应用统计学 第五章 方差分析

1

1

, .

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2

1 A1 A2 24 60 A3

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3

1A A1 A2 A31073 1107 1093 1009 1092 1029 1060 990 1080 1001 1109 1021 1002 1090 1022 1012 1074 1032 1009 1122 1029 1028 1001 1048

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4

A A1, A2, A3 yij

Ai 60 i=1, 2, 3, j=1, 2,, 10

A j

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5

21 A, r r A1, A2,…, Ar r

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6

1)

N( i, i 1, 2,…, r

2 i

)

2) 3)

:

2 1

=

2 2

=…=

2 r

=

2

yij

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7

,: H0 H1 H0 H0 A16 March 20121

= 2=…=,…,

r

1

1

,

2

r

H1 A r A A r A A r

8

1 i m yij r j yij i 1, 2,…, r m i n=r m j 1, 2,…, m, i j

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9

Aii ij

yijij

= yij

i

yij= 2 yij

i

+

ij

2

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10

yijij

i

ij

, i 1, 2,..., r, j 1, 2,..., m N (0,2

)

3

:i

1 ( r

1

...i

r

)

1 r

r i i 1

.:

i ai= i 16 March 2012

Ai

11

3yijr

ai ai 0

ij

,

i 1,2,..., r, j 1,2,..., m

(4)N(0,2

i 1 ij

)

1 H0 a1=a2=…=ar=0 5

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12

3

2m

Tij 1 r

yij Tii 1

yi . y

Ti m T r m T n

i 1, 2,

,r

T n16 March 2012

r m

13

2

A1 A2

y11 y12… y1m y21 y22… y2m

T1 T2

y1 y2

Ar

yr1 yr2… yrm

Tr T

yr

y

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14

yij yij yyij y ( yij yi. ) ( yi. y)

y

61 nr m ij i 1 j 1

i.

1 m

m ij j 1

,

1 r

r i. i 1

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15

yij

yi.

(

i

ij

) (

i

i

)

ij

i

7

yij– y i

yi.

y

(

i

i.

) (

)

ai

i.

8 i

yi.

y

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16

k y=(y1+…+ yk )/kQ ( y1 y)2

y1, y2,…, ykk

( yk

y)

2 i 1

( yi

y )2

k

16 March 2012