《数据分析》课程期末考试试题A卷

命题方式：单独命题

佛山科学技术学院2008-2009学年第一学期

《数据分析》课程期末考试试题A卷

专业、班级：姓名：学号：

题号—•~~三四五六七八九十十二总成绩

得分

说明：1.请仔细阅读题目，按要求在SAS软件系统编程运算：

2.将SAS程序及运算的有关结果作为解答copy到试卷的后面.

一、(12分)有关SAS的简答题：

1、SAS所采用的Windows操作系统中，SAS界面有哪三个部分？

日志框，编辑框，输出窗口

2、怎样输入非数值变量？

在非数值变量后面加“$”

3、与固定格式不同的自由格式输入数据应加上何种标记？

加上”@@”

4、写出三均值的计算公式。

入111

二、(15分)北京市GDP同比增长1978〜1995年的数据如下：

100.00107.57112.4296.21121.58107.21117.16116.19101.37

109.78112.83104.37105.40109.50111.60112.10113.50112.40

(D计算均值、方差、标准差、变异系数、偏度、峰度；

(2)计算中位数，上、下四分位数，四分位极差；

(3)做出直方图、QQ图、茎叶图、箱线图；

(4)进行正态性W检验(取a=0.05).

共3页第1页

三、(15分)已知数据如下:

X1x2x3x4(1)计算协方差矩阵，Pearson相关矩阵；

16.726.76.435.0(2)分析各指标间的相关性(取a=(H0)

18.228.03.229.7

16.726.72.134.9

18.126.74.331.5

16.726.03.032.7

18.130.27.034.9

20.230.54.834.4

20.229.55.536.2

21.531.55.836.5

18.830.65.435.4

21.627.85.434.1

21.329.55.835.8

四、(15分)已知某工厂产量y及工人数xl、

成本x2的有关数据如下:

序号yxlx2(1)求回归方程，给出各参数的实际解

11692653782释；

281983008(2)求出方差分析、参数估计的结果；

31923302450

41161952137

555532560

61622742450

71201803254

82233753802

91312052838

1067862347

五、(13分)已知数据如下:

X1x2x3x4x5x6x7

12.516.416.722.829.33.01726.6

7.89.910.212.617.60.84110.6

13.410.99.910.913.91.77217.8

19.119.819.029.739.62.44935.8

8.09.88.911.916.20.78913.7

9.74.24.24.66.50.8743.9

0.60.70.70.81.10.0561.0

13.99.49.39.813.32.12617.1

9.111.39.512.216.41.32711.6

对以上样本进行主成分分析，并求出相应的主成分.

共3页第2页

六、(15分)已知数据如下:

序号类别X1x2x3x4x5x6x7

36.057.137.7516.6711.682.3812.88

37.697.018.9416.1511.080.8311.67

38.696.018.8214.7911.441.7413.23

37.759.618.4913.159.761.2811.28

35.718.048.3115.137.761.4113.25

39.778.4912.9419.2711.052.0413.29

40.917.328.9417.6012.751.1414.80

33.707.5910.9818.8214.731.7810.10

35.024.726.2810.037.151.9310.39

52.417.709.9812.5311.702.3114.69

52.653.849.1613.0315.261.9814.57

55.855.507.459.559.522.2116.30

44.687.3214.5117.1312.081.2611.57

45.797.6610.3616.5612.862.7511.69

50.3711.3513.3019.2514.592.7514.87

64.348.0022.2220.0615.120.7222.89

(1)求出三个协方差矩阵；

(2)用距离判别求出线性判别函数，用交叉确认法计算误判率；

(3)判别待判样品属于哪一类.

七、(15分)利用上一题的数据(共16个)进行聚类分析：

(1)最短距离法，写出聚类过程，画出谱系图(取nclusters=4)；

(2)最长距离法，写出聚类过程，画出谱系图(取nclusters=4),求出四个聚类

统计量；

(3)快’速聚类法分3类的结果，在平面坐标系中画出分类图.

共3页第3页

-(1)SAS界面包括

输出框，日志框，编辑器

(2)在非数值变量后面家上“$”符号.

(3)自由格式输入数据应加上“@@”标记.

A

(4)三均值的计算公式M=1/4Q1+1/2M+1/4Q3

程序：

datat1;

inputx@@;

cards;

100.00107.57112.4296.21121.58107.21117.16116.19101.37

109.78112.83104.37105.40109.50111.60112.1113.50112.40

procunivariateplotnormal;

run;

proccapabilitygraphicsnormal;

histogramx/normal;

qqplotx/normal(....);

run;

N18权重总和18

均值109.510556观测总和1971.19

标准偏差6.36948929方差40.5703938

偏度-0.3324812峰度0.05978054

未校平方和216555.809校正平方和689.696694

变异系数5.81632451标准误差均值1.50130302

(1)由上图可知道

均值：109.510556方差：40.5703938

变异系数：5.81632451峰度:0.05978054

偏度:-0.3324812

(2)

中位数：巴库数110.69Q0J

上四分位数：IZ5O3112.831

下四分位数：幽Qi项画1

四分位极差：屿分位被差7.43000|

(3)做出直方图、QQ图、茎叶图、箱线图

直方图：

QQ图

25-

20-

15-

10-

05-

oo-

95'

-2-1012

茎叶图:

茎叶#

1221

11672

11002222348

105783

100143

981

——+------+-------+-------+

茎・叶乘以10**+1

箱线图:

一一最小值——

值观测

104.3712

105.4013

盒形图

I

+---+

(4)进行正态性w检验(取a=0.05).

检验——统计量-----------P值-------

w

pr<D

Shapiro-WiIkW0.978265pr>0.9304

KoImogorov-SmirnovD0.128559>0.1500

pr>W-sq

Cramer-vonMisesW-Sq0.044882pr>sq>0.2500

Anderson-DarIingA-Sq0.247567A->0.2500

由上图可以知道Wo=0.978265,P=0.9304>a=0.05；

故不能拒绝原假设Ho,所以是高度显著的。

三

datat2;

inputxl-x4;

cards

16.726.76.435.0

18.228.03.229.7

16.726.72.134.9

18.126.74.331.5

16.726.03.032.7

18.130.27.034.9

20.230.54.834.4

20.229.55.536.2

21.531.55.836.5

18.830.65.435.4

21.627.85.434.1

21.329.55.835.8

proccorrcovpearson;

run;

(1)计算协方差矩阵，Pearson相关矩阵;

协方差矩阵：

蚀方差矩阵，自由度二11

xlx2x3x4

xl3.5771969702.3332575761.2264393941.542196970

x22.3332575763.5244696971.5731060612.067348485

x31.2264393941.5731060612.1681060611.643257576

x41.5421969702.0673484851.6432575764.064469697

Pearson相关矩阵:

xlx2x3x4

xl1.000000.657120.440390.40445

0.02020.15190.1922

x20.657121.000000.569080.54622

0.02020.05350.0662

x30.440390.569081.000000.55356

0.15190.05350.0619

x40.404450.546220.553561.00000

0.19220.06620.0619

(2)分析各指标间的相关性(取a=0.10)

山Pearson相关矩阵的上三角矩阵看出rl3,rl4都大于0.10

故这些向量的相关性不是很强。

四：

data14;

inputnum$yxlx2;

cardsr

11692653782

281983008

31923302450

41161952137

555532560

61622742450

71201803254

82233753802

91312052838

1067862347

procregdata=t4;

modely=xl-x2/i;

run;

(1)求回归方程，给出各参数的实际解释

Parameter

VariableDFEstimate

Intercept14.14260

xl10.49482

x210.00890

由上图可以知道

000890

8）=4.14260,jgi=0.49482,j^^-

回归方程为y=4.14260+0.49482x1+0.00890x2;

工厂产量y及工人数xl、成本X2的有关数据如下

Bo为基本产量，当成本x2固定时，工人数xl每增加•个单位，产量y就增加0.49482个

单位，同理当成本xl固定时，成本x2每增加•个单位，产量y就增加0.00890个单位。

（2）求出方差分析、参数估计的结果

方差分析：

Analysisotvariance

SumofMean

SourceDFSquaresSquareFValuePr>F

Model227272136362935.52<.0001

Error732.516074.64515

CorrectedTotal927304

由方差分析图可以知道

32=4.64515

R2=SSM/SST=27272/27304=0.9988

F值为2935.52

参数估计

ParameterEstimates

ParameterStandard

VariableDFEstimateErrortValuePr>HI

Intercept14.142603.555111.170.2821

xl10.494820.0073467.43<.0001

x210.008900.001336.700.0003

第五题：

datat5;

inputxl-x7;

cards;

12.516.416.722.829.33.01726.6

7.89.910.212.617.60.84110.6

13.410.99.910.913.91.77217.8

19.119.819.029.739.62.44935.8

8.09.88.911.916.20.78913.7

9.74.24.24.66.50.8743.9

0.60.70.70.81.10.0561.0

13.99.49.39.813.32.12617.1

9.111.39.512.216.41.32711.6

procprincomp;

run;

EigenvaluesoftheCorrelationMatrix

EigenvalueDifferenceProportionCumu1ative

16.368806955.970882200.90980.9098

20.397924750.237540340.05680.9667

30.160384420.114957090.02290.9896

40,045427330.023012480.00650.9961

50.022414850.017666030.00320.9993

60.004748820.004455930.00071.0000

70.000292890.00001.0000

Eigenvectors

PrinlPrin2Prin3Prin4Prin5PrinGPrin7

0.3488240.6123630.6820500.1332460.136972-.013602-.037959

0.390078-.1767270.0020060.456233-.5905200.5062310.058911

0.391810-.169297-.1106550.344580-.130939-.813353-.090308

0.385562-.3496340.020863-.1028180.4029360.226117-.710356

0.383622-.3754840.096918-.0477990.4649390.0779110.691324

0.3537200.549207-.7153370.0297170.2041800.1272680.052690

0.3894910.0160380.031998-.801013-.441775-.0927680.040273

特征值：

xl=6.36880695,x2=0.39792475,x3=0.16038442,x4=0.04542733,x5=0.02241485,x6=0.00474882.

X7=0.00029289;

ProportionCumulative

0.90980.9098

0.05680.9667

0.02290.9896

0.00650.9961

0.00320.9993

0.00071.0000

0.00001.0000

贡献率和累计贡献率分别为：

各主成分分别为：由于W1已经达到了90%所以第一主成分为

wl=0.348824X1+0.390078X2+0.391810X3+0.385562X4+0.383622X5+0.353720X6+0.389491x

7

六：

datat6;

inputxy$xl-x7;

cards;

136.057.137.7516.6711.682.3812.88

137.697.018.9416.1511.080.8311.67

138.696.018.8214.7911.441.7413.23

137.759.618.4913.159.761.2811.28

135.718.048.3115.137.761.4113.25

139.778.4912.9419.2711.052.0413.29

140.917.328.9417.6012.751.1414.80

133.707.5910.9818.8214.731.7810.10

135.024.726.2810.037.151.9310.39

252.417.709.9812.5311.702.3114.69

252.653.849.1613.0315.261.9814.57

255.855.507.459.559.522.2116.30

244.687.3214.5117.1312.081.2611.57

245.797.6610.3616.5612.862.7511.69

250.3711.3513.3019.2514.592.7514.87

datat61;

inputxl-x7;

cards;

64.348.0022.2220.0615.120.7222.89

procdiscrimdata=t6testdata=t61

out=al

outstat=a2outcross=a3

testout=a4method=normal

listcrosslisttestlistall;

classxy;

varxl-x7;

priorsequal;

run;

(1)求出三个协方差矩阵；

S!=

Variablexlx2x3x4x5xGX?

xl136.3561056-12.7039611-32.1020333-43.9701278-4.7449722-0.278977861,7896722

x2-12.703961147.685705632.455933347,69347229.33239441.9405222-0.4706611

x3-32.102033332.455933363.950133377.496000029.5911333-1.5631000-11.4411667

x4-43.970127847.693472277.4960000131.109872263.98726111.9098222-6.8091944

x5-4.74497229.332394429.591133363.987261165.85563891.1910111-1.2275389

xG-0.27897781.9405222-1.56310001.90982221.19101113.45262220.3389556

X?61.7996722-0.4706611-11.4411667-6.8091944-1.22753890.938955637,3281722

S2=

Variablexlx2x3x4x5xGx7

xl18.54121667-3.74661667-8.57356667-11.76273000-2.189896670.522380007.93868333

x2-3.746616676.374656674.282866676.663030000.830496670.63964000-0.64302333

x3-8.573566674.282866676.958386678.268320001.92554667-0.42338000-3.00631333

x4-11.762730006.663030008.2683200012.818710004.331510000.26400000-4.10899000

x5-2.169896670.830496671.925546674.331510004.328416670.20144000-0.75466333

x60.522380000.63964000-0.423380000.264000000.201440000.309720000.29376000

X?7.93868333-0.64302333-3.00631333-4.10899000-0.754663330.293760003.61457667

S=

Variablexlx2x3x4x5xGx7

xl10.48893120-0.97722778-2.48938718-3.38231752-0.36499786-0.021459834.75382094

x2-0.977227783.668131202.496610263.668728630.717876500.14927094-0.03620470

x3-2.469387182.496610264.919241035.361230772.27624103-0.12023846-0.88008974

x4-3.382317523.668728635.9612307710.085382484.322097010.14690940-0.52378419

x5-0.364997880.717876502.276241034.922097015.065818380.09161824-0.09442607

x6-0.021459830.14927094-0.120238460.146909400.091616240.265586320.07222735

x74.75382094-0.03620470-0.88008974-0.52378419-0.094426070.072227352.87139786

(2)用距离判别求出线性判别函数，用交叉确认法计算误判率;

LinearDiscriminantFundionforxy

Variable12

Constant-206.18758-382.57458

xl16.6024023.14210

x2-2.77150-3.89531

x3-5.80267-5.94472

x414.1735917.23215

x5-8.00073-10.19191

xG7.4917412.60276

x7-22.87514-32.83581

由上图可以知道线性判别函数为：

W!=-206.18758+16.6024x1-2.77150x2-5.80267x3+14.17359x4-8.00073x5+7.4917

4x6-22.87514x7

W2=-382.57458+23.14210x1-3.89531x2-5.94472x3+17.23215x4-10.19191x5+12.6

0276x6-32.83581x7

PosteriorProbabi1ityofMembershipinxy

FromClassified

Obsxyintoxy12

1111.00000.0000

2111.00000.0000

3111.00000.0000

4111.00000.0000

5111.00000.0000

612*0.00001.0000

7111.00000.0000

8111.00000.0000

9111.00000.0000

10220.00001.0000

11220.00001.0000

12220.00001.0000

1321*1.00000.0000

14220.00001.0000

15220.00001.0000

*Misclassifiedobservation

用交叉确认法计算误判率P=2/15=13.33%

(3)判别待判样品属于哪一类.

PosteriorProbabiIityofMembershipinxy

Classified

Obsintoxy12

120.00001.0000

判别待判样品属于2类

七、（15分）利用上一题的数据（共16个）进行聚类分析：

（1）最短距离法，写出聚类过程，画出谱系图（取nclusters=4）；

（2）最长距离法，写出聚类过程，画出谱系图（取nclusters=4）,求出四个聚类统

计量；

（3）快速聚类法分3类的结果，在平面坐标系中画出分类图.

datat7;

inputxl-x7;

cards;

36.057.137.7516.6711.682.3812.88

37.697.018.9416.1511.080.8311.67

38.696.018.8214.7911.441.7413.23

37.759.618.4913.159.761.2811.28

35.718.048.3115.137.761.4113.25

39.778.4912.9419.2711.052.0413.29

40.917.328.9417.6012.751.1414.80

33.707.5910.9818.8214.731.7810.10

35.024.726.2810.037.151.9310.39

52.417.709.9812.5311.702.3114.69

52.653.849.1613.0315.261.9814.57

55.855.507.459.559.522.2116.30

44.687.3214.5117.1312.081.2611.57

45.797.6610.3616.5612.862.7511.69

50.3711.3513.3019.2514.592.7514.87

64.348.0022.2220.0615.120.7222.89

procclusterdata=t7method=sinstdnonormouttree=treel;

varxl-x7;

run;

proctreedata=treelgraphicshorizontalout=clnclusters=4;

run;

procprintdata=cl;

run;

procclusterdata=t7method=comstdnonormouttree=tree2;

varxl-x7;

run;

proctreedata=treezgraphicshorizontalout=c2nclusters=4;

run;

procprintdata=c2;

run;

procfastclusmaxc=3distancelistcluster=c

data=t6out=d;

run;

procplot;

plotx2*xl=c;

run;

(1)最短距离法，写出聚类过程，画出谱系图(取nclusters=4)；

ClusterHistory

Min

NCL-ClustersJoined—FREQDist

oB1onuQ2

15on2oD721.3976

14Dn—1.4581

o4oDb-2

13cB15oD1431.525

12cL12cU1—451.5721

11L—1.6783

o6oB.132

10cB11cL1371.8356

9LB—01.8609

cL9cL19

8—01.865

cL8oL.10

7cL7oB1211.9501

66oB—22.0097

cLo

5cL5oB1132.126

4B—

cL4oBQ42.6429

33ou52.707

cLB.15

2cL2oB1662.7151

1—5.0941

OBI----------------------------------1

0B3----------------------------------1

OBI4----------------------------------------

0B2-------------------------------------_

0B7-------------------------------------1

0B4----------------------------------------------

0B5--------------------------------------1

0B6----------------------------------------------

0B13----------------------------------------------

OB10-------------------------------------------------

0B12--------------------------------------------------

0B8------------------------------------------------------

0B11------------------------------------------------------------------