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1、实验课程： 数据分析 专 业： 数学与应用数学 班 级： 10070141 学 号： 1007014119 姓 名： 齐 凯 中北大学理学院实验一SAS系统的使用【实验目的】 1.了解SAS系统，熟练掌握SAS数据集的建立及一些必要的SAS语句。【实验内容】1. 将SCORE数据集的内容复制到一个临时数据集test。SCORE数据集NameSexMathChineseEnglishAlicef908591Tomm958784Jennyf939083Mikem808580Fredm848589Katef978382Alexm929091Cookm757876Bennief827984Hellen
2、f857484Winceletf908287Buttm778179Geogem868582Todm898484Chrisf898487Janetf8665872将SCORE数据集中的记录按照math的高低拆分到3个不同的数据集：math大于等于90的到good数据集，math在80到89之间的到normal数据集，math在80以下的到bad数据集。3将3题中得到的good，normal，bad数据集合并。【实验所使用的仪器设备与软件平台】1.SAS软件【实验方法或步骤】1.将SCORE数据集的内容复制到一个临时数据集test实验程序：建立score数据集data test;set score
3、;keep NameSex MathChinese English;run;2.拆分数据集实验程序：Data good normal bad;Set work.score; Select; When (math=90) output good; when (math=80&math90) output normal;when (math |t| 0.0526 Sign M 2.5 Pr = |M| 0.0625 Signed Rank S 7.5 Pr = |S| 0.0625正态性检验(检验) Tests for Normality Test -Statistic- -p Value- Sh
4、apiro-Wilk W 0.864196 Pr D 0.1500 Cramer-von Mises W-Sq 0.062815 Pr W-Sq 0.2500 Anderson-Darling A-Sq 0.374757 Pr A-Sq 0.2500 Quantiles (Definition 5) Quantile Estimate 100% Max 36.00000 99% 36.00000 95% 36.00000 90% 36.00000 75% Q3 27.95000 50% Median 11.94778 25% Q1 5.15000 10% 4.87519 5% 4.87519
5、1% 4.87519 0% Min 4.87519即中位数为11.94778，上、下四分位数分别为27.95、5.15，四分位极差为22.8，三均值为14.24889P值为0.2437，大于显著性水平0.05，故认为服从正态分布The SAS System 14:29 Monday, April 10, 2000 The UNIVARIATE Procedure Variable: price Extreme Observations -Lowest- -Highest- Value Obs Value Obs 4.87519 5 4.87519 5 5.15000 2 5.15000 2 1
6、1.94778 4 11.94778 4 27.95000 3 27.95000 3 36.00000 1 36.00000 1茎叶图、箱线图、正态分布QQ图 Multiply Stem.Leaf by 10*+1 Normal Probability Plot 37.5+ *+ | + | *+ | + | + | + * | +* 2.5+ *+ +-+-+-+-+-+-+-+-+-+-+ -2 -1 0 +1 +2直方图：2.The SAS System 14:29 Monday, April 10, 2000 17 The CORR Procedure 2 With Variables
7、: eps scale 1 Variables: price协方差矩阵:Simple StatisticsVariable N Mean Std Dev Median Minimum Maximum eps 36 9422 6018 7900 1338 34000 scale 36 0.03308 0.05235 0.03200 -0.08400 0.19400 price 36 11.94778 4.87519 11.35500 5.15000 27.95000Pearson相关矩阵:Spearman相关矩阵: 残差：2）给定显著性水平=0.05，回归关系的显著性:对回归关系的显著性进行F检
8、验:检验统计量的观测值为，检验的P值为0.0001，且大于显著性水平0.05，故认为三者之间有显著地线性回归关系。检验各自变量对因变量的影响的显著性:3）拟合残差关于拟合值的残差图及残差的正态QQ图的残差图:的残差图：的残差图：由以上各图可看出，残差均没有明显的趋势性变化，是较为满意的形式，从而说明eps和scale的线性回归模型拟合，从各方面考察都较为优良。实验三 美国50个州七种犯罪比率的数据分析【实验目的】通过使用SAS软件对实验数据进行主成分分析和因子分析，熟悉数据分析方法，培养学生分析处理实际数据的综合能力。【实验内容】表3给出的是美国50个州每100 000个人中七种犯罪的比率数据
9、。这七种犯罪是：Murder（杀人罪），Rape（强奸罪），Robbery（抢劫罪），Assault（斗殴罪），Burglary（夜盗罪），Larceny（偷盗罪），Auto（汽车犯罪）。表3 美国50个州七种犯罪的比率数据StateMurderRapeRobberyAssaultBurglaryLarcenyAutoAlabama14.225.296.8278.31135.51881.9280.7Alaska10.851.696.8284.01331.73369.8753.3Arizona9.534.2138.2312.32346.14467.4439.5Arkansas8.827.683.
10、2203.4972.61862.1183.4California11.549.4287.0358.02139.43499.8663.5Colorado6.342.0170.7292.91935.23903.2477.1Connecticut4.216.8129.5131.81346.02620.7593.2Delaware6.024.9157.0194.21682.63678.4467.0Florida10.239.6187.9449.11859.93840.5351.4Georgia11.731.1140.5256.51351.12170.2297.9Hawaii7.225.5128.064
11、.11911.53920.4489.4Idaho5.519.439.6172.51050.82599.6237.6Illinois9.921.8211.3209.01085.02828.5528.6Indiana7.426.5123.2153.51086.22498.7377.4Iowa2.310.641.289.8812.52685.1219.9Kansas6.622.0100.7180.51270.42739.3244.3Kentucky10.119.181.1123.3872.21662.1245.4Louisiana15.530.9142.9335.51165.52469.9337.7
12、Maine2.413.538.7170.01253.12350.7246.9Maryland8.034.8292.1358.91400.03177.7428.5Massachusetts3.120.8169.1231.61532.22311.31140.1Michigan9.338.9261.9274.61522.73159.0545.5Minnesota2.719.585.985.81134.72559.3343.1Mississippi14.319.665.7189.1915.61239.9144.4Missouri9.628.3189.0233.51318.32424.2378.4Mon
13、tana5.416.739.2156.8804.92773.2309.2Nebraska3.918.164.7112.7760.02316.1249.1Nevada15.849.1323.1355.02453.14212.6559.2New Hampshire3.210.723.276.01041.72343.9293.4New Jersey5.621.0180.4185.11435.82774.5511.5New Mexico8.839.1109.6343.41418.73008.6259.5New York10.729.4472.6319.11728.02782.0745.8North C
14、arolina10.617.061.3318.31154.12037.8192.1Ohio7.827.3190.5181.11216.02696.8400.4North Dakota0.99.013.343.8446.11843.0144.7Oklahoma8.629.273.8205.01288.22228.1326.8Oregon4.939.9124.1286.91636.435061388.9Pennsylvania5.619.0130.3128.0877.51624.1333.2Rhode Island3.610.586.5201.01489.52844.1791.4South Car
15、olina11.933.0105.9485.31613.62342.4245.1South Dakota2.013.517.9155.7570.51704.4147.5Tennessee10.129.7145.8203.91259.71776.5314.0Texas13.333.8152.4208.21603.12988.7397.6Utah3.520.368.8147.31171.63004.6334.5Vermont1.415.930.8101.21348.22201.0265.2Virginia9.023.392.1165.7986.22521.2226.7Washington4.339
16、.6106.2224.81605.63386.9360.3West Virginia6.013.242.290.9597.41341.7163.3Wisconsin2.812.952.263.7846.92614.2220.7Wyoming5.421.939.7173.9811.62772.2282.01、1） 分别用样本协方差矩阵和样本相关矩阵作主成分分析，二者的结果有何差异？ 2）原始数据的变化可否由三个或者更少的主成分反映，对所选取的主成分给出合理的解释。 3）计算从样本相关矩阵出发计算的第一样本主成分的得分并予以排序.2、从样本相关矩阵出发，做因子分析。【实验所使用的仪器设备与软件平台
17、】SAS软件【实验方法或步骤】实验程序：1. PROC IMPORT OUT= WORK.CRIME DATAFILE= C:Documents and SettingsAdministratorcrime.xls DBMS=EXCEL2000 REPLACE; GETNAMES=YES;RUN; proc PRINCOMP data=CRIME out=mcrime1 OUTSTAT=CRIME_put1 COVARIANCE ;run;proc PRINCOMP data=CRIME out=mcrime2 OUTSTAT=CRIME_put2 ;run;proc sort data= m
18、crime1 out=sort1;by descending prin1;proc print data=sort1;var state prin1;run; proc sort data= mcrime1 out=sort2;by descending prin1;proc print data=sort2;var state prin1;run;2proc factor data=crime score;run;【实验结果】The SAS System 17:01 Saturday, April 8, 2000 1 The PRINCOMP Procedure Observations 50 Variables 7 Simple Statistics