SAS选修作业要点

1、1：今有某种型号的电池三批，它们分别是a、日c三个工厂所生产的，为评比其质量，各随机抽取5只电池为样品，经试验得其寿命（ h）如下：abc404226 2839 50484534 3240 50383043试在显著性水平0.05下检验电池的平均寿命有无显著的差异，若差异是显著的，试求均差a- b,a-c和b-c的置信水平为95%勺置信区间。(1)程序 data 11;do i= 1 to 5;do g= 1 to 3; input x; | output ; end ; end ; cards ;40 26 39 42 28 50 48 34 40 45 32 50 38 30 43proc

2、anova ; class g; | model x=g;1 means g; |run ;sum ofsquaresmean squaref valuepr f615.6000000307.800000017.070.0003216.400000018.0333333832.0000000dependent variable: xsourcedfmodel2error12corrected total14r-square coeff var root msex mean0.73990410.888634.24656739.00000结果分析：因为p=0.0003 fmodel81.42857

3、1440.714285711.310.0003error2590.00000003.6000000corrected total27171.4285714r-square coeff var root msex mean0.47500018.706431.89736710.14286结果分析：因为p=0.0003 fmodel3403.3500000134.45000003.770.0321error16571.200000035.7000000corrected total19974.5500000r-square coeff var root msex mean0.4138833.1840

4、915.974948187.6500sourcedfanova ss mean square f value pr fa3403.3500000134.45000003.770.0321结果分析：因为p=0.0321 fmodel3108.333333336.111111112.500.0005error1234.66666672.8888889corrected total15143.0000000r-square coeff var root msex mean0.75757627.194771.6996736.250000sourcedfanova ss mean square f va

5、lue pr ftype3108.333333336.111111112.500.0005结果分析：因为p=0.0005 fmodel41480.823000370.20575040.88 fa41480.823000370.20575040.88.0001结果分析：因为p f0.04420.56570.5684do a= 1 to 3;do b= 1 to 4;do c= 1 to 2;input x; |output ;end ;end ;end;drop c;cards ;14 10 11 11 13 9 10 129 7 10 8 7 11 6 105 11 13 14 12 13 1

6、4 10proc anova ;class a b; | model x=a b a*b;run ;sourcemodelerrorcorrected totaldfsquaresmean squaref value1182.83333337.53030301.391265.00000005.416666723147.8333333sum ofpr f0.2895r-square coeff var root msex mean0.56031622.342782.32737310.41667结果分析：因为pr(a)=0.04420.05,pr(a*b)=0.56840.05,所以只有浓度的影响

7、是显著的。7:为了研究金属管的防腐的功能，考虑了4种不同的涂料涂层。将金属管设在3种不同性质的土壤中，经历了一定时间，测得金属管腐蚀的最大深度如下所示（以mr#）:土壤类型（因素b）1231.631.351.27涂层（因素a）1.341.3041.271.301.091.32试取显著性水平a =0.05检验在不同涂层下腐蚀的最大深度的平均值有无显著差异，在不同 土壤下腐蚀的最大深度的平均值有无显著差异。设两因素间没有交互作用效应。程序data t7; |do a= 1 to 4;do b= 1 to 3;input x; |output ;end ;end ;cards

8、;1.63 1.35 1.27 1.34 1.30 1.22 1.19 1.14 1.27 1.30 1.09 1.32proc anova ;class a b; |model x=a b;sum ofsourcedfsquaresmean squaref valuepr fmodel50.124100000.024820001.940.2209error60.076600000.01276667corrected total110.20070000r-squarecoeff varroot msex mean0.6183368.7929710.1129901.285000sourcedfa

9、nova ssmean squaref valuepr fa30.080700000.026900002.110.2007b20.043400000.021700001.700.2601结果分析：因为 pr(a)=0.20070.05,pr(b)=0.26010.05，所以因素 a、因素b的影响均不显著8：下表数据是退火温度 x(c)对黄铜诞性丫效应的试3结果，y是以延长度计算的。x(c)300400500600700800丫()405055606770画出散点图并求出丫x的现行回归方程。程序data t8 ;input x y; cards ;300 40 400 50 500 55 600

10、 60 700 67 800 70 proc reg ； model y=x ;proc plot ; | plot y*x;run ;sum ofmeansourcedfsquaressquaref valuepr fmodel1606.22857606.22857176.080.0002error413.771433.44286corrected total5620.00000root msedependent meancoeff var1.85549r-square0.977857.00000adj r-sq0.97223.25525parameter estimatesparamete

11、rstandardvariabledfestimateerrort valuepr |t|intercept124.628572.554419.640.0006x10.058860.0044413.270.0002结果分析：因为b=0.05886, a=24.62857,所以回归方程为y=24.62857+0.05886x9:在钢线碳含量对于电阻的效应的研究中，得到一下的数据:碳含量x ( %0.100.300.400.550.700.800.9520c时电阻y (科q )1518192122.623.826(1)画出散点图.(2)求线回归方程 错误！未找到引用源。.(3)求的方差(t之的无偏

12、估计.(4)检验假设 h0: b=0, h: bw0.(5)若回归效果显著，求 b的置信水平0.95的置信区间(6)求x=0.50处科(x)的置信水平为0.95的置信区间.(7)求x=0.50处观察值y的置信水平为0.95的预测区间程序datat9；input x y;cards ;0.10 15 0.30 18 0.40 19 0.55 21 0.70 22.6 0.8 23.8 0,95 26 0.50 .proc reg ;model y=x / clm cliprocplot ；ploty*x;run ;sourcedfsquaressquaref valuepr fmodel183,

13、8183183,818311940.48 |t|.0001xthe reg procedure12.550340.2849144.05.0001model: model1dependent variable: youtput statisticsobsdep varypredictedvaluestd errormean predict95% cl mean95% cl predictresidual115.000015.21340.148614.831415.595514.556615.8702-0.2134218.000017.72350.104717.454417.992617.1253

14、18.32170.2765319.000018.97850.088518.751119.205918.397919.55920.0215421.000020.86110.078620.659121.063120.289921.43220.1389522.600022.74360.090422.511222.976022.161023.3262-0.1436623.800023.99870.107423.722524.274823.397324.6000-0.1987726.000025.88120.140125.521126.241325.236926.52550.11888.20.23360

15、.079520.029220.437919.661620.8056.sum of residuals0sum of squared residuals0.21597predicted residual ss (press)0.50730结果分析：(2)因为 b=12.5503,a=13.9584所以 y=12.5503x+13.9584(3) &的方差0 2的无偏估计为0.0432 (4)因pr(b) |t|15.1 15.1 14.514.0 .proc reg ； model y=x/ cli proc plot ; | plot y*x; run ;sourcemodelerrorcor

16、rected totaldfsum ofsquaresmeansquaref valuepr f194.0998794.099872311.89.0001160.651240.040701794.75111root msedependent meancoeff var0.20175r-square0.993114.72222adj r-sq0.99271.37037parameter estimatesparameter standard-0.104050.312000.988050.02055-0.330.743148.08 fmodel1136.41142136.4114230.200.0

17、001error1358.725914.51738corrected total14195.13733root mse2.12541r-square0.6991dependent mean26.68667adj r-sq0.6759coeff var7.96432parameter estimatesvariabledfestimateintercept1-3.85494x11.83396parameterstandarderror t value pr |t|5.58491-0.690.50220.333745.500.0001结果分析:因为 a=1.83396,b=-3.85494所以 y

18、= 1.83396x-3.8549412:下面列出了自1952年2004年各界奥林匹克运动会男子10000米赛跑的冠军的成绩(时间按min计)：年份(x)1952195619601964196819721976成绩(y)29.328.828.528.429.427.627.7年份(x)1980198419881992199620002004成绩(y)27.727.827.427.827.127.327.1(1)求y关于x的线性回归方程 错误！未找到引用源。.(2)检验假设 h): b=0, h: bw 0(显著性水平 a =0.05).(3)求2008年冠军成绩的预测值.程序data t12;

19、input x y;cards ;1952 29.3 1956 28.8 1960 28.5 1964 28.4 1968 29.4 1972 27.6 1976 27.71980 27.7 1984 27.8 1988 27.4 1992 27.8 1996 27.1 2000 27.3 2004 27.12008 .proc reg ;model y=x/ p ;run ;sum ofmeansourcedfsquaressquaref valuepr fmodel15.586475.5864733.47 |t|intercept1105.4826413.394297.88.0001x1-

20、0.039180.00677-5.79 fmodel10.785260.7852612.400.0038error130.823080.06331corrected total141.60833sum ofmeanroot mse0.25162r-square0.4882dependent mean6.96667adj r-sq0.4489coeff var3.61180parameter estimatesvariabledfinterceptparameterstandardestimate1.89615error1.44124t value1.32pr |t|0.211095% conf

21、idence limits-1.217465.009760.538460.152903.520.00380.208150.86877结果分析:(1)因为 b=0.53846,a=1.89615所以 y=0.53846x+1.89615(2) b的置信区间为(0.208,0.869 )。14:棚寄生是一种寄生在大树上部树枝上的寄生植物，它喜欢寄生在年轻大树上。下面给出在一定条件下完成的试验中采集的数据：大树的年龄x(年)3491540每株大树上棚 寄生的株数y281015613336221412224109(1)作出(xi,y i)的散点图.(2)令zi =lny i,作出(xi,zi)的散点图

22、.(3)以模型y=aebx ,ln n(0, j)拟合数据，其中a, b,2与x无关。试求曲线回归方程错误！未找到引用源。程序data t14;input x y; z=log(y);y1=log(y);cards ;3 28 3 33 3 22 4 10 4 36 4 24 9 15 9 22 9 10 15 6 15 14 15 9 40 1 40 1proc plot ;plot y*x;proc plotplotz*x;procmodelreg ;y1=x;run ;sourcedfsum ofsquaresmeansquaref valuepr fmodel115.4601715.46017110.86 |t|intercept13.479870.1425624.41.0001x1-0.086730.00824-10.53 fmodel238.9371419.468579.540.0033error1224.476192.03968corrected total1463.41333root mse1.42817r-square0.6140dependent mean30.03333adj r-sq0.5497coeff var4.75530parameter estimatesparameter st