第十三章方差分析卡方分布與

卡方分布與F分布北京師范大學(xué)心理學(xué)院要解決的問題在日常生活中，我們經(jīng)常關(guān)注總體方差，甚至超過高于總體平均數(shù)。教育均衡問題生產(chǎn)過程中的質(zhì)量管理投資風(fēng)險(xiǎn)的控制抽樣分布的作用在推估總體平均值時(shí)，基于樣本平均數(shù)的抽樣分布在用樣本方差來推估總體方差時(shí)，必須知道樣本方差的抽樣分布如果要比較兩個(gè)總體的方差是否相等時(shí)，必須知道樣本方差的聯(lián)合抽樣分布卡方分布（chi-square

distribution)卡方分布的定義卡方分布的密度函數(shù)卡方分布的形狀卡方分布的性質(zhì)卡方分布的臨界值卡方分布定理一卡方分布定理二卡方分布的定義設(shè)Z1,Z2,…Zυ相互獨(dú)立且都為標(biāo)準(zhǔn)正態(tài)隨機(jī)變量，則稱變量c

2

=

Z

2

+

Z

2

++

Z

21

2

v所服從的分布為自由度為υ的χ2分布For

n

=2:2YZ1s2(Y

-m

)2=

1

Y

2Ys

2(Y

-

m

)2Z

2

=

2

Y

=

Z2

+

Z21

2n=2c2卡方分布的密度函數(shù)定義：如果隨機(jī)變量U的概率密度函數(shù)f(u)為則稱隨機(jī)變數(shù)U

服從自由度為n的卡方分配，記為21u2n2nn2-1

-f

(u)

=u

e

,

0

￡

u

<

￥G(

)2U

~

c。2

(v)卡方分布的形狀卡方分布的性質(zhì)（1）卡方曲線所圍的面積和為1卡方分布為在大于等于0(正數(shù))范圍的正偏分布不同的自由度決定不同的卡方分布0.160.140.120.100.080.060.040.020.000

10

20

30

40

50

6070f

(c2

)c

2v

=

5v

=

10v

=

30卡方分布的性質(zhì)（2）卡方分布只有一個(gè)參數(shù)即自由度，為ν。卡方分布的平均數(shù)與方差為：卡方分布隨著自由度增加而逐漸趨于對(duì)稱，當(dāng)自由度趨近于無窮大時(shí)，卡方分布趨近于正態(tài)分布v2E(c

)

=

vvVar(c

2

)

=

2vc2~

N

(n,2n

)vn

fi

￥,卡方分布的性質(zhì)（3）卡方分布的加法定理兩個(gè)獨(dú)立的卡方隨機(jī)變量相加所得的隨機(jī)變量仍滿足卡方分布，其自由度為其自由度之和。n1

+v2v1

v2(U

+W

)

~

c

2U

~

c

2

,W

~

c

2

且U

,W獨(dú)立，則卡方分布的臨界值9.23640f

(c

2

)c2v

=

5α=0.10Χ2α=Table

VII:

Chi-SquareProbabilities0.10.050.0250.010.005df2.7063.8415.0246.6357.87914.6055.9917.3789.21010.59726.2517.8159.34811.34512.83837.7799.48811.14313.27714.86049.23611.07012.83315.08616.750510.64512.59214.44916.81218.548612.01714.06716.01318.47520.278713.36215.50717.53520.09021.955814.68416.91919.02321.66623.589915.98718.30720.48323.20925.1881017.27519.67521.92024.72526.7571118.54921.02623.33726.21728.3001219.81222.36224.73627.68829.8191321.06423.68526.11929.14131.3191422.30724.99627.48830.57832.801159.2364α=0.10f

(c

2

)c2v

=

5查表練習(xí)df=200.0250.0250.95χ2.025=?0χ2.975=?df0.9950.990.9750.950.90.10.050.0250.010.0051------0.0010.0040.0162.7063.8415.0246.6357.87920.010.020.0510.1030.2114.6055.9917.3789.2110.59730.0720.1150.2160.3520.5846.2517.8159.34811.34512.83840.2070.2970.4840.7111.0647.7799.48811.14313.27714.8650.4120.5540.8311.1451.619.23611.0712.83315.08616.7560.6760.8721.2371.6352.20410.64512.59214.44916.81218.54870.9891.2391.692.1672.83312.01714.06716.01318.47520.27881.3441.6462.182.7333.4913.36215.50717.53520.0921.95591.7352.0882.73.3254.16814.68416.91919.02321.66623.589102.1562.5583.2473.944.86515.98718.30720.48323.20925.188112.6033.0533.8164.5755.57817.27519.67521.9224.72526.757123.0743.5714.4045.2266.30418.54921.02623.33726.21728.3133.5654.1075.0095.8927.04219.81222.36224.73627.68829.819144.0754.665.6296.5717.7921.06423.68526.11929.14131.319154.6015.2296.2627.2618.54722.30724.99627.48830.57832.801165.1425.8126.9087.9629.31223.54226.29628.8453234.267.672

10.085

24.769

27.5879.39

10.865

25.989

28.869.02517185.6976.2656.4087.0157.5648.2318χ2

=34.1730.19131.52633.40934.80535.71837.156196.8447.6338.90710.11711.65127.20430.14432.85236.19138.582207.4348.269.59110.85112.44328.41231.4134.1737.56639.997218.0348.89710.28311.59113.2429.61532.67135.47938.93241.401228.6439.54210.98212.33814.04130.81333.92436.78140.28942.796df0.9950.990.9750.950.90.10.050.0250.010.0051------0.0010.0040.0162.7063.8415.0246.6357.87920.010.020.0510.1030.2114.6055.9917.3789.2110.59730.0720.1150.2160.3520.5846.2517.8159.34811.34512.83840.2070.2970.4840.7111.0647.7799.48811.14313.27714.8650.4120.5540.8311.1451.619.23611.0712.83315.08616.7560.6760.8721.2371.6352.20410.64512.59214.44916.81218.54870.9891.2391.692.1672.83312.01714.06716.01318.47520.27881.3441.6462.182.7333.4913.36215.50717.53520.0921.95591.7352.0882.73.3254.16814.68416.91919.02321.66623.589102.1562.5583.2473.944.86515.98718.30720.48323.20925.188112.6033.0533.8164.5755.57817.27519.67521.9224.72526.757123.0743.5714.4045.2266.30418.54921.02623.33726.21728.3133.5654.1075.0095.8927.04219.81222.36224.73627.68829.819144.0754.665.6296.5717.7921.06423.68526.11929.14131.319154.6015.2296.2627.2618.54722.30724.99627.48830.57832.801165.1425.8126.9087.9629.31223.54226.29628.8453234.267.672

10.085

24.769

27.5879.39

10.865

25.989

28.869.97517185.6976.2656.4087.0157.5648.2318χ2

=9.59130.19131.52633.40934.80535.71837.156196.8447.6338.90710.11711.65127.20430.14432.85236.19138.582207.4348.269.59110.85112.44328.41231.4134.1737.56639.997218.0348.89710.28311.59113.2429.61532.67135.47938.93241.401228.6439.54210.98212.33814.04130.81333.92436.78140.28942.796卡方分布定理一X

~

N

(m,s

2

)自正態(tài)總體中抽取n個(gè)隨機(jī)樣本(X1,X2,…Xn)，則212

2

=

ssnsni=1

iX

-

m

X -

mX

-

m

+=

c2

++

c2

=

c21

1

nnn2s

2si=1(x

-

m)2i

1==

為自由度為n的卡方分布

X

-

m

卡方分布定理二2-X

~

N

(m,s

2

)自正態(tài)總體中抽取n個(gè)隨機(jī)樣本(X1,X2,…Xn)，令X為樣本平均數(shù)，S

2為樣本方差n=

i=1

(

X

-

X

)2(n

-1)S

2~

cn

1s

2為自由度為n

-1的卡方分布s

2(n

-1)S

2s

2為自由度為n

-1的卡方分布(n

-1)S

2s

2證明s

2s

2=(n

-1)(

X

-

X

)2(n

-1)S

2n(n

-1)

i=1

=n(

X

-

X

)2

i=1

s

2]2s

2=ni=1[(

X

-

u)

-(

X

-

u)2

]2s

2=ni=1[(

X

-

u)

-

2(

X

-

u)(

X

-

u)

+

(

X

-

u)s

2s

2s

2+-=(

X

-

u)22(

X

-

u)(

X

-

u)(

X

-

u)2

X

,u常數(shù)s

2s

2s

2n(

X

-

u)2+-=2(

X

-

u)(

X

-

u)(

X

-

u)2為自由度為n

-1的卡方分布(n

-1)S

2證明s

2

s

2

s

2s

2n(

X

-

u)2-

+=(n

-1)S

22(

X

-

u)(

X

-

u)s

2(

X

-

u)2(

X

-

u)

=

X

-

u

=

nX

-

nu

=

n(

X

-

u)注意(X

-X

)=0但(X

-u)?0s

2s

2s

2s

2+2(

X

-

u)n(

X

-

u)

n(

X

-

u)2-=(n

-1)S

2(

X

-

u)22=-=

n

X

-

u

(

X

-

u)2n(

X

-

u)2(

X

-

u)2-

ss

2s

2s

22=

n

X

-

u

(n

-1)S

2(

X

-

u)2+

ss

2s

2為自由度為n

-1的卡方分布(n

-1)S

2s

2證明2=

n

X

-

u

(n

-1)S

2(

X

-

u)2+

ss

2s

2為自由度為n的卡方分布n2s

i=1

s

2(x

-

m)2=

X

-

m

n

i=11~

c

2為自由度為1的卡方分布2

s

n

X

-

u

根據(jù)卡方分布的加法定理2n-1(n

-1)S

2~

cs

2單一總體方差的統(tǒng)計(jì)推斷2n-1(n

-1)S

2~

cs

2從一個(gè)標(biāo)準(zhǔn)差為σ的正態(tài)分布中取n個(gè)樣本總體方差的區(qū)間估計(jì)

總體方差的顯著性檢驗(yàn)0f

(c

2

)c2總體方差的置信區(qū)間2n-1(n

-1)S

2~

cs

2sa￡￡

c

a

)

=

1

-a222n-1,2n-1,1-2(n

-1)S

2P(cα/2α/21-αa2c2n-1,1-2c2n-1,a)

=

1

-ac2￡

s

2

￡c22

2n-1,1-an-1,a(n

-1)S

2(n

-1)S

2P(c

2

c

2￡

s

2

￡n-1,a

n-1,1-a2

2(n

-1)S

2

(n

-1)S

2總體方差的1-a置信區(qū)間總體方差的置信區(qū)間（例子）在一次全區(qū)統(tǒng)考中，某校40名學(xué)生成績(jī)的方差為

144分，問全區(qū)學(xué)生成績(jī)的方差95%的置信區(qū)間是多少？解：df=40-1=39該區(qū)學(xué)生成績(jī)的方差95%的置信區(qū)間（94.7，230.2）2

2c

2c

2￡

s

2

￡n-1,1-an-1,a(n

-1)S

2(n

-1)S

224.459.339

·14439

·144￡

s

2

￡0.025(0.05

/

2)=

c

2

=

59.3c

20.975(1-0.05

/

2)=

c

2

=

24.4c

2總體方差的顯著性檢驗(yàn)建立假設(shè)(two

tails

test)(left

tail

test)(right

tail

test)H

:

s

2

￡

s

2

H

:

s

2

>

s

20

0

1

0H

:

s

2

?

s

2

H

:

s

2

<

s

20

0

1

0H

:s

2

=

s

2

H

:

s

2

?

s

20

0

1

02s

2(n

-1)S

2總體方差假設(shè)檢驗(yàn)的統(tǒng)計(jì)量c

=0假設(shè)為真時(shí)的總體方差例題某工廠生產(chǎn)10mm的螺釘，假設(shè)所生產(chǎn)的螺絲釘直徑為常態(tài)分布且期望值為10mm，雖然每支螺釘?shù)闹睆讲灰欢〞?huì)剛好等于10mm，生管部門希望將方差控制在0.09mm以內(nèi)，抽取12支螺釘樣本來檢驗(yàn)，得出以下數(shù)據(jù)，根據(jù)這些樣本數(shù)據(jù)，在α=5%的水平下，推論所生產(chǎn)的鏍釘是否合乎品管的要求？10.0510.0010.029.9710.0710.039.9810.109.959.9910.0010.080f

(c

2

)c2例題寫出假設(shè)>

0.092

2H0

:s

￡

0.09

H1

:

s0.0902= =

2.988s

2c

=(n

-1)S

2

(12

-1)(.0022)計(jì)算統(tǒng)計(jì)檢驗(yàn)量S

2

=0.0022χ20.05=19.675找出臨界值α=0.05,df=12-1,

χ2=?比較檢驗(yàn)統(tǒng)計(jì)量與臨界值2.988<19.675F

分布（

F

distribution

）F分布的密度函數(shù)F分布的形狀F分布的臨界值F分布的定理一F分布的定理二兩個(gè)總體方差的統(tǒng)計(jì)推論兩總體方差是否相等可以用利F分布作估計(jì)與檢驗(yàn)如果隨機(jī)變量X

有密度函數(shù)f(x)為1

2112

222

2

(

1

2

2

2n1

n1nnn

nn

n2n1

+n2-1-n

+nG(

)f

(x)

=)

x

(1+x) ,

x>

0G()G(

)則稱隨機(jī)變量X服從自由度為n1，n2的F分布（Ronald

Fisher），計(jì)為X

~

F

(v1,

v2

)F

分布的形狀Degrees

of

freedomfor

thenumerator（分子自由度，第一個(gè)自由度）Degrees

of

freedomfor

thedenominator（分母自由度，第二個(gè)自由度）F分布的性質(zhì)性質(zhì)1:

F曲線所圍的面積和為1性質(zhì)2:F曲線橫軸數(shù)值從0向右伸展至無限，與橫軸沒有相交性質(zhì)3:F曲線為正偏分布，其形狀由兩個(gè)自由度ν1,ν2決定,不同自由度有不同F(xiàn)分布F分布的性質(zhì)性質(zhì)4:22(n

>

2)n

-

2n2E(F

)

=(n2

>

4)n1

(n2

-

2)(n2

-

4)2n

2

(n

+n

-

2)Var(F

)=

2

1

2

定理一1

2如果U與W自由度分別為ν

及ν

的獨(dú)立卡方分布，則212

1

v1

,v2v2v1~

FvvW

vU

vc

2c

2=v1U

~

c

222vW

~

c兩個(gè)獨(dú)立的卡方隨機(jī)變量分別除以其自由度后，兩者相除可得F隨機(jī)變量，其中分子的自由度決定F分布的第一個(gè)自由度，分母的自由度決定F分布的第二個(gè)自由度。定理二1

2假設(shè)X

與X

獨(dú)立，且11n

-1S

222=

1

1

=

2

2

n

-1

(

X

-

X

)2

(

X

-

X

)2S

22111X

~

N

(m

,s

)2222X

~

N

(m

,s

)，分別從這兩個(gè)總體中抽取獨(dú)立隨機(jī)樣本n1與n222

1

1

1

-1,n2

-1~

FnS

2

s

2S

2

s

2證明 定理二，根據(jù)卡方分布的定理：個(gè)樣本，令：2

2=

1

1

1

-1,n2

-1~

FnS2

s

2S

2

s

221

1

1

為自由度為(n1

-1)的卡方分布(n

-1)S

2s22121n

-1(n1

-1)S1~

cs222n2

-1(n

-1)S~

c

2s同理

2

2

根據(jù)定理一，兩個(gè)獨(dú)立的卡方隨機(jī)變量分別除以其自由度后，兩者相除可得F隨機(jī)變量。211

1(n1

-1)(n

-1)S

2s22

2

2

2(n

-1)(n

-1)S

2s定理二，根據(jù)定理二：22

1

1

1

-1,n2

-1~

FnS

2

s

2S

2

s

21

2n1

-1,n2

-1

1 2

~

FS

2

S

2s

2

s

2我們可以根據(jù)這個(gè)知識(shí)，從兩樣本的方差的比較來推論兩總體方差是否相等。F分布的臨界值右側(cè)臨界值左側(cè)臨界值v2

/

v112345…910…120∞1161.4199.5215.7224.6230.2…240.5241.9…253.3254.3218.5119.0019.1619.2519.30…19.3819.40…19.4919.50310.139.559.289.129.01…8.818.79…8.558.5347.716.946.596.396.26…6.005.96…5.665.6356.615.795.415.195.05…4.774.74…4.404.36……104.964.103.713.483.33…3.022.98…2.582.54114.843.983.593.363.20…2.902.85…2.452.40……∞3.843.002.602.372.21…1.881.83…1.221F分布表(α=0.05)分子自由度分母自由度找出df

=(5,

10)

F0.05=?找出df

=(10,

5)

F0.05=?找出df

=(10,

5)

F0.05=?S14.74Ff

(F)a

=

0.05df

=

(10,

5)F的倒數(shù)仍為F隨機(jī)變量，v1

,v2v2

2v1

1~

Fvvc2c2F

=v2

,v1v1

1v2

2~

FvvFc2c21

=1v2

,v1v1

1v2

2~

FvvFc2c2n1

,n2=兩個(gè)獨(dú)立的卡方隨機(jī)變

量分別除以其自由度后，兩者相除可得F隨機(jī)變量。自由度的分子分母對(duì)調(diào)F的左側(cè)臨界值1a

2

1F

(n

,n

)F

(n1

,n2

)

=1-a，F(xiàn)表中為右側(cè)概率為α的F值，因?yàn)樽髠?cè)的F值可用

F分布的倒數(shù)值求得，左側(cè)概率為α的F值表為：F1-a

(v1

,

v2

)F-Table

α=0.025F0.025(9,8)=4.36F0.975(9,8)=

=1/

F0.025(8,9)=1/4.1=0.24兩總體方差比的區(qū)間估計(jì)要對(duì)總體方差作區(qū)間估計(jì)必須假設(shè)兩總體為正態(tài)分布且獨(dú)立，2221

1

2

=

1

1

1

-1,n2

-1~

FnS

2

S

2

S

2S

2

s

2s

2s

2

s

2F1

2v

,v

,1-a

/20

1

2Fv

,v1

2Fv

,v

,a

/2F1

-

af

(F)的概率區(qū)間(1

-

a

)

F1-P(F

a

(v1,

v2

)

￡

F

￡

Fa

(v1,

v2

))

=1-a2

2(1

-a

)F

的概率區(qū)間1-2

2P(F

a

(v1,

v2

)

￡

F

￡

Fa

(v1,

v2

))

=1-aa

1

22221

221-aS

2

/

s

2S

2

/

s

2P(F

(v

,

v

)￡

1 1

￡

F

(v

,

v

))

=1-aa

1

22211

221-a 1

￡

F

(v

,

v

))

=1-as

2

S

2s

2

S

2P(F

(v

,

v

)￡

2

111

2222F

(v

,

v

)F

(v

,

v

)S

2S

22

a

1

22s

2

S

2s

2

S

21-a￡

1

1

￡

1

總體方差比的置信區(qū)間兩總體方差比的區(qū)間估計(jì)（例子）差之比的置信區(qū)間，能否說二總體方差相等：解：計(jì)算自由度v1=9，v2=14F0.025(9,14）=3.21

F0.975(9,14)=

1/3.77=0.27故二總體方差之比，在0.26—3.14之間，作此推論正確的概率為0.95。111

2F

(v

,

v

)

F

(v

,

v

)S

2S

22

a

1

2

2

22

2s

2

S

2s

2

S

21-a

1

￡

1

￡

1

2n

2

-1Sn1

-1已知n1=10，S2 =5，n2=15， =6，求二總體方5

16

0.276

3.215

12·￡

1

￡·s

2s

2兩總體方差的假設(shè)檢驗(yàn)比較兩方差的比值22s

2s

2

s

2s

2fi

H0

:

1

=1,

H0

:

1

?

1H

:s

2

=

s

20

1

22222s

2s

2

s

2s

2?

s

2

fiH0

:

1

?1,

H1

:

1

<1H0

:s12222s

2s

2