第十三章方差分析卡方分布与

卡方分布与F分布北京师范大学心理学院要解决的问题在日常生活中，我们经常关注总体方差，甚至超过高于总体平均数。教育均衡问题生产过程中的质量管理投资风险的控制抽样分布的作用在推估总体平均值时，基于样本平均数的抽样分布在用样本方差来推估总体方差时，必须知道样本方差的抽样分布如果要比较两个总体的方差是否相等时，必须知道样本方差的联合抽样分布卡方分布（chi-square

distribution)卡方分布的定义卡方分布的密度函数卡方分布的形状卡方分布的性质卡方分布的临界值卡方分布定理一卡方分布定理二卡方分布的定义设Z1,Z2,…Zυ相互独立且都为标准正态随机变量，则称变量c

2

=

Z

2

+

Z

2

++

Z

21

2

v所服从的分布为自由度为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卡方分布的密度函数定义：如果随机变量U的概率密度函数f(u)为则称随机变数U

服从自由度为n的卡方分配，记为21u2n2nn2-1

-f

(u)

=u

e

,

0

£

u

<

¥G(

)2U

~

c。2

(v)卡方分布的形状卡方分布的性质（1）卡方曲线所围的面积和为1卡方分布为在大于等于0(正数)范围的正偏分布不同的自由度决定不同的卡方分布0.160.140.120.100.080.060.040.020.000

10

20

30

40

50

6070f

(c2

)c

2v

=

5v

=

10v

=

30卡方分布的性质（2）卡方分布只有一个参数即自由度，为ν。卡方分布的平均数与方差为：卡方分布随着自由度增加而逐渐趋于对称，当自由度趋近于无穷大时，卡方分布趋近于正态分布v2E(c

)

=

vvVar(c

2

)

=

2vc2~

N

(n,2n

)vn

fi

¥,卡方分布的性质（3）卡方分布的加法定理两个独立的卡方随机变量相加所得的随机变量仍满足卡方分布，其自由度为其自由度之和。n1

+v2v1

v2(U

+W

)

~

c

2U

~

c

2

,W

~

c

2

且U

,W独立，则卡方分布的临界值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查表练习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

)自正态总体中抽取n个随机样本(X1,X2,…Xn)，则212

2

=

ssnsni=1

iX

-

m

X -

mX

-

m

+=

c2

++

c2

=

c21

1

nnn2s

2si=1(x

-

m)2i

1==

为自由度为n的卡方分布

X

-

m

卡方分布定理二2-X

~

N

(m,s

2

)自正态总体中抽取n个随机样本(X1,X2,…Xn)，令X为样本平均数，S

2为样本方差n=

i=1

(

X

-

X

)2(n

-1)S

2~

cn

1s

2为自由度为n

-1的卡方分布s

2(n

-1)S

2s

2为自由度为n

-1的卡方分布(n

-1)S

2s

2证明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常数s

2s

2s

2n(

X

-

u)2+-=2(

X

-

u)(

X

-

u)(

X

-

u)2为自由度为n

-1的卡方分布(n

-1)S

2证明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为自由度为n

-1的卡方分布(n

-1)S

2s

2证明2=

n

X

-

u

(n

-1)S

2(

X

-

u)2+

ss

2s

2为自由度为n的卡方分布n2s

i=1

s

2(x

-

m)2=

X

-

m

n

i=11~

c

2为自由度为1的卡方分布2

s

n

X

-

u

根据卡方分布的加法定理2n-1(n

-1)S

2~

cs

2单一总体方差的统计推断2n-1(n

-1)S

2~

cs

2从一个标准差为σ的正态分布中取n个样本总体方差的区间估计

总体方差的显著性检验0f

(c

2

)c2总体方差的置信区间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总体方差的1-a置信区间总体方差的置信区间（例子）在一次全区统考中，某校40名学生成绩的方差为

144分，问全区学生成绩的方差95%的置信区间是多少？解：df=40-1=39该区学生成绩的方差95%的置信区间（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总体方差的显著性检验建立假设(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总体方差假设检验的统计量c

=0假设为真时的总体方差例題某工厂生产10mm的螺钉，假设所生产的螺丝钉直径为常态分布且期望值为10mm，虽然每支螺钉的直径不一定会刚好等于10mm，生管部门希望将方差控制在0.09mm以内，抽取12支螺钉样本来检验，得出以下数据，根据这些样本数据，在α=5%的水平下，推论所生产的镙钉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例題写出假设>

0.092

2H0

:s

£

0.09

H1

:

s0.0902= =

2.988s

2c

=(n

-1)S

2

(12

-1)(.0022)计算统计检验量S

2

=0.0022χ20.05=19.675找出临界值α=0.05,df=12-1,

χ2=?比较检验统计量与临界值2.988<19.675F

分布（

F

distribution

）F分布的密度函数F分布的形状F分布的临界值F分布的定理一F分布的定理二两个总体方差的统计推论两总体方差是否相等可以用利F分布作估计与检验如果随机变量X

有密度函数f(x)为1

2112

222

2

(

1

2

2

2n1

n1nnn

nn

n2n1

+n2-1-n

+nG(

)f

(x)

=)

x

(1+x) ,

x>

0G()G(

)则称随机变量X服从自由度为n1，n2的F分布（Ronald

Fisher），计为X

~

F

(v1,

v2

)F

分布的形状Degrees

of

freedomfor

thenumerator（分子自由度，第一个自由度）Degrees

of

freedomfor

thedenominator（分母自由度，第二个自由度）F分布的性质性质1:

F曲线所围的面积和为1性质2:F曲线横轴数值从0向右伸展至无限，与横轴没有相交性质3:F曲线为正偏分布，其形状由两个自由度ν1,ν2决定,不同自由度有不同F分布F分布的性质性质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与W自由度分别为ν

及ν

的独立卡方分布，则212

1

v1

,v2v2v1~

FvvW

vU

vc

2c

2=v1U

~

c

222vW

~

c两个独立的卡方随机变量分别除以其自由度后，两者相除可得F随机变量，其中分子的自由度决定F分布的第一个自由度，分母的自由度决定F分布的第二个自由度。定理二1

2假设X

与X

独立，且11n

-1S

222=

1

1

=

2

2

n

-1

(

X

-

X

)2

(

X

-

X

)2S

22111X

~

N

(m

,s

)2222X

~

N

(m

,s

)，分别从这两个总体中抽取独立随机样本n1与n222

1

1

1

-1,n2

-1~

FnS

2

s

2S

2

s

2证明 定理二，根据卡方分布的定理：個樣本，令：2

2=

1

1

1

-1,n2

-1~

FnS2

s

2S

2

s

221

1

1

为自由度为(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

根据定理一，两个独立的卡方随机变量分别除以其自由度后，两者相除可得F随机变量。211

1(n1

-1)(n

-1)S

2s22

2

2

2(n

-1)(n

-1)S

2s定理二，根据定理二：22

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我们可以根据这个知识，从两样本的方差的比较来推论两总体方差是否相等。F分布的临界值右侧临界值左侧临界值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的倒数仍为F随机变量，v1

,v2v2

2v1

1~

Fvvc2c2F

=v2

,v1v1

1v2

2~

FvvFc2c21

=1v2

,v1v1

1v2

2~

FvvFc2c2n1

,n2=两个独立的卡方随机变

量分别除以其自由度后，两者相除可得F随机变量。自由度的分子分母对调F的左侧临界值1a

2

1F

(n

,n

)F

(n1

,n2

)

=1-a，F表中为右侧概率为α的F值，因为左侧的F值可用

F分布的倒数值求得，左侧概率为α的F值表为：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=0.24两总体方差比的区间估计要对总体方差作区间估计必须假设两总体为正态分布且独立，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)的概率区间(1

-

a

)

F1-P(F

a

(v1,

v2

)

£

F

£

Fa

(v1,

v2

))

=1-a2

2(1

-a

)F

的概率区间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

总体方差比的置信区间两总体方差比的区间估计（例子）差之比的置信区间，能否说二总体方差相等：解：计算自由度v1=9，v2=14F0.025(9,14）=3.21

F0.975(9,14)=

1/3.77=0.27故二总体方差之比，在0.26—3.14之间，作此推论正确的概率为0.95。111

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