ ∈ Θ0 ←→ H1 : θ ∈ Θ1, (1.1)

3(5.1.1)ª¥, eΘ0½Θ1��¹ëê�mΘ¥���:, K¡�{üb� (simple hypothe-

sis);ÄK, ¡�EÜb� (composite hypothesis). ~X, ��ÄgN(a, σ20), σ2

0®�, Këê�m

�Θ = {a : −∞ < a < +∞}. -H0 : a = a0 ←→ H1 : a 6= a0,KH0�{üb�, H1�EÜb�.

2X, Óþ¯K, -H′

0 : a ≤ a0 ←→ H′

1 : a > a0, K"b�H′

0Úéáb�H′

1��EÜb�.

�!b�u����V�n

3~5.1.1¥, duù1�¬êþé�, �ePX�Ä��100��¬¥�g¬ê, K�±Cq

@�X ∼ B(100, p). XJ"b�0 < p ≤ 0.01´�(�, K

P (X ≥ 3) ≤ 1−2∑i=0

Ci1000.01i0.99100−i = 0.079

=XJ@�ù1�¬´Ü��, K100��¬¥k3�g¬½ö�õg¬��U5�k7.9%, ù

�VÇ'��, Uì�VÇ�n, Ø��U3�g¢�¥Òu), �·�  *ÿ�. Ïdk

nd~¦"b�´Ø�(�.

A^�VÇ�n�U�NþL�·�é"b�´Ä¤á���íä.

n!Ľ�!u�¼êÚu�ÚOþ

·�EÏL~f5`²ù�Vg.

~~~5.1.2 �X = (X1, · · · , Xn)�loNX ∼ N(a, 1)¥Ä���Å��. �Äu�¯K:

H0 : a = a0 ←→ H1 : a 6= a0, (1.2)

d?, a0��½�~ê.

ù«u���«�*þ��{´: k¦a����Oþ,·���X = 1n

∑ni=1Xi´a���

`û�O. e|X − a0|��,·�Ò��uĽH0;��, XJ|X − a0|��, ·�Ò@�Ä�(

2

Page 3: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

J�H0��C,Ï ��u�ÉH0. äN/`,·��(½��êA,é��X = (X1, . . . , Xn),�

ÑX, �|X − a0| > A�ÒĽH0;�|X − a0| ≤ A�Ò�ÉH0.·�¡

D = {X = (X1, . . . , Xn) : |X − a0| > A} (1.3)

�Ľ�,½��áý� (reject region).=,Ľ�´d���m X¥��¦|X − a0| > A�

@��X = (X1, . . . , Xn)�¤. kĽ�,�duò���m X ©¤Ø���üÜ

©X1 = X −DÚX2 = D,��k��X,�X ∈ X1�,Ò�ÉH0; �X ∈ X2 = D�,ÒĽH0.

·�¡X1��É� (acceptance region).��A½e5,KĽ�(½�É�) �Ò(½. Ï

d, 3d¯K¥u��À�Xe��«{K:

T :

{� |X − a0| > A�, áý H0

� |X − a0| ≤ A�, �É H0

þª¥�T,�½�«{K,��k��,·�Ò�±3�ÉH0½Ä½H0 ùü�(Ø¥ÀJ

��. ·�¡ù��«{KT�u�¯K(1.2)���u�.

�BuêÆþ?n,·�Ú\Xeu�¼êϕ(x)�Vg,ϕ(x)�u�T ´��éA�.3

~5.1.2¥

ϕ(x) =

{1 � |X − a0| > A�

0 � |X − a0| ≤ A�(1.4)

·�kXe½Â:

½½½ÂÂÂ5.1.1 d(5.1.4)�Ñ�u�¼ê ϕ(x)´½Â3���m Xþ, ��u[0,1]�¼ê.

§L«�k��X�,ĽH0�VÇ.

d½Â��,eϕ(x) = 1,K±VÇ�1ĽH0,�ϕ(x) = 0, K±VÇ�0ĽH0(=±VÇ

�1�ÉH0).eϕ(x)��0,1ùü��, Kù«u�¡���Åzu� (non-randomized test).d

�,Ľ���^u�¼êL«Xe: D = {X = (X1, . . . , Xn) : ϕ(x) = 1}.

eé,��X,k0 < ϕ(x) < 1,K¡ϕ(x)��Åzu� (randomized test).X3~5.1.1¥,�X =

(X1, . . . , Xn)���,�∑100i=1Xi < c �@�ù1�¬Ü�,�ÉH0;�

∑100i=1Xi > c�,@�ØÜ

�, áýH0.�∑100i=1Xi = c�,e5½áýH0,��ú��áý��U5�, ¯º. ��, e

�ÉH0,ï�(ûA)�ÉØÜ��¬��U5�,�ú�¯º.3V�ð±Øe��¹e,e�ò

¥�Y´V�Ñ�±�É�: ½e��ê0 < r < 1,5½�∑100i=1Xi = c�,±VÇ�r��gÁ

�,�âÁ�(J5û½áý�´�Éù1�¬. X�r = 1/2,K�ÏL��qM15û½.5½

eÑy�¡KáýH0, ÄK�ÉH0.ù�,�Ñy∑100i=1Xi = c,V�Ñk1/2��U,�ÑégCØ

|�û½,V�Ñú�Ün, �±�É.XJ�r = 1/3,Á��±ÏL3k2�x¥Ú1�ç¥�Ý

f¥¹¥5û½, e¹�ç¥(u)�VÇ�1/3)KáýH0,e¹�x¥(u)�VÇ�2/3) K�

ÉH0.ù«�Åzu�¼ê�L�

ϕ(x) =

1 e

∑100i=1Xi > c

r e∑100i=1Xi = c

0 e∑100i=1Xi < c.

(1.5)

3

Page 4: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

3~5.1.2¥�(½u�,7L½Ñ(5.1.3)½(5.1.4)ª¥�A, d?A ¡��.� (critical

value).�½ec��I�é�u�ÚOþ�©Ù. 3d~¥u�ÚOþ´T = X. Ó�3

~5.1.1¥, u�¼ê(5.1.5) ¥�c¡��.�,u�ÚOþ´T =∑100i=1Xi.(½u�ÚOþ�©

Ù´)ûb�u�¯K�'�. �u�ÚOþ�°(©ÙéJé��, eÙ4�©Ù'�{

ü,·��^4�©Ù�O°(©Ù,¼�b�u�¯K�Cq).

o!üa�Ø�õ�¼ê

ÚOíä´±����â�,du����Å5,·�ØU�yÚOíä�{�ýé�(5,

�U±�½�VÇ��yù«íä���5.3b�u�¯K¥�UÑye�ü«�/¬�

�Ø:

PPPPPPPPPPb�

ûüáýH0 �ÉH0

H0�ý �� Ø��

H1�ý Ø�� ��

1. "b� H0�5´é�, du����Å5, ��*�á\Ľ�D, �Ø/òH0Ľ

, ¡�ïý. ù����Ø¡�1�a�Ø (Type I error).

2."b� H0�5Øé, du����Å5, ��*�á\�É�D, �Ø/òH0�É,

¡���. ù����Ø¡�1�a�Ø( Type II error).

X3~5.1.1¥(½��Åu�Xe:

ϕ(x) =

{1 e

∑100i=1Xi > 3;

0 e∑100i=1Xi ≤ 3.

XJoN�ý¢g¬Ç�p = 0.005 < 0.01,du����Å5,Ä�(Jw«∑100i=1Xi = 5,=�

�á\Ľ�,ù�·��1�a�Ø. ��k�UoN�ý¢g¬Çp = 0.03 > 0.01, du��

��Å5,Ä�(Jw«∑100i=1Xi = 1,=��á\�É�. ù�·��1�a�Ø.

A�5¿,3z�äN|Ü,·��¬�üa�Ø¥���. �u�(½�,�üa�Ø�V

Ç�Ò(½. ·�F"�üa�Ø�VÇ���Ð,�ù�:éJ��. 3����n�½�

cJe,�öØ�o�. ùÒXÓ«m�O¯K¥��ÝÚ°Ý�öØ�o���. @o,N��

O��üa�Ø�VÇQ?�d, ÚÑõ�¼ê�Vg.

½½½ÂÂÂ5.1.2 �ϕ(x)´H0 : θ ∈ Θ0 ←→ H1 : θ ∈ Θ1���u�¼ê,K

βϕ(θ) = Pθ{^u� ϕĽ H0

}= Eθ[ϕ(X)], θ ∈ Θ

¡�ϕ�õ�¼ê (power function),�¡��¼ê½³¼ê.

eϕ(x)���Åzu�,Ľ��D,K

βϕ(θ) = Pθ(X = (X1, . . . , Xn) ∈ D

)

4

Page 5: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

Ïdõ�¼êL«���©Ùëê�θ�,ĽH0�VÇ. é~5.1.1,�u�¼ê��Åzu

�(1.5)�, |^∑ni=1Xi ∼ b(n, θ), 0 < θ < 1��u�õ�¼ê�

βϕ(θ) = Eθ[ϕ(x)] = P

( n∑i=1

Xi > c

)+ rP

( 100∑i=1

Xi = c

)

=

100∑k=c+1

(100

k

)θk(1− θ)100−k + r

(100

c

)θc(1− θ)100−c .

±e?Ø¥b½ϕ(x)����Åz�u�¼ê,Ø�AO�², Ø@�ϕ(x)��Åzu�¼ê.

��u�ϕ(x)�õ�¼ê�,Ò�±O��üa�Ø�VÇ. e±α∗ϕ(θ) Úβ∗ϕ(θ)©OP�

1�!�a�Ø�VÇ,K�1�a�Ø�VÇ�L«�

α∗ϕ(θ) =

{βϕ(θ) � θ ∈ Θ0

0 � θ ∈ Θ1,

�1�a�Ø�VÇ�L«�

β∗ϕ(θ) =

{0 � θ ∈ Θ0

1− βϕ(θ) � θ ∈ Θ1.

�I�`²��:´: Xc¤ã, �üa�Ø�VÇ��dõ�¼êû½, lù�:þw,

XJü�u�kÓ�õ�¼ê, Kdüu�35�þ����Ó.

o!u�Y²Ú���1�a�ØVÇ��K

c¡`L, ·�F"��u��üa�Ø�VÇÑé�, �Ø4~�/, ��`53�

½�����é?Ûu�Ñ�Ø�. ~X, �¦�1�a�Ø�VÇ~�, Ò� �áý�, ¦

�É�O�, ù7,���1�a�ØVÇO�, ��½,. Ïd, Neyman-PearsonJÑ�

^�K, Ò´���1�a�ØVÇ��K. =3�y�1�a�Ø�VÇØ�L�½ê�α

(0 < α < 1,Ï~����ê) �u�¥, Ïé�1�a�ØVǦ�U��u�. eP

Sα = { ϕ : α∗ϕ(θ) = βϕ(θ) ≤ α, �θ ∈ Θ0},

SαL«d¤k�1�a�Ø�VÇÑØ�Lα�u�¼ê�¤�a. ·���ÄSα¥�u�.

3Sα¥]À/�1�a�Ø�VǦ�U��u�0, ù«{K¡����1�a�ØVÇ�

{K.

�âNeyman-Pearson�K, 3�b�H0�ý�, ·��Ñ�Øû½(=ĽH0) �

VÇÉ���. ùL², �b�H0É��o, Ø�u�´�Ľ. ¤±3äN¯K

¥, ·� òkrº!ØU�´Ä½�·K���b�H0, rvkrº�!Ø

U�´�½�·K��éáb�. Ïd�b�H0Úéáb�H1�/ ´Ø²��,

ØU�pN�.

5

Page 6: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

��1�a�ØVÇ�éX�,��Vg´u�Y², Ù½ÂXe:

½½½ÂÂÂ5.1.3 �ϕ´(1.1)���u�, 0 ≤ α ≤ 1. XJϕ�1�a�Ø�VÇoØ�Lα

(½�d/`, eϕ÷v: βϕ(θ) ≤ α,é��θ ∈ Θ0) , K¡α´u�ϕ���Y², ϕ¡�wÍ5

Y²�α�u�, {¡Y²�α�u�.

Uù�½Â,u��Y²Ø��.eα�u�ϕ�Y², α < α′ < 1,Kα′�´u�ϕ�Y².

�;�ù�¯K, k�¡��u����Y²�Ùý¢Y². �Ò´

u�ϕ�ý¢Y² = sup{ βϕ(θ), θ ∈ Θ0} (1.6)

�uY²�ÀJ,S.þrα��'���IOz,Xα = 0.01, 0.05, 0.10�. IOz´��

BEL.

Y²�À�, éu��5�ké�K�. ØJ),XJY²À�é$, @o·�NN�1�

a�Ø�VÇé�, ���ù�:³7�� �Ľ�, ù�ÒO\�1�a�Ø��

U5. ��,eY²À�p,KĽ�*�, ¦�É� �, l �1�a�Ø�VÇ�A�òü

$. ù�w5,Y²�ÀJØ´��êƯK, ´��7Ll¢S�Ý5�Ä�¯K. ��`5

k±eA�Ï�K�Y²�À½.

1.���u��9ü�|Ã�, Y²�À½~´V��Æ�(J. ±~5.1.1�~,ûA�ó

�?À, u�Ùg¬Ç´Ä�L0.01,eY²À�$,K�Uk�õ�g¬�ûA�É;��,eY

²½�p, Kòk�õ�Ü�¬�ûAáÂ. ÏdY²½����9ûAÚó�V�|Ã,Ad

V�û½. Xc¤ã,k���æ��Åz��{,¦V�|Ã��²ï.

2.ü«�Ø��J��35�þké��ØÓ. XJ1�a�Ø��J35�þéî­, ·

�Òå¦3Ün���S¦þ~��ù«�Ø��U5,ù��A�Y²Ò���$�. ~

X,����)��«#��OÎ�£�,«;¾, Sü�Á�,�é#Î�Ô���Ñu

�. duÎ�®²�Ï�K¦^,k�½���. #�ÿ�²�Ï�K¦^, ���JØÐ�,ò

�9¾<�)·S�, E¤��J¬éî­. ¤±3?1u��, ò�b�H0��”Î�Ø'

#��”,�¦u�Y²α½����, ù�¦H0�Ľ��U5��~�. ù�Ò�y:

/�b��Ľ!#���É�u�0ò´�~î��.

3.��`5,Á�ö3Á�cé¯K��¹oØ´�ä��. ¦é¯K�)¦¦é"b

� ´ÄU¤áÒk�½�w{, ù«w{�UK��¦éY²�ÀJ. '�`��ÔnÆ[

�â,«nØí½�ÅCþXAk©ÙF, ¦��òù�nØGÃu�. é²w,XJ¦éù�

nØék&%,¦ò�~��u@�b�U¤á, ù��kékå�yââ�U¦¦@�ùb�

Øé. �A/,¦òru�Y²��$�.

3¢S¯K¥, "b��Ľ, ~~¿�Xí��«nؽ^#�{5�O��¦^�

IO�{. 3�õê�¹e,<�F"ù���k�����â. lùp�±w�, Neyman-

Pearson���1�a�Ø��K,3"b��ÀJ¥ké��¢S¿Â, ûØüX´�êÆ

¯K. Ó�,�?�Ún)3b�u�¯K¥,"b�?3âÑ/ ��Ï.

���`²��:´:eY²αé�,�b�H0ج�´�Ľ. XJ��*�á\Ľ

�, ·��Ñ/Ľ�b�H00�(ØÒ'���(Ï�, d�·��¬�1�a�Ø, �ÙV

Çé�). ��, �αé��,XJ��*�á\�É�,·��Ñ/�É�b�H00�(Ø�7

��. ù�UL²:3¤À½�Y²evk¿©�â@�H0ؤá,ûØ¿�Xk¿©�â`²

§�((Ï�d�·��¬�1�a�Ø, �ÙVÇ�Ué�) .

6

Page 7: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

Ê!¦)b�u�¯K���Ú½

1. �â¯K��¦JÑ"b�H0Ú�Jb�H1;

2. �ÑĽ��/ª, (½u�ÚOT (X), Ù¥�.�A�½.

3. À�·�Y², |^u�ÚOþ�©Ù¦Ñ�.�A.

4. d��X�Ñu�ÚOþT (X)�äN�, �\�Ľ�¥, ��.��'�, �Ñ�ɽ

öáý�b�H0�(Ø.

2 ��oNëê�b�u�

��©Ù´�~��©Ù, 'u§�ëê�b�u�´¢S¥~���¯K, Ïd�´�­

���au�¯K. �!ò©e�A«�¹5?Ø��oNëê��*u��{: ü���o

Nþ�Ú���u�; ü���oNþ��Ú��'�u�; 4�©Ù���©Ù�k'���

u�.

3?Ø��©ÙoNëê�b�u�¯K�, §2.4¥�½n2.2.3ÚíØ2.4.2–íØ2.4.53¦

u�ÚOþ�©Ù¥å��©­���^.

�!ü���oNþ��u�

�X = (X1, . . . , Xn)�l��oNN(µ, σ2)¥Ä��{ü�Å��, �½u�Y²α, ¦e

�nau�¯K:

(1) H0 : µ = µ0 ←→ H1 : µ 6= µ0;

(2) H ′0 : µ ≤ µ0 ←→ H ′1 : µ > µ0;

(3) H ′′0 : µ ≥ µ0 ←→ H ′′1 : µ < µ0;

Ù¥µ0Úu�Y²α�½.

·�¡u�¯K(1)�V>u� (two-side test), ¡u�¯K(2)Ú(3)�ü>u� (one-side

test).

ü���oN������u�¯K�'��®��¹�~�, ·�ò­:?Øù��/.

'uþ�®���u��{, ·�ò3�¡���`².

Äk�Äu�¯K(1), =

H0 : µ = µ0 ←→ H1 : µ 6= µ0

·�^�*�{�Eu��Ľ�. ·���X = 1n

∑ni=1Xi ´µ�à�O, �ä

kûÐ5�. �*þw|X − µ0|��, H0�Ø�¤á. Ïdu��Ľ���Xe/ª:

{X = (X1, . . . , Xn) : |X − µ0| > A}, A�½. �σ2���, díØ2.4.2��,3µ = µ0^�e,

T =

√n(X − µ0)

S∼ tn−1. (2.1)

Ïd�T =√n(X−µ0)

S ��u�ÚOþ,KĽ���d/ª���{(X1, . . . , Xn) : |T | > c

}, c�½.

7

Page 8: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

du�Y²�α,��

P (|T | > c |H0) = P(∣∣√n(X − µ0)/S

∣∣ > c∣∣H0

)= α,

��c = tn−1(α/2).ÏddĽ�

D1 ={

(X1, . . . , Xn) :∣∣√n(X − µ0)/S

∣∣ > tn−1(α/2)}

(2.2)

(½�u��u�¯K(1)�Y²�α�u�.

éu�¯K(2), �T =√n(X − µ0)/S��u�ÚOþ,Ïd�*þĽ��/ª�

{(X1, . . . , Xn) : T > c}, c�½.

�¦u�(2′)äkY²α,, =�¦

P (T > c |H ′0) = P(√n(X − µ0)/S > c

∣∣µ ≤ µ0

)≤ α, (2.3)

5¿�

P(√n(X − µ0)/S > c

∣∣µ ≤ µ0

)= P

(√n(X − µ)/S > c+

√n(µ0 − µ)

S

∣∣µ ≤ µ0

)≤ P

(√n(X − µ)/S > c,

∣∣µ ≤ µ0

)Ïd�IP

(√n(X − µ)/S > c,

∣∣µ ≤ µ0

)= αKª(2.3)¤á. ¤±c = tn−1(α).�u�¯

K(2′)�Y²�α�u��Ľ�´

D∗2 ={

(X1, . . . , Xn) : T =√n(X − µ0)/S > tn−1(α)

}. (2.4)

aq�{��u�¯K(3)�Y²�α�u��Ľ�.

D3 ={

(X1, . . . , Xn) :√n(X − µ0)/S < −tn−1(α)

}ù«Äuu�ÚOþÑlt©Ù�u��{¡����tu� ,T¡����tu�ÚOþ.

5555.2.1 ��®����oNþ��u��{�Xe`²: 3N(µ, σ2)¥, �σ2®�,

�H0¤á, =µ = µ0�k

U =√n(X − µ0)/σ ∼ N(0, 1) (2.5)

Ïd�U=√n(X −µ0)/σ ��u�ÚOþ, ^�����σ2���/�Ó��{(¤ØÓ�Ò´

^u�ÚOþU �O@��u�ÚOþT ) ��u�¯K(1)-(3)�Y²�α�u��Ľ�. Ä

½�¥��.�òtn−1© êU¤�A�IO��©Ù�© ê. �[(J�L5.2.1.

ù«Äuu�ÚOþÑlN(0, 1)©Ù�u��{¡����Uu�.

L5.2.1 ü���oNþ��b�u�H0 H1 u�ÚOþ9٩٠Ľ�

σ2 µ = µ0 µ 6= µ0 U =√n(X − µ0)/σ |U | > uα/2

® µ ≤ µ0 µ > µ0 U |µ = µ0 ∼ N(0, 1) U > uα

� µ ≥ µ0 µ < µ0 U < −uασ2 µ = µ0 µ 6= µ0 T =

√n(X − µ0)/s |T | > tn−1(α/2)

� µ ≤ µ0 µ > µ0 T |µ = µ0 ∼ tn−1 T > tn−1(α)

� µ ≥ µ0 µ < µ0 T < −tn−1(α)

8

Page 9: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

~~~5.2.1  ¬�^gÄC-ÅC-Þ ¬,z-IO­þ�500�,zUmóIu�Åì�

ó�G¹. 8Ä�10-,ÿ�Ù­þ(ü :�)

495, 510, 505, 498, 503, 492, 502, 512, 497, 506.

b½-Þ­þXÑl��©ÙN(µ, σ2),®�σ = 6.5, ¯Åì´Äó��~?(�u�Y²α =

0.05)

))) u�¯K�

H0 : µ = 500←→ H1 : µ 6= 500

�K¥σ2®�,�Ľ�dL5.2.1¥1�1�Ñ,=

D ={

(X1, . . . , Xn) : |√n(X−µ0)

σ | > µα2

}Ù¥n = 10, α = 0.05,�L�u0.025 = 1.96,d����X = 502, Ïd

|U | = |√n(X−µ0)

σ | = |√

10(502−500)6.5 | = 0.973 < 1.96

¤±3wÍ5Y²α = 0.05e@�vkvnd`²gÄC-Åó�Ø�~,��ÉH0.

~~~5.2.2 ,<�¤)��/<�|ärÝXÑl��©ÙN(µ, σ2), 8lT�)��/

<¥�ÅÄ�6¬ÿ�|ärÝXe(ü :kg/cm2)

32.56, 29.66, 31.64, 30.00, 31.87, 31.03.

¯ù�1/<�|ärÝ�Ä@�Ø$u32.50 kg/cm2? (�u�Y²α = 0.05)

))) u�¯K�

H0 : µ ≥ 32.50←→ H1 : µ < 32.50

�K¥σ2��,�æ^tu�{,Ľ�dL5.2.1¥���1�Ñ,=

D = {(X1, . . . , Xn) :√n(X−µ0)

S < −tn−1(α)},

Ù¥n = 6, α = 0.05,�L�t5(0.05) = 2.015,dêâ��X = 31.13, S = 1.123,Ïdk

T =√

6(31.13−32.50)1.123 = −3.36 < −2.015

�ĽH0,=@�/<r��32.50 kg/cm2.

�!ü���oN���u�

�X1, . . . , Xn�g��oNN(µ, σ2)¥Ä��{ü�Å��, �ã�?Øe�nau�¯

K:

(4) H0 : σ2 = σ20 ←→ H1 : σ2 6= σ2

0 ;

(5) H ′0 : σ2 ≤ σ20 ←→ H ′1 : σ2 > σ2

0 ;

(6) H ′′0 : σ2 ≥ σ20 ←→ H ′′1 : σ2 < σ2

0 ;

Ù¥σ20Úu�Y²α�½.

·�ò­:?Ø, þ����ü���oN����u��{. �þ�®��XÛ?n, ·

�ò3�¡���`². Äk�Äu�¯K(4), =

H0 : σ2 = σ20 ←→ H1 : σ2 6= σ2

0

9

Page 10: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

duþ�µ��, ·���S2 = 1n−1

∑ni=1 (Xi − X)2 ´σ2���à�O, �äkûÐ

5�. �*þwS2/σ20��½öS

2/σ20���,H0Ø�¤á. Ïdu��Ľ���Xe/ª:

{(X1, . . . , Xn) : S2/σ20 < A1 ½ S2/σ2

0 > A2}, A1, A2 �½. 3�½σ2 = σ2

0�^�e, d½

n2.2.3��

(n− 1)S2/σ20 ∼ χ2

n−1. (2.6)

��u�ÚOþ�χ2 = (n− 1)S2/σ20 . Ïd, Ľ���d/ª���

D ={

(X1, . . . , Xn) : (n− 1)S2/σ20 < c1½ (n− 1)S2/σ2

0 > c2}

Pθ = (µ, σ2),�(½c1, c2,-

α = Pθ((n− 1)S2/σ2

0 < c1½ (n− 1)S2/σ20 > c2

∣∣H0

)÷vþª�c1, c2�éfkéõ,�3�éc1, c2´�`�,�O��E,, �¦^Ø�B. (

½c1, c2���{ü¢^��{´:-

Pθ((n− 1)S2/σ2

0 < c1∣∣H0

)= α/2, Pθ

((n− 1)S2/σ2

0 > c2∣∣H0

)= α/2

dþãüªÚ(2.6)´��.�c1 = χ2n−1(1 − α/2), c2 = χ2

n−1(α/2).¤±u�¯K(4)�Y²

�α��É��

D4 ={

(X1, . . . , Xn) : χ2n−1(1− α/2) ≤ (n− 1)S2/σ2

0 ≤ χ2n−1(α/2)

}d�É��L�ª'Ľ�{ü,¦^þ��B, �d?æ^�É��OĽ�.

^��aqu(�) ¥¦u�¯K(2)!(3)��{�©O¦�u�¯K(5)Ú(6)�Y²�α�

Ľ�Xe:

D5 ={

(X1, . . . , Xn) : (n− 1)S2/σ20 > χ2

n−1(α)},

D6 ={

(X1, . . . , Xn) : (n− 1)S2/σ20 < χ2

n−1(1− α)}.

5555.2.2 �þ�µ®��, ��σ2�u��{{ãXe: �µ®��,σ2���äkûÐ5

��à�O´S2∗ = 1

n

∑ni=1 (Xi − µ)2.�σ2 = σ2

0�,díØ2.4.1 ��

nS2∗/σ

20 =

n∑i=1

(Xi − µ)2/σ20 ∼ χ2

n. (2.7)

Ïd�u�ÚOþ�χ2∗ = nS2

∗/σ20 ,^§�Oχ

2 = (n− 1)S2/σ20 .æ^��aquc¡µ���/

�?Ø�{, ��u�¯K(4)-(6)�Y²�α�Ľ�,=5¿3Ľ�¥ò(½�.��χ2©

Ù�gdÝdn− 1U¤n=�. �[(J�L5.2.2.

ù«Äuu�ÚOþÑl�½gdÝ�χ2©Ù�u��{¡�χ2u�.

L5.2.2 ü���oN���b�u�

10

Page 11: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

H0 H1 u�ÚOþ Ľ�

9Ù©Ù

µ σ2 = σ20 σ2 6= σ2

0 χ2∗ = ns2

∗/σ20 nS2

∗/σ20 < χ2

n(1− α/2)

® ½ nS2∗/σ

20 > χ2

n(α/2)

� σ2 ≤ σ20 σ2 > σ2

0 χ2∗|σ2

0 ∼ χ2n nS2

∗/σ20 > χ2

n(α)

σ2 ≥ σ20 σ2 < σ2

0 nS2∗/σ

20 < χ2

n(1− α)

µ σ2 = σ20 σ2 6= σ2

0 χ2 = (n−1)s2

σ20

(n− 1)S2/σ20 < χ2

n−1(1− α/2)

� ½ (n− 1)S2/σ20 > χ2

n−1(α/2)

� σ2 ≤ σ20 σ2 > σ2

0 χ2|σ20 ∼ χ2

n−1 (n− 1)S2/σ20 > χ2

n−1(α)

σ2 ≥ σ20 σ2 < σ2

0 (n− 1)S2/σ20 < χ2

n−1(1− α)

~~~5.2.3 ,ó�)���«[ã|êÑl��©Ù,ÙoNIO��1.2. yl,F)�

��1�¬¥Ä�16Ã?1|êÿþ,ÿ���IO��2.1,¯ã�þ!Ý´ÄC�?(α = 0.05)

))) u�¯K�

H0 : σ2 = 1.44←→ H1 : σ2 6= 1.44.

u�����

D = {(X1, . . . , Xn) : χ2n−1(1− α/2) < (n− 1)S2/σ2

0 < χ2n−1(α/2)}

d?n = 16, α = 0.05.�L��χ215(0.975) = 6.262, χ2

15(0.025) = 27.488. d®�êâ�

�S2 = 2.12 = 4.41, σ20 = 1.22 = 1.44.Ïdk

χ2 = (n−1)S2

σ20

= 15×4.411.44 = 45.94 > 27.488.

�ĽH0,=@��ã�þ!ÝC�.

n!ü���oNþ���u�

�X = (X1, . . . , Xm)�g��oNN(µ1, σ21) ¥Ä��{ü�Å��, Y = (Y1, . . . , Yn)�

g��oNN(µ2, σ22) ¥Ä��{ü�Å��, �Ü��X1, . . . , Xm;Y1, . . . , YnÕá. �ã�?

Øe�nau�¯K:

(7) H0 : µ2 − µ1 = µ0 ←→ H1 : µ2 − µ1 6= µ0;

(8) H ′0 : µ2 − µ1 ≤ µ0 ←→ H ′1 : µ2 − µ1 > µ0;

(9) H ′′0 : µ2 − µ1 ≥ µ0 ←→ H ′′1 : µ2 − µ1 < µ0;

Ù¥µ0Úu�Y²α�½.

±þb�u�¯K�¦þ����&«m�Behrens-Fisher¯K (�§4.2,n) ��, ا

�A«AÏ�/®¼��÷)û, ���¹�8ÿvk��{ü!°(�){. e¡©A«

�¹©O?Ø.

1. �σ21Úσ

22®��, þ���u�¯K

Äk�Äü>u�¯K(7), =

H0 : µ2 − µ1 ≤ µ0 ←→ H1 : µ2 − µ1 > µ0

11

Page 12: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

duY − X´µ2 − µ1���à�O, �äkûÐ5�. �*þw|Y − X − µ0|���,

H0�Ø�¤á, �Ľ���Xe/ª{(X1, . . . , Xm;Y1, . . . , Yn) : |Y − X − µ0| > A}, A�½.

duY − X ∼ N(µ2 − µ1,√σ2

1/m+ σ22/n), ��µ2 − µ1 = µ0�k

U =Y − X − µ0√σ2

1/m+ σ22/n

∼ N(0, 1). (2.8)

Ïd, �u�ÚOþ�U = (Y − X − µ0)/√

σ21/m+ σ2

2/n, KĽ��d/ª�{(X1, . . . , Xm;Y1, . . . , Yn) :

∣∣∣∣ Y − X − µ0√σ2

1/m+ σ22/n

∣∣∣∣ > c

}, c�½.

Pθ = µ2 − µ1, �(½c,-

α = Pθ (|U | > c|H0) = P

(∣∣∣∣ Y − X − µ0√σ2

1/m+ σ22/n

∣∣∣∣ > c∣∣∣H0

)= 2− 2Φ(c)

dd�(½�.�c = uα/2.Ïdu�¯K(3)�Y²�α�u�Ľ��

D7 =

{(X1, . . . , Xm;Y1, . . . , Yn) :

∣∣∣∣ Y − X − µ0√σ2

1/m+ σ22/n

∣∣∣∣ > uα/2

},

��aq(�)¥�¦u�¯K(2)!(3)�{, ��u�¯K(8)!(9)�Y²�α�u��Ľ

��

D8 =

{(X1, . . . , Xm;Y1, . . . , Yn) :

Y − X − µ0√σ2

1/m+ σ22/n

> uα

}D9 =

{(X1, . . . , Xm;Y1, . . . , Yn) :

Y − X − µ0√σ2

1/m+ σ22/n

< −uα}

ù«Äud(2.8)�Ñ�u�ÚOþU�u��{,¡�ü��Uu�.

2. �σ21 = σ2

2 = σ2���, þ���u�

eσ21 = σ2

2 = σ2®�, Kdc¡ff?Ø�ü��Uu���, d�u�ÚOþUC�

U =Y − X − µ0√σ2/m+ σ2/n

=Y − X − µ0

σ

√mn

m+ n. (2.9)

3σ21 = σ2

2 = σ2���, þãL�ª¥�σ2~^

S2w =

1

n+m− 2

((m− 1)S2

1 + (n− 1)S22

)=

1

n+m− 2

( m∑i=1

(Xi − X)2 +

n∑j=1

(Yj − Y )2

)(2.10)

5�O, Ù¥

S21 =

1

m− 1

m∑i=1

(Xi − X)2, S22 =

1

n− 1

n∑j=1

(Yj − Y )2

12

Page 13: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

©O���X1, . . . , XmÚY1, . . . , Yn�����. �µ2−µ1 = µ0�,òúª(2.9)¥�σ^(2.10)¥

�sw�O, ��e��ÚOþTw,díØ2.4.3��

Tw =Y − X − µ0

Sw

√mn

m+ n∼ tn+m−2. (2.11)

Ïd·��Tw = Y−X−µ0

Sw

√mnm+n��u�ÚOþ.

�(�)¥���tu��{aq, 3ü��þ���u�¯K¥, ^u�ÚOþTw �O@�

����tu�ÚOþT,^���Ó�?Ø�ª, ��u�¯K(7)-(9)�Y²�α �Ľ�, �

�5¿3Ľ�¥ò(½�.��t©Ù�gdÝdn − 1U�n + m − 2=�. �[(J�e�

L5.2.3.

ù«Äuu�ÚOþÑltn+m−2©Ù�u��{,¡�ü��tu�.

L5.2.3 ü���oNþ���b�u�

H0 H1 u�ÚOþ Ľ�

9Ù©Ù

σ21 µ2 − µ1 = µ0 µ2 − µ1 6= µ0 U = Y−X−µ0√

σ21/m+σ2

2/n|U | > uα

2

σ22

® µ2 − µ1 ≤ µ0 µ2 − µ1 > µ0 U |µ0 ∼ N(0, 1) U > uα

� µ2 − µ1 ≥ µ0 µ2 − µ1 < µ0 U < −uασ2

1 µ2 − µ1 = µ0 µ2 − µ1 6= µ0 Tw=Y−X−µ0

Sw

√mnm+n |Tw| >

σ22 tn+m−2(α2 )

� µ2 − µ1 ≤ µ0 µ2 − µ1 > µ0 Tw|µ0 ∼ tn+m−2 T > tn+m−2(α)

� µ2 − µ1 ≥ µ0 µ2 − µ1 < µ0 S2w=

(m−1)S21+(n−1)S2

2

n+m−2 T<−tn+m−2(α)

~~~5.2.4 �ïÄ�~¤cIåÉ�ù[�²þê��O, u�,/�~¤cIf156<,

åf74<, O�Iåù[��²þêÚ��IO�©O�

I: X = 465.13� /mm3, S1 = 54.80� /mm3

å: Y = 422.16� /mm3, S2 = 49.20� /mm3

b½�~¤cIåù[�ê©OÑl��©Ù, ����Ó. u��~¤c<ù[�ê´Ä�5

Ok'. (α = 0.01)

))) �X1, . . . , Xm i.i.d. ∼ N(µ1, σ2); Y1, . . . , Yn i.i.d. ∼ N(µ2, σ

2), �b½ùü|��Õ

á. u�¯K�

H0 : µ2 − µ1 = µ0 ←→ H1 : µ2 − µ1 6= µ0

u��Ľ��

D ={

(X1, . . . , Xm, Y1, . . . , Yn) :∣∣∣ Y−XSw

√mnm+n

∣∣∣ > tn+m−2(α2 )}

d?

m = 156, n = 74, X = 465.13, Y = 422.16, S1 = 54.80, S2 = 49.20,

S2w =

1

n+m− 2

[(m− 1)S2

1 + (n− 1)S22

]= 2816.6, Sw = 53.07,

13

Page 14: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

�L�t228(0.005) = 2.576 .d

|T | =∣∣∣ Y−XSw

√mnm+n

∣∣∣ =∣∣∣ 422.16−465.13

53.07

√156×74156+74

∣∣∣ = 5.74 > 2.576,

�ĽH0,=@��~¤c<�ù[�ê�5Ok'.

3. ���Nþm = n�þ���u�—¤é'�¯K

c¡?Ø�^uü���oNþ���u�¥, b½5gü���oN���´�

pÕá�. �3¢S¯K¥, k�ÿ�¹Øo´ù�. �Uùü���oN���´5gÓ

��oNþ�­E*, §�´¤éÑy�, �´�'�. ~X�*�«S����

J, P¹n ���¾<Ñ�c�z�Z��mX1, X2, . . . , XnÚÑ^dS���z�Z��

mY1, Y2, . . . , Yn.ùp(Xi, Yi)´1i�¾<ØÑ^S��ÚÑ^S��z��Z��m. §�´

k'X�, ج�pÕá. ,��¡, X1, X2, . . . , Xn´n�ØÓ��¾<�Z��m, du�<

N�Ã�¡�^�ØÓ, ùn�*�ØU@�´5gÓ����oN���. Y1, Y2, . . . , Yn�

´��. ù��êâ¡�¤éêâ. ù��êâ^ü��tu�ÒØÜ·, Ï�XiÚYi´Ó3

1i�¾<�þ*����Z��m, ¤±Zi = Yi −Xi Ò�Ø<�N�Ã�¡��É, =

�eS����J. eS��Ã�, Zi��É=d�ÅØ�Úå, �ÅØ��@�Ñl��©

ÙN(0, σ2).��b½Z1, . . . , Zn�gN(µ, σ2)¥Ä��{ü�Å��, µÒ´S���²þ�J.

S��´Äk�, Ò8(�u�Xeb�

H0 : µ = 0←→ H1 : µ 6= 0

Ï�Z1, . . . , Zn�@�´5g��oNN(µ, σ2)�{ü��, ��^'uü���oNþ�

�tu��{. u��Ľ��

D = {(Z1, Z2, . . . , Zn) : |T | > tn−1(α2 )}

d?α�u�Y², T =√n Z/S�u�ÚOþ, Ù¥ZÚS2�Z1, Z2, . . . , Zn ���þ�Ú��

��.

~~~5.2.5 8kü�ÿþá�¥,«7á¹þ�1̤AÚB,��½§���þkÃwÍ

�É, éT7á¹þØÓ�9�á��¬?1ÿþ, ��9é*�Xe:

µ(%) : 0.20, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80, 0.90, 1.00.

ν(%) : 0.10, 0.21, 0.52, 0.32, 0.78, 0.59, 0.68, 0.77, 0.89.

¯�â¢�(J, 3α = 0.01e, UÄ�äùü�1̤��þkÃwÍ�É?

))) ò1̤AÚBé9��¬�ÿ½�P�X1, X2, . . . , X9,ÚY1, Y2, . . . , Y9. duù9�

�¬7á¹þØÓ, ÏdX1, X2, . . . , X9 ØUw¤5gÓ�oN. Y1, Y2, . . . , Y9���. �I^

¤é'�. P

Zi = Yi −Xi, i = 1, 2 . . . , 9.

eùü1̤�þ��, ÿþ���zéêâ��É=d�ÅØ�Úå. �ÅØ��@�Ñl

��©ÙN(0, σ2).��b½Z1, . . . , Zn�gN(µ, σ2)¥Ä���Å��, �u�

H0 : µ = 0←→ H1 : µ 6= 0, α = 0.01

dL5.2.1��du��Ľ��

{(Z1, · · · , Zn) : |T | > tn−1(α/2)}

14

Page 15: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

d?n = 9.dK¥êâ��:

Z = 19

∑9i=1 Zi = 0.06, S2 = 1

8

∑9i=1 (Zi − Z)2 = 0.01505, S = 0.12268

�L� tn−1(α2 ) = t8(0.005) = 3.3554.du

|T | =∣∣∣√nZS ∣∣∣ = 3×0.06

0.12268 = 1.47 < 3.3554

�Ãvyâw«ü�¤ìkwÍ�É, Ïd�ÉH0.

o!ü���oN��'�u�

�X = (X1, . . . , Xm)´l��oNN(µ1, σ21)¥Ä��{ü�Å��,Y = (Y1, . . . , Yn)´l

��oNN(µ2, σ22)¥Ä��{ü�Å��, �Ü��(X,Y) = (X1, . . . , Xm;Y1, . . . , Yn)Õá.

?Øe�nab�u�¯K:

(10) H0 : σ22/σ

21 = 1←→ H1 : σ2

2/σ21 6= 1;

(11) H ′0 : σ22/σ

21 ≤ 1←→ H ′1 : σ2

2/σ21 > 1;

(12) H ′′0 : σ22/σ

21 ≥ 1←→ H ′′1 : σ2

2/σ21 < 1;

u�Y²α�½.

PXÚS21�X1, . . . , Xm���þ�Ú����; YÚS2

2�Y1, . . . , Yn���þ�Ú���

�. Ù¥

S21 = 1

m−1

∑mi=1 (Xi − X)2, S2

2 = 1n−1

∑nj=1 (Yj − Y )2.

e¡òX­?ص1Úµ2�����'�u��{,�µ1Úµ2®�´��'�u�XÛ?n,

·�ò3�¡�Ñ��`².

Äk?Øu�¯K(1), =

H0 : σ22/σ

21 ≤ 1←→ H1 : σ2

2/σ21 > 1

duS21ÚS

22©O´σ

21Úσ

22�à�O,¿äkûÐ5�. �*þw, S2

2/S21��½öS

22/S

21�

��, H0Ø�¤á. ���Ľ��/ª�

{(X1, . . . , Xm;Y1, . . . , Yn) : S22/S

21 < c1½ S2

2/S21 > c2}, c1, c2�½.

3S22/S

21 = 1�^�e, díØ2.4.4��

F = S22/S

21 ∼ Fn−1,m−1. (2.12)

Ïd�u�ÚOþ�F = S22/S

21 . Pθ = (µ1, µ2, σ

21 , σ

22),�(½Ä½�¥��.�c1, c2,-

α = Pθ(S2

2/S21 < c1 ½ S2

2/S21 > c2|H0

)÷vþª�¦�c1Úc2kéõ, Ù¥�3�éc1, c2�`, �O�E,, ¦^Ø�B. (½c1, c2�

��{ü¢^��{´: -

Pθ(S2

2/S21 < c1|H0

)= α/2, Pθ

(S2

2/S21 > c2|H0

)= α/2.

dþãüªÚ(2.12)´��.�c1 = Fn−1,m−1(1 − α/2), c2 = Fn−1,m−1(α/2), ¤±u�¯

K(10)�Y²�α��É��

15

Page 16: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

D10 = {(X,Y) : Fn−1,m−1(1− α/2) 6 S22/S

21 6 Fn−1,m−1(α/2)}

d�É��L�ª'Ľ�{ü, ¦^�B, �d?æ^�É��OĽ�.

^��aqu(�) ¥¦u�¯K(2)! (3)��{�©O¦�u�¯K(11)Ú(12)�Y²

�α�Ľ�Xe:

D11 = {(X1, . . . , Xm;Y1, . . . , Yn) : S22/S

21 > Fn−1,m−1(α)}

D12 = {(X1, . . . , Xm;Y1, . . . , Yn) : S22/S

21 < Fn−1,m−1(1− α)}

ùp�5¿�:�´: �α´���ê, Xα = 0.01, 0.05�, lF−©Ù�© êLþ�Ø�Fn−1,m−1(1− α)�ê�, �|^1�ÙSK¥®y²�¯¢

Fn,m(1− α) = 1/Fm,n(α), (2.13)

¦¯K¼�)û. ~XlLþ�Ø�F5,10(1 − 0.01)��, ����F10,5(0.01)��, |^ú

ª(2.13) ��F5,10(1− 0.01) = 1/F10,5(0.01),l �¦�¤��ê�.

5555.2.3 �µ1Úµ2®��, ��'�u��{{ãXe: �µ1Úµ2®��, σ21Úσ

22äkû

Ð5��à�O©O´

S21∗ =

1

m

m∑i=1

(Xi − µ1)2, S22∗ =

1

n

n∑i=1

(Yi − µ2)2

�σ21/σ

22 = 1�, |^íØ2.4.1ÚF©Ù�½Â, N´y²

F∗ = S22∗/S

21∗ ∼ Fn, m . (2.14)

Ïd,�u�ÚOþF∗�OF = S22/S

21 , ��aquµ1Úµ2���/�?Ø, ���u

�¯K(10)-(12)�Y²�α�Ľ�,��5¿3Ľ�¥, ò(½�.��F©Ù�gdÝ

dn− 1,m− 1©OU�n,m=�. �[(J�e�L5.2.4.

ù«Äuu�ÚOþÑlF©Ù�u��{, ¡�Fu�.

L5.2.4 ü���oN��'�b�u�

H0 H1 u�ÚOþ9٩٠Ľ�

µ1 σ22 = σ2

1 σ22 6= σ2

1 F∗ = s22∗/S

21∗ F∗ < Fn,m(1− α/2)

µ2 F∗|σ22=σ21∼ Fn,m ½ F∗ > Fn,m(α/2)

® σ22 ≤ σ2

1 σ22 > σ2

1 S21∗=

1m

∑mi=1(Xi − µ1)2 F∗ > Fn,m(α)

� σ22 ≥ σ2

1 σ22 < σ2

1 S22∗= 1

n

∑ni=1(Yi − µ2)2 F∗ < Fn,m(1− α)

µ1 σ22 = σ2

1 σ22 6= σ2

1 F = s22/S

21 F<Fn−1,m−1(1−α/2)

µ2 F |σ22=σ21∼ Fn−1,m−1 ½ F>Fn−1,m−1(α/2)

� σ22 ≤ σ2

1 σ22 > σ2

1 S21= 1

m−1

∑mi=1 (Xi − X)2 F > Fn−1,m−1(α)

� σ22 ≥ σ2

1 σ22 < σ2

1 S22= 1

n−1

∑nj=1 (Yj − Y )2 F<Fn−1,m−1(1− α)

~~~5.2.6 ÿ�ü1������6�>fìá>{�þ�X = 0.14, Y= 0.139, ��IO

�©O�SX = 0.0026, SY = 0.0024,b�ùü1ìá�>{©OÑlN(µ1, σ21), N(µ2, σ

22), þ

������, �ü|��Õá, ¯ùü1>fì��>{´Ä�Ó? (α = 0.05)

16

Page 17: 2011 c F - USTCstaff.ustc.edu.cn/~zwp/teach/Math-Stat/lec9.pdf · 2011-04-11 · i= c;V’Ñk1=2˙„U,›ÑØgCØ |˙߉,V’ÑœˆÜn, „–ˆÉ.XJ r= 1=3;` „–ˇL3k2⁄x¥Ú1⁄ç¥˙Ý

): ù�¯KL¡w´éü���oNþ���u�,�·�Ø��´Äkσ21 = σ2

2 ,ÏdÄ

k¦ü���oN��´Ä�Ó�u�. XJu�@�σ21 = σ2

2 , ,�2�ü��tu�. XJ²

u�Ľσ21 = σ2

2 , K·�ØU^ü���tu��{u�þ��, ùÒC¤Behrens-Fisher¯

K, ò33�!��)û.

Äk�Äe�u�¯K:

(1) H0 : σ21 = σ2

2 ←→ H1 : σ21 6= σ2

2 , α = 0.05 .

dL5.2.4��du���É�´

{(X,Y) : Fm−1,m−1(1− α/2) ≤ S2X/S

2Y ≤ Fm−1,n−1(α/2)},

d?m = n = 6, S2X/S

2Y = 0.00262/0.00242 = 1.17.dα = 0.05 , �F©ÙL�F5,5(0.025) =

7.15. du

1/7.15 = F5,5(1− 0.025) < F = S2X/S

2Y = 1.17 < F5,5(0.025) = 7.15,

�@�vkv�yâĽH0,Ïd�ÉH0.

3�Éþãu��, ·��±b½σ21 = σ2

2 ,?�Ú�Äe�u�¯K:

(2) H ′0 : µ1 = µ2 ←→ H ′1 : µ1 6= µ2, α = 0.05 .

dL5.2.3��du��Ľ��

{(X,Y) : |Tw| > tn+m−2(α/2)}

d?m = n = 6, X = 0.14, Y = 0.139, SX = 0.0026, Sy = 0.0024,Ïdk

S2w = 1

10 [5× 0.00262 + 5× 0.00242] = 6.26× 10−6, Sw = 0.0025.

dα = 0.05,�t−©ÙL�t10(0.0025) = 2.228 . du

|T | =√

nmn+m

∣∣∣ Y−XSw

∣∣∣ =√

3×∣∣ 0.14−0.139

0.0025

∣∣ = 0.6928 < 2.228

�vk¿v�ndĽü1>fì��>{��Ó,Ïd�ÉH ′0

17